Electromagnetic Induction PDF

Electromagnetic Induction 1 v 60 S N Magnetic Flux. E3 The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to the magnetic flux linked with that area. ^ dA   B For elementary area dA of a surface flux linked d   B dA cos  or d   B ...

Electromagnetic Induction 1 v 60 S N Magnetic Flux. E3 The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to the magnetic flux linked with that area. ^ dA   B For elementary area dA of a surface flux linked d   B dA cos  or d   B  d A    So, Net flux through the surface   B  d A  BA cos θ  D YG U ID For N-turns coil  = NBA cos (1) Unit and Dimension Magnetic flux is a scalar quantity it’s S.I. unit is weber (wb), CGS unit is Maxwell or Gauss × N  m Joule Volt  Coulomb    Volt  sec = Ohm × cm2; 1 wb  10 8 Maxwell. Other units : Tesla × m2  Amp Amp Amp Coulomb = Henry × Amp. It’s dimensional formula [] = [ML2T–2A–1] (2) Maximum and Zero flux If  = 0o, i.e. plane is held perpendicular to the direction of magnetic field then flux from the surface is maximum and if   90 o i.e. plane is held parallel to the direction of magnetic field then flux linked with the surface is zero. B ^ dA B ^ dA  max  BA  0  = 90o  = 0o U Note :  In case of a body present in a field, either uniform or non-uniform, outward flux is ST taken to be positive while inward negative and Net flux linked with a closed   surface is zero i.e.   B  ds  0  B ^ dA ^ dA  in  R 2 B Total = 0  out  R 2 B Specific example  Let at a place B  B0ˆi (with usual notations). Then flux for the following cases y y y A x z x z x z 2 Electromagnetic Induction  A  Aˆi     B  A  (B0ˆi )  Aˆi  B0 A  A  A kˆ     B  A  (B0ˆi )  Akˆ  0  A  Aˆj     B  A  (B0ˆi )  Aˆj  0 (3) Variation of magnetic flux We know that magnetic flux linked with an area A is  = BA cos i.e.  will change if either B, A 60 or  will change Flux changes Flux not changes v N S S B A        l        A  v A        S B N  Flux changes as area (swept by rod) changes D YG  B ID  N x U  x Flux changes as B  changes    v E3 v  B A Flux changes as  changes x             v A             In all these three cases flux  will not change because B, A and  doesn’t change with time U Faraday’s Experiment and Laws. (1) First experiment ST A coil is arranged to link some of the magnetic flux from a source S. If relative motion occurs between coil and source S such that flux linked with the coil changes, a current is induced in it. Source S v Source N S Flux increasing v N Flux decreasing (2) Second experiment Two coils are arranged so that a steady current flows in one and some of its magnetic flux links with the other. If the current in the first coil changes a current is induced in the second. Electromagnetic Induction 3 Induced current I coil II coil 60 Main current increasi ng (3) Faradays first law E3 Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes (or a moving conductor cuts the magnetic flux) an emf is produced in the circuit (or emf induces across the ends of the conductor) called induced emf. The induced emf persists only as long as there is change or cutting of flux. (4) Faradays second law The induced emf is given by rate of change of magnetic flux linked with the circuit i.e. e   d. dt ST U D YG U ID N d ; Negative sign indicates that induced emf (e) opposes the change of flux. dt (i) Other forms : We know that  = BA cos ; Hence  will change if either, B, A or  will change N ( 2  1 ) NA(B 2  B1 ) cos  NBA (cos  2  cos  1 ) d So e   N    dt t t t Note :  Term B 2  B1  rate of change of magnetic field, it’s unit is Tesla/sec t e N d (ii) Induced current : If circuit is closed, then induced current is given by i   . ; R R dt where R is the resistance of circuit dq (iii) Induced charge : If dq charge flows due to induction in time dt then i  ; dt N dq  i dt    d i.e. the charge induced does not depend on the time interval in which flux R through the circuit changes. It simply depends on the net change in flux and resistance of the circuit. (iv) Induced power : It exists when the circuit is open or closed 2 e2 N 2  d  P  ei   i2 R   . R R  dt  For N turns e   It depends on time and resistance (5) Induced electric field It is non-conservative and non-electrostatic in nature. Its field lines are concentric circular dB closed curves. A time varying magnetic field always produced induced electric field in all dt   space surrounding it. Induced electric field is directly proportional to induced emf so e  Ein  d l  here Ein  induced electric field ……(i)  Also Induced emf from Faraday laws of EMI e   d ……(ii) dt 4 Electromagnetic Induction   d From (i) and (ii) e  Ein.d l   This is known as integral form of Faraday’s laws of EMI. dt  dB/dt in cylindrical space Concentric circular field lines of induced electric field existing everywhere inside and outside of cylindrical space    B(t  )         P     r 60                   a         d dB dB i.e. E(2r)  a 2 Eind l  e  A dt dt dt where r  a or E  ID So E3 A uniform but time varying magnetic field B(t) exists in a circular region of radius ‘a’ and is directed into the plane of the paper as shown, the magnitude of the induced electric field (Ein) at point P lies at a distance r from the centre of the circular region is calculated as follows. a 2 dB 1 ; E in  2r dt r (6) Change in induced parameter (e, i and q) with change in  Suppose a coil having N turns, area of each turn is A placed in a transverse magnetic field B U such that it’s plane is perpendicular to the direction of magnetic field i.e. initially 1 = 0o. If R is Change D YG the resistance of entire circuit and 1 = NBA cos 0o = NBA, is initial flux linked with the coil then. Final flux (2) Coil turn through 180o (end to end) Time taken Induced emf Induced current e i R Induce d charge (t) – NBA – 2NBA t 2 NBA t 2 NBA Rt 2 NBA R Zero – NBA t NBA t NBA Rt NBA R Zero – NBA t NBA t NBA Rt NBA R U ST  in   ( 2  1 ) Turn through 90o Taken out of the field Change flux e  t q  it Concepts If a bar magnet moves towards a fixed conducting coil, then due to the flux changes an emf, current and charge induces in the coil. If speed of magnet increases then induced emf and induced current increases but induced i1 i2 v2 (> v1) charge remains same. v1 S N S N Electromagnetic Induction 5 Induced parameter : e1, i1, q1 Induced parameter : e2 (> e1), i2( > i1), q2 (= q1) Can ever electric lines of force be closed curve ? Yes, when produced by a changing magnetic field.  It should be kept in mind that the total induced emf in a loop is not confined to any particular point but it is distributed around the loop in direct proportion to the resistance of it's parts. 60  Example s A coil of area A = 0.5 m2 is situated in a uniform magnetic field B = 4.0 wb/m2 and area vector makes an angle of 60o with respect to the magnetic field as shown in figure. The value of the magnetic flux through the area A would be equal to E3 Example: 1 (a) 2 weber A (b) 1 weber 60 (c) 3 weber Solution: (b) o ID (d) B 3 weber 2 n̂ Angle between normal to the plane of the coil and direction of magnetic field is  = 60o Example: 2 U  Flux linked with coil  = BA cos = 4.0 × 0.5 × cos 60o   = 1 weber A coil of N turns and area A is rotated at the rate of n rotations per second in a magnetic D YG field of intensity B, the magnitude of the maximum magnetic flux will be (a) NAB (b) nAB (d) 2nNAB (c) NnAB Solution: (a) Since  = NBA cos ; For  to be maximum; cos = max = 1 so max = NBA. Example: 3 A square coil of 10–2 m2 area is placed perpendicular to a uniform magnetic field of intensity 103 wb/m2. The magnetic flux through the coil is [MP PMT 1990, 2001] (a) 10 weber By using  = BA cos ; here  = 0o (c) 105 weber Consider the following figure, a uniform magnetic field of 0.2 T is directed along the positive x-axis. What is the magnetic flux through top surface ofY the figure ST Example: 4 Back 10 cm (a) Zero (b) 0.8 m-wb 9 cm (c) 1.0 m-wb (d) – 1.8 m-wb Solution: (c) (d) 100 weber   = BA = 103 × 10–2 = 10 weber U Solution: (a) (b) 10–5 weber Z 60° 30° 10 cm 4 cm X Fron t Magnetic flux  = BA cos for the top surface, the angle between normal to the surface and the x-axis is   60 o   = 0.2  (10  10  10–4)  cos 60o = 10–3 wb = 1m-wb Example: 5 A coil of area 100 cm2 has 500 turns. Magnetic field of 0.1 weber/metre2 is perpendicular to the coil. The field is reduced to zero in 0.1 sec. The induced emf in the coil is 6 Electromagnetic Induction (a) 1 V (b) 5 V e  (c) 50 V (d) Zero N (B2  B1 )A 500 (0  0.1)  100  10 4 ; e  5V. t 0.1 By using Example: 6 A coil has 1000 turns and 500 cm2 as it's area. The plane of the coil is placed at right angles to a magnetic induction field of 2  10–5 wb/m2. The coil is rotated through 180o in 0.2 sec. The average emf induced in the coil in mV is (a) 5 e (c) 15 NBA (cos  2  cos  1 ) t Initially 1 = 0o and finally 2 = 180o so e   10 mV. A coil having 500 square loops each of side 10 cm is placed normal to a magnetic field which increases at a rate of 1T/s. The induced emf in volt is (a) 0.1 (b) 0.5 By using e   (c) 1 (d) 5 B  B1 N (B 2  B1 ) A cos  T ; Given  = 0o, N = 500, A = 100 × 10–4 m2, 2 1 t t sec U Solution: (d) 1000  2  10 5  500  10 4 (cos 180 o  cos 0 o )  10  2 V = 0.2 ID Example: 7 (d) 20 E3 Solution: (b) (b) 10 60 Solution: (b)  e   500  1  10 2  cos 0 o   5 V, | e |  5 V The magnetic field of 2 × 10–2 Tesla acts at right angle to a coil of area 100 cm2 with 50 turns. The average emf induced in the coil is 0.1 V when it is removed from the field in time t. The value of t is D YG Example: 8 (a) 0.1 s Solution: (a) (b) 0.01 s U N (B 2  B1 ) A cos  ; t 0.1   50  (0  2  10 2 )  10 2  cos 0 o t  t  0.1 s ST (b) 0.4 C (c) 2 C (d) 4 C Given N = 500, A = 0.1 m2, 1 = 0o, 2 = 180o, B = 0.2 T, t = 0.1 sec, R = 50 By using q  Example: 10 (d) 20 s A circular coil of 500 turns of a wire has an enclosed area of 0.1 m2 per turn. It is kept perpendicular to a magnetic field of induction 0.2 T and rotated by 180 o about a diameter perpendicular to the field in 0.1 sec. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of 50 ohms (a) 0.2 C Solution: (b) (c) 1 s Given B1 = 2 × 10–2 T, B2 = 0,  = 0o, N = 50, e = 0.1 V and A = 100 × 10– 4 m2 By using e   Example: 9 [CBSE 1992; CPMT 2001] N N. d    BA (cos  2  cos  1 ) ; q  0.4 C. R R Flux  (in weber) in a closed circuit of resistance 10 ohm varies with time t (in sec) according to the equation   6 t 2  5 t  1. What is the magnitude of the induced current at t = 0.25 s ? (a) 1.2 A (b) 0.8 A (c) 0.6 S (d) 0.2 A Electromagnetic Induction 7 1 d 1 e 1 d 1 ; i (6 t 2  5 t  1)   (12 t  5) ; i   (12  0.25  5)  0.2 A  10 dt 10 R R dt 10 Solution: (d) By using i  Example: 11 The variation of induced emf (E) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as E N E (a) 60 S E (b) E (c) (d) E3 t Solution: (b) t t t As the magnet moves towards the coil, the magnetic flux increases (nonlinearly). Also there A square loop of side ‘a’ and resistance R is placed in a transverse uniform magnetic field B. If it suddenly changes into circular form in time t then magnitude of induced charge will be (a) (b) Ba 2 R 1   1   4   (c) Ba 2 R  1   1  4    (d) Ba 2 R 4    1    It’s area A1 = a2 ; and flux linked 1 = BA1 Initially D YG Solution: (d) Ba 2 (4  1) R U Example: 12 ID is a change in polarity of induced emf when the magnet passes on to the other side of the coil. a Finally r 4a = 2r Induced B( A 2  A1 ) Ba 2  4   2  1       1 t t t t   so induced charged | q |  A circular coil and a bar magnet placed near by are made to move in the same direction. The coil covers a distance of 1 m in 0.5 sec and the magnet a distance of 2 m in 1 sec. The induced emf produced in the coil (a) Zero (b) 1 V (c) 0.5 V (d) Cannot be determined from the given information ST Example: 13 | e|.t R Ba 2  4   1 R π  U  emf | e |  2 4a2  2a  It’s area A2 = r2      and flux linked 2 = BA2    Solution: (a) Speed of the magnet v1  2  2m / s 1 1  2m / s Speed of the coil v 2  0.5 N S v1 v2 Relative speed between coil and magnet is zero, so there is no induced emf in the coil. 8 Electromagnetic Induction A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would (a) Halved Solution: (b) l r2 (d) Quadrupled ; where R = resistance, r = radius of wire, l = length of wire  number of turns N P N 2r 2  Nr 2  1  1. P2 l E3 (if area of each turn is constant)  P  Example: 15 (c) Doubled d e  dB    NA  ; Here e = induced emf =   dt R  dt  Power P   R (b) The same 2 60 Example: 14 A conducting circular loop is placed in a uniform magnetic field B = 40 mT with its plane perpendicular to the field. If the radius of the loop starts shrinking at a constant rate 0.2 mm/s, then the induced emf in the loop at an instant when its radius is 1.0 cm is (b) 0.2 V (c) 1.0 V (d) 0.16 V ID (a) 0.1 V dA d dr    dr    B (r 2 )   B 2r   2Br     e  2    40  10 3  10 2  (0.2  10 3 )  0.16 V. dt dt dt    dt  e  B Example: 16 A solenoid has 2000 turns wound over a length of 0.314 m. Around its central section a coil of 100 turns and area of cross-section 1  10–3 m2 is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 sec, the emf induced in the coil is equal to (a) 6 × 10–4 V Magnetic (c) 6 × 10–2 V (b) 12.8 mV field at (d) 12.8 V the centre of the solenoid is given by 2000   2  16  10 3 T. This magnetic field is perpendicular 0.314 D YG Solution: (b) U Solution: (d)  0 Ni  4  3.14  10 7 l to the plane of the coil B   0 ni   Magnetic flux linked with coil   N BA  Induced emf e  2  100  16  10 3  1  10 3  12.8 mV. 0.25 U  e d  ( B  B) 2 N BA d dB   N A   ( N BA )   N A  dt dt dt dt dt ST Tricky example: 1 A square coil ABCD lying in x-y plane with it’s centre at origin. A long straight wire passing through origin carries a current i = 2t in negative z-direction. The induced y current in the coil is A B (a) Clockwise x (b) Anticlockwise (c) Alternating C D (d) Zero Solution : (d) Magnetic lines are tangential to the coil as shown in figure. Thus net magnetic flux y A passing through the coil is always zero or Bthe induced current will be zero. x C D Electromagnetic Induction 9 Tricky example: 2 60 In the following figure, the magnet is moved towards the coil with a speed v and induced emf is e. If magnet and coil recede away from one another each moving with speed v, the induced emf in the coil will be (a) e S N (b) 2e E3 v (c) e/2 coil (d) 4e  d  e Solution : (b)    dt  In first case N v U Lenz’s law. v ID  d   d   2  2e     dt  relative velocity2v  dt  I case S ST U D YG This law gives the direction of induced emf/induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon law of conservation of energy. To understand the Lenz’s law consider the followings. (1) Motion of bar magnet towards a coil When N-pole of a bar magnet moves towards the coil, the flux associated with loop increases and an emf is induced in it. Since the circuit of loop is closed, induced current also flows in it. Cause of this induced current, is approach of north pole and therefore to oppose the cause, i.e., to repel the approaching north pole, the induced current in loop is in such a direction so that the front face of loop behaves as north pole. Therefore induced current as seen by observer O is in anticlockwise direction. (figure (i)) v S N Observer v S N N Observer (i) (ii) Induced magnetic field In other words when N-pole of bar magnet moves towards the coil, inward magnetic lines of force (i.e. (×)) linked with coil (as viewed from left) increases. To oppose this change some dots () must be produced i.e. direction of induced current is anticlockwise. (figure (ii)) In this example, If the loop is free to move the cause of induced emf in the coil can also be termed as relative motion. Therefore to oppose the cause, the relative motion between the approaching magnet and the loop should be opposed. For this, the loop will itself start moving in the direction of motion of the magnet. 10 Electromagnetic Induction Note :  It is important to remember that whenever cause of induced emf is relative motion, 60 the new motion is always in the direction of motion of the cause.  In the above discussion, If once the coil is of Cu and once of brass and magnet approaches the coil with same velocity in both the case, then induced current in Cu will be greater (because of lesser resistance) and more energy conversion takes place in case of Cu coil. (2) The various positions of relative motion between the magnet and the coil When the north pole of magnet approaches the coil Anticlockwise direction N G Observer Clockwise direction S Observer N U north Repulsive force Cross (×), Increases south Attractive force Cross (×), Decreases Clockwise direction As a pole south Repulsive force Dots () Increases Anticlockwise direction As a pole north Attractive force Dots () Decreases G ST S Magnetic field linked with the coil and it’s progress as viewed from left G When the south pole of magnet approaches the coil N Type of magnetic force opposed As a pole D YG When the north pole of magnet recedes away from the coil As a pole U S Behaviour of face of the coil E3 Direction of induced current ID Position of magnet Observer When the south pole of magnet recedes away from the coil N Observer S G Electromagnetic Induction 11 Some Standard Cases for Questions Based on Direction. (1) Relative motion between co-axial circular coils (i) When a current carrying coil moves towards/away from a stationary coil B Observer G A Anticlockwi se Stationary coil Observer Clockwise G Stationary coil Induced current in coil B is in the same direction to the main current in coil A E3 Induced current in coil B is opposite to the main current in coil A B 60 A (ii) When two current carrying coils carries currents in the same direction and B B A U A Moves away from each other ID Moves towards each other Observer Observer D YG Induced current in both the coils opposite to that of main current so current through each coil decreases Induced current in both the coils assist the main current so current through each coil increases (iii) When two current carrying coils carries currents in the opposite direction and Moves towards each other B A U Observer Moves away from each other ST Induced current in coil A is clockwise and that in coil B is anti-clockwise i.e. in both the coils induced current flows in the direction of main current. Hence current through both the coil increases B A Observer Induced current in coil A is anti-clockwise and that in coil B is clockwise i.e. in both the coils induced current flows in the direction opposite to main current. Hence current through both the coil decreases (2) When the inductive circuits are closed or opened If two coils A and B (primary and secondary) are arranged as shown in the figure and if the primary circuit is closed or opened then the direction of induced current in secondary will be as follows (i) Current increases in coil A by pressing the key (ii) Current decreases in coil A by opening the key A Current increasi ng Observer B A Inverse induced emf and current B increase Flux d K increase d Direction of induced current in the secondary coil is opposite to that in the primary coil Current decreasi ng B Direct induced emf and current B decrease Flux d decrease d Direction of induced current in the secondary coil is same as that in the primary coil Observer K 12 Electromagnetic Induction (3) Increasing and decreasing of current in current carrying coil (ii) When current decreases by 60 (i) When current increases by pressing the key opening the key Induced current + B E3 Main current Induced current Main current K – B + Direction of induced current in the coil will be in a direction opposite to that of main current. K – ID Direction of induced current in the coil will be same as that of the main current Concepts  To apply Lenz's law, you can remember RIN (when the loop lies on the plane of paper). In RIN, U R stands for D YG right, I stands for increasing and N for north pole (anticlockwise). It means, if a loop is placed on the right side of a straight current carrying conductor and the current i in the conductor is increasing, then induced current in the loop is anticlockwise (  Induced current i (increasin g) ) No flux cutting No EMI (No current induces) i wire U Horizontal field lines Horizontal loop ST Example s Example: 17 Consider a metal ring kept on a horizontal plane. A bar magnet is held above the ring with its length along the central axis of the ring. If the magnet is now dropped freely, the acceleration of the falling magnet is (g is acceleration due to gravity) Kerala (Engg.) 2001; MP PET 1990, 99, 2001; MP PMT 2001] (a) More than g S (b) Equal to g N a (c) Less than g (d) Depends on mass of magnet Solution: (c) When the magnet is allowed to fall vertically along the axis of loop with its north pole towards the ring. The upper face of the ring will become north pole in an attempt to oppose Electromagnetic Induction 13 the approaching north pole of the magnet. Therefore the acceleration in the magnet is less than g. Note :  If the coil is broken at any point then induced emf will be generated in it but no induced current will flow. In this condition the coil will not oppose the motion of magnet and the magnet will fall freely with acceleration g. (i.e. a = g) N A bar magnet is falling freely inside a long copper tube and a solenoid as shown in figure (i) and (ii) respectively then acceleration of magnet inside the copper tube and solenoid are respectively (acceleration due to gravity = g) (a) g, g S ID (b) Greater than g, lesser than g N (c) Greater than g, g (d) Zero, lesser than g S N If bar magnet is falling vertically through the hollow region of long vertical copper tube U Solution: (d) a=g E3 Example: 18 60 S D YG then the magnetic flux linked with the copper tube (due to 'non-uniform' magnetic field of magnet) changes and eddy currents are generated in the body of the tube by Lenz's law the eddy currents opposes the falling of the magnet which therefore experience a retarding force. The retarding force increases with increasing velocity of the magnet and finally equals the weight of the magnet. The magnet then attains a constant final terminal velocity i.e. magnet ultimately falls with zero acceleration in the tube. The resistance of copper solenoid is much higher than that of copper tube, hence the induced current in it, due to motion of magnet, will be much less than that in the tube. Consequently the opposition to the motion of magnet will be less and the magnet will fall with an acceleration (a) less than g. (i.e. a < g). A current carrying solenoid is approaching a conducting loop as shown in the figure. The U Example: 19 ST direction of induced current as observed by an observer on the other side of the loop will be (a) Anticlockwise (b) Clockwise v Observer (c) East (d) West Solution: (b) The direction of current in the solenoid is anti-clockwise as seen by observer. On displacing it towards the loop a current in the loop will be induced in a direction so as to oppose the approach of solenoid. Therefore the direction of induced current as observed by the observer will be clockwise. v N N S Observer 14 Electromagnetic Induction Example: 20 Two coils P and Q are lying a little distance apart coaxially. If an anticlockwise current i is suddenly set up in the coil P then the direction of current induced in coil Q will be (a) Clockwise 60 i (b) Towards north P Q Observer (c) Towards south E3 (d) Anticlockwise Solution: (a) Since current setup in the coil P is anticlockwise which increases the dot’s linked with coil Q hence induced current in coil Q will be clockwise. Example: 21 A rectangular loop is drawn from left to right across a uniform magnetic field perpendicular into the plane of the loop               (b) The direction of current in position 2 is clockwise         1 (c) The direction of current in position 3 is anticlockwise                 ID   (a) The direction of current in position 1 is clockwise U 2 3 4 (d) The direction of current in position 4 is clockwise No current is induced in position 1, anticlockwise current is induced in position 2 because it is a case of increase of flux, no current in position 3 as there is no change of flux, clockwise current is produced in position 4 because it is a case of decrease of flux. Example: 22 A small loop lies outside a circuit. The key of the circuit is closed and opened alternately. The closed loop will show D YG Solution: (d) (a) Clockwise pulse followed by another clockwise pulse (b) Anticlockwise pulse followed by another anticlockwise pulse U (c) Anticlockwise pulse followed by a clockwise pulse + S – (d) Clockwise pulse followed by an anticlockwise pulse When key is closed dots are linked with closed loop (i.e. increases from zero to a certain value) so induced current will be clockwise when key is opened dots linked with loop decreases (from a certain value to zero) so induced current will be anticlockwise in direction. Example: 23 Consider the arrangement shown in figure in which the north pole of a magnet is moved away from a thick conducting loop containing capacitor. Then excess positive charge will arrive on ST Solution: (d) (a) Plate a N (b) Plate b a b C S Electromagnetic Induction 15 (c) On both plates a and b (d) On neither a nor b plates When north pole of the magnet is moved away, then south pole is induced on the face of the loop in front of the magnet i.e. as seen from the magnet side, a clockwise induced current flows in the loop. This makes free electrons to move in opposite i.e. direction, to plate b to a inside the loop. Thus excess positive charge appear on plate b. Example: 24 A square loop of side 1m is placed in a perpendicular magnetic field. Half of the area of the loop inside the magnetic field. A battery of emf 10 V and negligible internal resistance is connected in the loop. The magnetic field changes with time according to relation B = 0.01 – 2t Tesla. The resultant emf in the loop will be E3 60 Solution: (b) (a) 1 V (b) 11 V ID (d) 9 V dB   2 Tesla / sec , dt U Given B = 0.01 – 2t Tesla ;  B                 d d dB 1   (BA )   A   (1 2 )  ( 2)  e = 1V dt dt dt 2    C 10V                   B C A D YG Induced emf e     A (c) 10 V Solution: (d)  10V D Since magnetic field (×) decreasing so according to Lenz’s law direction of induced current in upper part of square will be clockwise i.e. from A to C or in other words emf induces in a direction opposite to the main emf so resultant emf = 10 – 1 = 9V. U Tricky example: 3 ST A short magnet is allowed to fall along the axis of a horizontal metallic ring. Starting from rest, the distance fallen by the magnet in one second may be (a) 4 m (b) 5 m (c) 6 m (d) 7 m Solution : (a) We know that in this case acceleration of falling magnet will be lesser than g. If ‘g’ would have been acceleration, then distance covered  1 2 gt  5 m. 2 Now the distance covered will be less than 5 m. hence only option (a) is correct. Tricky example: 4 A conducting wire frame is placed in a magnetic field which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced current   C  in wires AB and CD are    A B            D  16 Electromagnetic Induction (a) B to A and D to C (b) A to B and C to D (c) A to B and D to C 60 (d) B to A and C to D Solution : (a) Inward magnetic field () increasing. Therefore, induced current in both the loops should be anticlockwise. But as the area of loop on right side is more, induced emf in E3 d dB     A. this will be more compared to the left side loop  e   . Therefore net dt dt   U ID current in the complete loop will be in a direction shown below. Hence only option (a) is correct. D YG Dynamic (Motional) EMI Due to Translatory Motion. When a conducting rod moves in a magnetic field, it cuts the magnetic field lines, this process is called flux cutting. Due to this a potential difference developed across the ends of the rod called Dynamic (motional) emf.  ST U Consider a conducting rod of length l moving with a uniform velocity v perpendicular to a  uniform magnetic field B , directed into the plane of the paper. Let the rod be moving to the right as shown in figure. The conducting electrons also move to the right as they are trapped within the rod. × × × P × × × × × × × × – – F × Q × × ++ × × × × × e × × × × × × × v    Conducting electrons experiences a magnetic force Fm  e(v  B). In the present situation they experiences force towards Q, so they move from P to Q within the rod. The end P of the rod becomes positively charged while end Q becomes negatively charged, hence an electric field is set up within the rod which opposes the further downward movement of electrons i.e. an equilibrium is reached and in equilibrium electric force = magnetic force i.e. eE = evB or E = vB V  Induced emf e  El  Bvl [ E  ] l Electromagnetic Induction 17 Important cases If the rod is moving perpendicular to the magnetic field but it’s direction of motion is making an angle      with it’s length. angle  with the direction of magnetic field    v l v sin        An arbitrary shaped conducting rod translating in a uniform magnetic field.    l sin    e  Bvl sin θ         v      E3  This situation is equivalent to a straight rod of length l sin perpendicular to it’s direction of motion so induced emf across the rod Q l  B  This rod can be replaced by a straight conductor whose length is equal to the projected length of the conductor on to a plane perpendicular to the direction of motion (dotted rod) so induced emf between P and Q e  Bvl ID This situation is equivalent to a straight conductor moving perpendicular to the magnetic field with a induced emf e  B(v sin  )l e  Bv(l sin )  e  Bvl sin θ    Vector form of motional emf : e  (v  B).l D YG Note :   B   P v  U v cos  60 If the rod does not translate in a plane perpendicular to the magnetic field or in other words rod is moving in a direction which is making an  While solving the problems, flux cutting conducting rod can be treated as a single cell.  U   B         Q v P  l  ST P    e = Bvl Q 16 Electromagnetic Induction (1) Induced current If conducting rod moves on two parallel conducting rails as shown in following figure then phenomenon of induced emf can also be understand by the concept of generated area (The area swept of conductor in magnetic field, during it’s motion) As shown in figure in time t distance travelled by conductor = vt Area generated A = lvt × ×P × × × × l R × × × × × Q E3 × Flux linked with this area  = BA = Blvt Hence induced emf | e |  × × × × × × × 60 × v v × × × × × × vt d Bvl e  Bvl induced current i  ; i  dt R R ID Direction of induced current can be found with the help of Flemings right hand rule. Fleming’s right hand rule : According to this law, if we stretch the right hand thumb and two nearby fingers perpendicular to one another and first U finger points in the direction of magnetic field and the thumb in the direction conductor Magnetic field Current induced D YG of motion of the conductor then the central finger will point in the direction of Motion of the induced current. Note :  Here it is worthy to note that the rod PQ is acting as a source of emf and inside a source of emf direction of current is from lower potential to higher potential; so the point P of the rod is at higher potential than Q though the current in the rod PQ is from Q to P. (2) Magnetic force on conductor U Now current is set up in circuit (conductor). As we know when a current ST carrying conductor moves in a magnetic field, it experiences a force Fm = Bil (maximum) whose direction can be find with the help of Flemings left hand rule. So, here conductor PQ experiences a magnetic force Fm = Bil in opposite B 2 vl 2  Bvl  F  direction of it’s motion and Fm  Bil  B ; l  m R  R  (As a result of this force (Fm) speed of rod decreases as time passes.) × × × × × × i × R × × × × P× × Fm × × × v × × × Q × × × × × Electromagnetic Induction 17 Note :  To move the rod with uniform velocity some external mechanical force is × × ×  | Fext i × × R × B 2 vl 2 | R × × × × P× × × × × × × × × × × Fexternal Fm × × Q × 60 required and this is Fext = – Fm (3) Power dissipated in moving the conductor delivered by external source is given as Pmech  Pext  E3 For uniform motion of rod PQ, the rate of doing mechanical work by external agent or mech. Power dW B 2v 2l 2 B 2 vl 2  Fext. v   v  Pmech  dt R R ID Also electrical power dissipated in resistance or rate of heat dissipation across resistance is given as 2 Pthermal  H  Bvl   i2 R   .R ; t  R  Pthermal  B 2v 2l 2 R U Note :  It is clear that Pmech. = Pthermal which is consistent with the principle of conservation of energy. (4) Motion of conductor rod in a vertical plane : If conducting rod released from rest (at t = 0) as shown in D YG figure then with rise in it’s speed (v), induces emf (e), induced current (i), magnetic force (Fm) increases but it’s weight remains constant. × Rod will achieve a constant maximum (terminal) velocity vT if Fm  mg So  B 2 v T2 l 2  mg R mgR vT  2 2 B l × × × × × × × × × × × mg × × × × R l t=0 × × × Fm × × × × mg U (5) Motion of conducting rod on an inclined plane : When conductor start sliding from the top of an inclined plane as shown, it moves perpendicular to it’s length but at an angle (90 – ) with the direction of Q ST v sin (90 – ) magnetic field. Hence induced emf across the ends of conductor e  Bv sin( 90   )l  Bvl cos  Bvl cos  So induced current i  R (90 – ) B v P v  v cos (90 – ) R B (directed from Q to P). The forces acting on the bar are shown in following figure. The rod will move down with constant velocity Fm cos only if Fm cos   mg cos(90   )  mg sin  Fm R  (90–) mg cos (90–) mg  18 Electromagnetic Induction Bil cos   mg sin  60 mgR sin θ  Bv l cos   B T  l cos   mg sin   v T  2 2 R B l cos 2 θ   (6) Motion of a conducting rod in earth’s magnetic field : Suppose a conducting rod of length l, executes translatory motion with speed v in earth’s magnetic field with Position II Position III E3 Position I When conductor is held horizontal When conductor is held horizontal When conductor is held vertical and with it’s length along E-W direction with it’s length along N-S direction then it moves – and then it moves – and then it moves – I BH  E W I BV BV E W BH  I N E W BV S S D YG U S N ID BH  N Towards Vertically Towards Towards Vertically Towards Towards Vertically East or North or up or down East or North or up or down East or North or up or down West South West South West South In this Vertical Conductor Conductor Conductor Conductor Conductor Conductor Conductor condition Component cuts, cuts, the is moving moves in cut’s the moves in moves conductor (BV) is cut perpendicul vertical along its magnetic horizontal magnetic along it’s is moving by the arly component length so meridian i.e. component meridian so length so along it’s conductor horizontal perpendicul e 0 No perpendicul e 0 e 0 length, so perpendicul component arly so component arly so generated arly so (BH) so e  BV vl is cut by the e  BH vl ST U Towards area A = 0 hence e = 0 e  BV vl e  B H vl conductor so e = 0 (7) Movement of train in earth's magnetic field : When a train moves on rails, then a potential difference between the ends of the axle of the wheels is induced because the axle of the wheels of the train cuts the vertical component BV of earth’s magnetic field and so the magnetic flux linked with it changes and the potential difference or emf is induced. e = BVlv where l is the length of the axle and v is the speed of the train. (8) Motion of aeroplane in earth's magnetic field : A potential difference or emf across the wings of an aeroplane flying horizontally at a definite height is also induced because aeroplane cuts the vertical component Electromagnetic Induction 19 BV of earth’s magnetic field. Thus induced emf e = BVlv volt where l is the length of the wings of an aeroplane and v is the speed of the aeroplane. 60 (9) Orbital satellite : If the orbital plane of an artificial satellite of metallic surface is coincident with equatorial plane of the earth, then no emf will be induced. If orbital plane makes an angle with the equatorial plane, then emf will be induced on it. E3 (10) Translatory motion of metallic frame in uniform/non-uniform magnetic field (i) Metal frame of different shape moves in uniform magnetic field  B P Higher potential l sin   O v eOQ Q D YG Q Lower potential B 2R L Higher potential L v U B ST 2BvR N e LN  2 BvR N B B v v enet = 0 enet = 0 L v 2R N Lower potential l e QP  VP  VQ  2 Bv (l sin  ) l sin l For part QO eQO = VO – VQ = Bv l sin   U O For part OP eOP = VP – VO = Bv l sin   ID l P eOP B l v R enet = 0 a B l v eab = 0 b R P v ePQ = 0 Q (ii) Moving metal frame in non-uniform magnetic field e 1  B1 av e 2  B 2 av For loop enet = (e1 – e2) a e net  av(B1  B 2 ) a i x v B1 e1 B2 e2 e1 e2 e1 > e2 20 Electromagnetic Induction Now B1   0 iav  1 μ 0 ia 2 v 1    2  x x  a  2 π ( x ) ( x  a) 60 e net     In motional emf B, v and l are three vectors. If any two vector are parallel – No flux cutting.             l         B B v v l ID    B || l   v || l so e = 0 so e = 0 v   v || B so e = 0 or Normal to generated area makes  an angle 90o with B so  = 0; e = 0 so  = 0; e = 0 D YG so e = 0 l or Normal to generated area makes  an angle 90o with B U or Generated area A = 0  E3 Concepts  0 i 0 i and B 2  2x 2 (x  a) A piece of metal and a piece of non-metal are dropped from the same height near the surface of the earth. The non-metallic piece will reach the ground first because there will be no induced current in it.  If an aeroplane is landing down or taking off and its wings are in the east-west direction, then the potential difference or emf will be induced across the wings. If an aeroplane is landing down or taking off and its wings are in the north-south direction, then no potential difference or emf will be induced. When a conducting rod moving horizontally on equator of earth no emf induces because there is no vertical component of U  earth's magnetic field. But at poles BV is maximum so maximum flux cutting hence emf induces. When a conducting rod falling freely in earth's magnetic field such that it's length lies along East - West direction then induced ST  emf continuously increases w.r.t. time and induced current flows from West - East. Examples Example: 25 A two metre wire is moving with a velocity of 1m/sec perpendicular to a magnetic field of 0.5 weber/m2. The emf induced in it will be (a) 0.5 volt (b) 0.1 volt (c) 1 volt Solution: (c) e  Bvl  0.5  1  2  1volt Example: 26  A Cu rod PQ is drawn out normally through the magnetic field B then (d) 2 volt P Q  B Electromagnetic Induction 21 (a) Equal potential on both P and Q occurs and will be positive (b) Equal potential on both P and Q occurs and will be negative 60 (c) The potential at P will be greater then at Q (d) The potential at P will be lesser than at Q Solution: (c) By Fleming's right hand rule direction of induced current in rod PQ is from Q to P, hence P is at higher Example: 27 E3 potential. An electric potential difference will be induced between the ends of the conductor shown in fig. when (b) Q (c) L Solution: (d) D YG (d) M [AIIMS 1982] M Q L N S P U (a) P ID conductor moves in the direction When conductor moves either in the direction P, Q or L it will not cut the magnetic lines of force, so emf will induced only when conductor moves in the direction M Example: 28 A metallic square loop ABCD is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced   [IIT-JEE (Screening) 2001]   A     (b) In BC, but not in AD     (c) Neither in AD nor in BC  D   U (a) In AD, but not in BC ST  B  v C   (d) In both AD and BC Solution: (d) Both AD and BC are straight conductors moving in a uniform magnetic field and emf will be induced in both. This will cause electric field in both but no net current flows in the circuit. Example: 29 A square metallic wire loop of side 0.1 m and resistance of 1 is moved with a constant velocity in a magnetic field of 2 wb/m2 as shown in figure. The magnetic field is perpendicular to the plane of the loop, loop is connected to a network of resistances. What should be the velocity of loop so as to have a steady current of 1mA in loop                l  v              B P Q 3 3 3 3 3 22 Electromagnetic Induction (a) 1 cm/sec (b) 2 cm/sec 60 (c) 3 cm/sec (d) 4 cm/sec Solution: (b) Equivalent resistance of the given Wheatstone bridge circuit (balanced) is 3 so total resistance in circuit is So induced current i  e Bvl 2  v  (10  10 2 )   10 3   v  2cm / sec. R R 4 A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic  induction B (fig.). At the position MNQ the speed of the ring is v and the potential difference development ID Example: 30 E3 R  3  1  4 . The emf induces in the loop e  Bvl. across the ring is U (a) Zero (b)  B  Bv R 2 and M is at higher potential 2 D YG (c) RBv and Q is at higher potential [IIT-JEE 1996]   N                 M Q (d) 2RBv and Q is at higher potential Solution: (d) Suppose in time t vertical distance travelled by ring is v × t so change in area  e d dA  B dt dt U  |e| = 2RBv ST and by Flemings right hand rule. Q is at higher potential. Example: 31         M            N 2R dA  2 Rvt dt Q vt A metal aeroplane having a distance of 50 meter between the edges of its wings is flying horizontally with a speed of 360 km/hour. At the place of flight, the earth’s total magnetic field is 4.0 × 10 –5 weber/meter2 and the angle of dip is 30o. The induced potential difference between the edges of its wings will be (a) 0.1 V Solution: (a) (b) 0.01 V (c) 1 V (d) 10 V The flying of the aeroplane is shown in the adjoining figure. Its wings are cutting flux-lines due to the vertical component of earth’s magnetic field. So, a potential difference V (say) in induced between the edges of its wings. Electromagnetic Induction 23 If the earth’s total magnetic field be B and the angle of dip , then the vertical component of earth’s magnetic field is BV = B sin = (4.0 × 10–5) × W N sin 30o = 2.0 × 10–5 wb/metre2. S E 60 We know that when a conductor of length l meter moves with velocity v BV meter/second perpendicular to a magnetic field of B weber/meter2, then the potential difference induced in the conductor is given by V  BVvl volt 360  1000  100 meter / second 60  60 E3 Here l = 50 meter, v = 360 km/hour  and B = BV = 2.0 × 10–5 weber/meter2 ; V  (2.0  10 5 )  100  50  0.1 volt Example: 32 The two rails of a railway track, insulated from each other and the ground, are connected to a ID millivoltmeter. What is the reading of the millivoltmeter when a train travels at a speed of 20 m/sec along the track, given that the vertical component of earth’s magnetic field is 0.2 × 10 –4 wb/m2 and the rails are (a) 4 mV (b) 0.4 mV (c) 80 mV (d) 10 mV When a train runs on the rails, it cuts the magnetic flux lines of the vertical component of earth’s magnetic D YG Solution: (b) [CPMT 1981] U separated by 1 metre field. Hence a potential difference is induced between the ends of it’s axle. Distance between the rails l = 1m. Speed of train v  36 km 36  1000   10 m / sec hour 3600 By using e = Bvl ; e = 0.2 × 10–4 × 20 × 1 = 4 × 10–4 volt = 0.4 mV 4 The horizontal component of the earth’s magnetic field at a place is 3 × 10 –4 T and the dip is tan 1  . A 3 U Example: 33 metal rod of length 0.25 m placed in the north-south position and is moved at a constant speed of 10 cm/s ST towards the east. The emf induced in the rod will be (a) Zero Solution: (d) (b) 1V (c) 5 V (d) 10 V Rod is moving towards east, so induced emf across it’s end will be e = BVvl BV = vertical component of Earth’s magnetic field = BH tan (BH – horizontal component of earth’s magnetic field;  – angle of dip), B V  3  10  4  4  4  10  4 T 3  e  4  10 4  (10  10 2 )  0.25  10 5 V  10 V Example: 34 A conducting rod AB of length l = 1 m is moving at a velocity v = 4 m/s making an angle 30o with it’s length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motion. Then      v       A     30o   B  24 Electromagnetic Induction (a) VA – VB = 8 V (b) VA – VB = 4 V 60 (c) VB – VA = 8 V (d) VB – VA = 4 V The emf induced across the rod AB is e = Bvl Here v v = sin 30o = component of velocity perpendicular to E3 Solution: (b) 1 Hence e  Bvl sin 30 o  (2)(4 )(1)    4 V 2 length.   The free electrons of the rod shift towards right due to the force q(v  B). Thus the left side of the rod is at Example: 35 VA – VB = 4V ID higher potential. or A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic field of U 1 wb/m2, as shown in fig. If all the four sides are of 1m length each, then the induced emf between points A and D YG D is (a) 0 × × × × O (b) 1.41 volt (c) 0.71 volt × × × × B× × × × × × 90o × × × C × × A × × × × v D × × × × (d) None of the above There is no induced emf in the part AB and CD because they are moving along their length while emf U Solution: (b) ST   induces between B and C i.e. between A and D can be calculated  asfollows × × B× × × × × × × × O × × × × 90o × × × C × × A × × × × v D l  × ×  × ×  O  B 90o   A   v 2l   C   l  D  Induced emf between B and C = Induced emf between A and B = Bv ( 2 l)  1  1  1  2  1.41 volt. Example: 36 Two long parallel metallic wires with a resistance R forms a horizontal plane. A conducting rod AB is on the wires as shown here. The space has a magnetic field pointing vertically upwards. The rod is given an initial A B R v0 B Electromagnetic Induction 25 velocity v0. There is no friction and no resistance in the wires and the rod. After a time t, the velocity of the rod will be v such that [MP PMT 1995] 60 (a) v > v0 (b) v < v0 (c) v = v0 Solution: (b) E3 (d) v = – v0 When rod AB starts it’s motion, current induces in it from A to B, due to which rod experiences a magnetic force towards left (Flemings left hand rule) which opposes the motion of the rod. Hence v < v0. A player with 3m long iron rod runs towards east with a speed of 30 km/hr. Horizontal component of ID Example: 37 earth’s magnetic field is 4 × 10–5 wb/m2. If he is running with rod in horizontal (East-west) and vertical positions, then the potential difference induced between the two ends of the rod in two cases will be U (a) Zero in vertical position and 1 × 10–3 V in horizontal position (b) 1 × 10–3 V in vertical position and zero is horizontal position D YG (c) Zero in both cases (d) 1 × 10–3 V in both cases Solution: (b) In horizontal position rod is moving along it’s length so e = 0 In vertical position, horizontal component of earth's magnetic field is cut by rod so induced emf e = BHvl  Example: 38 e  4  10 5  30  1000  3  10  3 V 3600 At a place the value of horizontal component of the earth’s magnetic field H is 3 × 10–5 weber/m2. A U metallic rod AB of length 2 m placed in east-west direction, having the end A towards east, falls vertically downwards with a constant velocity of 50 m/s. Which end of the rod becomes positively charged and what ST is the value of induced potential difference between the two ends (a) End A, 3 × 10–3 mV Solution: (b) (b) End A, 3 mV (c) End B, 3 × 10–3 mV (d) End B, 3 mV According to Flemings right hand rule direction of induced current in rod AB is from B to A i.e. end A becomes positively charged. i.e. end A becomes positively charged. emf induces across the ends of the rod e = Hvl = 3 × 10–5 × 50 × 2 = 3 × 10–3 volt = 3mV. Example: 39 The current carrying wire and the rod AB are in the same plane. The rod moves parallel to the wire with a velocity v. Which one of the following statements is true about induced emf in the rod 26 Electromagnetic Induction v 60 (b) A and B will be at the same potential B (c) There will be no induced emf in the rod E3 (d) Potential at A will be higher than that at B Solution: (d) According to Fleming's right hands rule direction of induced current in rod is from B to A i.e. end A is at higher potential. As shown in the figure a metal rod makes contact and complete the circuit. The circuit is perpendicular to ID Example: 40 i A (a) End A will be at lower potential with respect to B the magnetic field with B = 0.15 Tesla. If the resistance is 3, force needed to move the rod as indicated with a constant speed of 2m/sec is (b) 3.75 × 10 N × × R × × × × × × × × × × × × × × × × v = 2 m/s 50 cm × × × B = 0.15 T (into page) D YG –2 × × U (a) 3.75 × 10–3 N × (c) 3.75 × 102 N (d) 3.75 × 01–4 N Solution: (a) Force needed to move the rod is F  B 2 vl 2 (0.15 ) 2  2  (0.5) 2   3.75  10 3 N R 3 U Tricky example: 5  A sphere frame of metallic wire is moving in a uniform magnetic field (B) acting perpendicular to the paper inward as shown. LP and QN are also metallic wires then find the potential difference ST L between L and N l P l (a) Zero l (b) Bvl B v Q N (c) 2Bvl (d) 3Bvl Solution : (d) Higher potential L Bvl Higher potential Bvl We know that a flux cutting conductor can be treated as a single cell of emf e = Bvl. Hence the Bvl Bvl Bvl given figure can be redrawn as follows Lower potential Bvl N Bvl Lower potential Electromagnetic Induction 27 E3 60  VLN = 3Bvl Tricky example: 6 A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2 m/s in a uniform shown in figure. Then × × A B× U (a) qA = + 80 C and qB = – 80 C ID P connected magnetic field B  4.0 T directed into the paper. A capacitor of capacity as × C = 10 × F is × × × × × × × × × × D YG × × × Q (b) qA = – 80 C and qB = + 80 C (c) qA = 0 = qB × v (d) Charge stored in the capacitor increases exponentially with time Solution : (a) i ×P × 80 Q = CV = C (Bvl) = 10  10– 6  4  ×2  1 = C Higher potential × × × × × × B× × × × × A rule induced current According to Fleming's right hand flows from Q to P. Hence P is at higher v × × A is × positively × potential and Q is at lower potential. charged and B is negatively charged. × Therefore ST U Q Lower potential Motional EMI Due to Rotational Motion. (1) Conducting rod A conducting rod of length l whose one end is fixed, is rotated about the axis passing through it’s fixed end and perpendicular to it’s length with constant angular velocity . Magnetic field (B) is perpendicular to the plane of the paper.            Q     B  l     P   28 Electromagnetic Induction emf induces across the ends of the rod e  1 2 Bl 2 Bl   Bl 2   2 T Note 60 where  = frequency (revolution per sec) and T = Time period. :  If above metallic rod rotated about its axis of rotation, then induced potential difference e OP  e OQ i.e. e PQ  0 E3 It is clear parts OP and OQ are identical hence ID (2) Cycle wheel  B N Q Similarly e LN  0 (VL  VN ) l/2 l/2 between any pair of identical located points of rod, is always zero. L O l/4 l/4  P axis A conducting wheel each spoke of length l is rotating with angular U velocity  in a given magnetic field as shown below in fig. Due to flux cutting each metal spoke becomes identical cell of emf e D YG (say), all such identical cells connected in parallel fashion e net  e (emf of 1 single cell). Let N be the number of spokes hence e net  Bwl 2 ;  2 2        O        B l                Here e net  N o i.e. total emf does not depends on number of spokes ‘N’. Note :  Here magnetic field (may be component of Earth’s magnetic field) some times, depends on plane of motion of wheel. If wheel rotates in horizontal plane, then B = BV used; If wheel U rotates in vertical plane, then B = BH used (BH-horizontal component of earth’s magnetic field ST while BV-vertical component) (3) Faraday copper disc generator During rotational motion of disc, it cuts away magnetic field lines.  B  O r  B  e e e e e e e e Electromagnetic Induction 29 A metal disc can be assumed to made of uncountable radial conductors when metal disc rotates in E3 60 transverse magnetic field these radial conductors cuts away magnetic field lines and because of this flux cutting 1 all becomes identical cells each of emf ‘e’ where e  Br 2 , as shown in following fig. and periphery of disc 2 becomes equipotential. All identical cells connected in parallel fashion, So net emf for disc enet = e  1 Br 2  B(r 2 ) 2 ID Note :  If a galvanometer is connected between two peripheral points or diametrical opposite ends it’s reading will be zero. U (4) Semicircular conducting loop For the given figure a semi-circular conducting loop (ACD) of radius ‘r’ with centre at O, the plane of loop D YG being in the plane of paper. The loop is now made to rotate with a constant angular velocity , about an axis passing through O and perpendicular to the plane of paper. The effective resistance of the loop is R. A Region I  No magnetic fieldC  r O  r        S      r      = t B   ST U D Region II   In time t the area swept by the loop in the field i.e. region II A dA r 2  1 1  r(r )  r 2 t ; 2 dt 2 2 Flux link with the rotating loop at time t   BA Hence induced emf in the loop in magnitude | e |  i | e | B r 2  R 2R Periodic EMI. d dA B r 2 B  dt dt 2 and induced current 30 Electromagnetic Induction Suppose a rectangular coil having N turns placed initially in a magnetic field such that magnetic field is perpendicular to it’s plane as shown.  = 2  – Angular speed 60  B  – Frequency of rotation of coil n̂ R – Resistance of coil E3 For uniform rotational motion with , the flux linked with coil at any time t   NBA cos   NBA cos t (as  = t)    0 cos ωt where 0 = NBA = flux amplitude or maximum flux R ID (This relation shows that the flux changes in periodic nature) (1) Induced emf in coil Induced emf also changes in periodic manner that’s why this phenomenon called periodic EMI d  NBA  sin  t  e  e 0 sin ωt where e0 = emf amplitude or max. emf  NBA    0  dt U e At i0  any D YG (2) Induced current time t, e0 NBA   0    R R R Note :  i e e0  sin  t  i0 sin  t R R where i0 = current amplitude or max. For rotating coil, induced emf and linked flux keep a phase difference of  current i.e. when 2 U  plane of coil is perpendicular to magnetic field B so flux linked will be max. and induced emf e  0  and when plane of coil parallel to B flux linked min = 0 and induced emf will be maximum e max  e 0 ST  Frequency of induced any parameter = Frequency of rotation of coil =   Both emf and current changes their value w.r.t. time according to sine function hence they called as sinusoidal induced quantities. (3) Special cases (i) A rectangular coil rotates at a constant speed about one of its sides AB. The side AB is parallel to a long, straight current carrying conductor. A The current carrying conductor is in the plane of the page and the magnetic field due to it at coil is perpendicular to the plane of the paper. The B Electromagnetic Induction 31 emf induced in the coil rotating in this field is minimum when the coil is perpendicular to the field that is in the plane of the conductor. The emf will be maximum, when the coil is perpendicular to the plane of the conductor. Current amplitude given as i0  r 2 × × × × × × × × × × × × × × × × × × × × G × × × E3 Area of loop  BA  B(2 ) (r 2 )  2 r 2 B    R R 2 R 60 (ii) A stiff wire bent into a semicircle of radius ‘r’ is rotated at a frequency  in a uniform field of magnetic  induction B as shown in figure. If resistance of the entire circuit is R then × × × × × × × × × 2 × × ID (Frequency of induced current = frequency of rotation of loop = ) Examples A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per U Example: 41 second. If the horizontal component of earth's magnetic field is 0.2  10 4 T , then the e.m.f. developed D YG between the two ends of the conductor is (a) 5 mV (b) 50  V (c) 5  V (d) 50 m V 1 1 B H l 2    0.2  10  4  (1) 2  5  5  10 5 V  50  V 2 2 Solution: (b) Induced emf e  Example: 42 A rectangular coil of 300 turns has an average area of 25 cm  10 cm. The coil rotates with a speed of 50 cps in a uniform magnetic field of strength 4  10 2 T about an axis perpendicular to the field. The peak U value of the induced emf is (in volt) (a) 3000  (b) 300  (c) 30  (d) 3  Peak value of emf = e0 = NBA = 2 NBA = 2  50  300  4  10–2  (25  10–2  10  10–2) = 30  volt Example: 43 A wheel with ten metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a place normal ST Solution: (c) to the earth's magnetic field at the place. If the magnitude of the field is 0.04 G, the induced emf between the axle and the rim of the wheel is equal to [AMU 2002] (a) 1.256  10–3 V Solution: (d) e (b) 6.28  10–3 V (c) 1.256  10–4 V 1 2 120 Bl   Bl 2    (0.04  10  4 )  (0.5) 2  3.14   6.28  10 6 V. 2 60 (d) 6.28  10–6 V 32 Electromagnetic Induction Example: 44 A copper disc of radius 0.1 m rotates about its centre with 10 revolutions per second in a uniform magnetic field of 0.1 Tesla. The emf induced across the radius of the disc is 2 V 10 (c) 10  mV The induced emf between centre and rim of the rotating disc is E Example: 45 (b) V 1 1 BR 2   0.1  2  10  (0.1) 2  10   10  3 volt 2 2 (d) 20  mV E3 Solution: (c)  10 60 (a) A rectangular coil having dimensions 10 cm × 5 cm has 100 turns. It is moving at right angles to a field of 5 Tesla at angular speed of 314 rad/sec. The emf at instant when flux passing the coils is half the maximum value is Solution: (c)    0 cos   Example: 46 0 2 785 V 2   0 cos  ;   60 o (c) 785 3 V 2 (d) 0 e  e 0 sin   e 0   NBA  314  100  5  50  10 4  785 V  e = 785 sin 60o  U  (b) ID (a) 785 V 785 3 V 2 A loop of area 0.1 m2 rotates with a speed of 60 rev/sec with the axis of rotation perpendicular to a D YG magnetic field B = 0.4 T. If there are 100 turns in the loop, the maximum voltage induced in the loop is (a) 15.07 V (b) 150.7 V (c) 1507 V [MP PMT 1995 (d) 250 V Solution: (c) Maximum voltage e0 = NBA = 2NBA = 2 × 3.14 × 60 × 100 × 0.4 × 0.1 = 1507 V Example: 47 A square loop of side a is rotating about its diagonal with angular velocity  in a perpendicular magnetic field as shown in the figure. If the number of turns in it is 10 then the magnetic flux linked with the loop at U any instant will be (a) 10 Ba cos  t                     2 ST (b) 10 Ba  (c) 10 Ba2 (d) 20 Ba2 Solution: (a) The magnetic flux linked with the loop at any instant of time t is given by  = BAN cos  t or  = 10 Ba2 cos  t Here N = 10, A = a2 Example: 48 A very small circular loop of area A and the resistance R and negligible inductance is initially coplanar and concentric with a much larger fixed loop of radius x. A constant current i is passed in the bigger loop and Electromagnetic Induction 33 the smaller loop is rotated with constant angular velocity  about a diameter then induced current in the smaller loop as a function of time will be (a)  0 iA (b)  0 iA (d) 0 0i 2x. So    0 iA 2x cos  t ; e   (a) 5 V               x  i A d  0 i e   A sin  t  i   0 sin  t. dt 2x R 2 xR (c) 5 3 volt (d) None of these U (b) 5 2 V 1 sec will be 300 e  10 sin 314 t  10 sin(3.14  100 ) t  10 sin 100  t  1  10 3 sec ; e  10 sin  10 sin 60 o   5 3 V. 300 3 2 D YG Putting t  Example: 50  In periodic motion of a coil in an uniform magnetic field if induced emf at any instant t is given by e  10 sin 314 t then induced emf at t  Solution: (c)  At any instant t flux linked with smaller loop   BA cos wt where B = magnetic field produced by larger loop at it’s centre  Example: 49   ID Solution: (b)  E3 sin  t 2 xR  0 iA sin 2 t 2 xR (c)  60 2 xR  sin  t In the previous question at what time t instantaneous induced emf will be half of maximum induced emf (a) 1 sec 300 (b) 1 sec 400 (c) 1 sec 500 (d) 1 sec 600 1 10 1 1   10 sin 314 t   sin 100 t  100 t  sin 1    , t  sec 2 2 600 2 6 Solution: (d) Given e  10 sin 314 t  10 sin 100 t ; Example: 51 In a region of uniform magnetic induction B = 10– 2 Tesla, a circular coil of radius 30 cm and resistance 2 U ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is [CBSE 1990] ST (a) 42 mA Solution: (c) i0  (b) 30 mA (c) 6 mA B A N  10 2    (30  10 2 ) 2  1  2  200   6  10  3 A  6 mA. 2 R   60 (d) 200 mA Electromagnetic Induction 29 Static EMI. Inductance is that property of electrical circuits which opposes any change in the current in the circuit. 60 Inductance is inherent property of electrical circuits. It will always be found in an electrical circuit whether we want it or not. The circuit in which a large emf is induced when the current in the circuit changes is said to have greater inductance. A straight wire carrying current with no iron part in the circuit will have lesser value of E3 inductance while if the circuit contains a circular coil having many number of turns, the induced emf to oppose the cause will be greater and the circuit is therefore said to have greater value of inductance. Inductance is called electrical inertia : Inductance is analogous to inertia in mechanics, because we know ID that due to inertia a body at rest opposes any attempt which tries to bring it in motion and a body in motion opposes any attempt which tries to bring it to rest. Inductance of an electrical circuit opposes any change of current in the circuit thus it is also called electrical inertia. U (1) Self-Induction Whenever the electric current passing through a coil or circuit changes, the magnetic flux linked with it will D YG also change. As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil or the circuit which opposes the change that causes it. This phenomenon is called ‘self induction’ and the emf induced is called back emf, current so produced in the coil is called induced current. U Induced current Rheostat Key ST Main current increasing Induced current Rheostat Key Main current decreasing (i) Coefficient of self-induction : If no magnetic materials are present near the coil, number of flux linkages with the coil is proportional to the current i. i.e. N   i or N   Li (N is the number of turns in coil and N – total flux N linkage) where L  = coefficient of self induction. i If i = 1amp, N = 1 then, L =  i.e. the coefficient of self induction of a coil is equal to the flux linked with the coil when the current in it is 1 amp. 30 Electromagnetic Induction By Faraday’s second law induced emf e   N d di di. Which gives e  L ; If  1 Amp / sec then |e|= L. dt dt dt Hence coefficient of self induction is equal to the emf induced in the coil when the rate of change of 60 current in the coil is unity. Note :  Here we must note that if we are asked to calculate the induced emf in an inductor, then we have di. But when we are asked to calculate the voltage (V) across the inductor then dt V | e |  E3 e  L di L dt (ii) Units and dimensional formula of ‘L’ weber Tesla  m 2 N m Joule Coulomb  volt volt  sec       ohm  sec 2 2 Amp Amp amp Amp Amp Amp 2 ID S.I. unit : But practical unit is henry (H). It’s dimensional formula [L] = [ML2T–2A–2] U Note :  1 henry = 109 emu of inductance or 109 ab-henry. (iii) Dependence of self inductance (L) : ‘L’ does not depend upon current flowing or change in current D YG flowing but it depends upon number of turns ( N), Area of cross section (A) and permeability of medium (). (Soft iron has greater permeability. Hence greater self inductance L) ‘L’ does not play any role till there is a constant current flowing in the circuit. ‘ L’ comes in to the picture only when there is a change in current. (iv) Magnetic potential energy of inductor : In building a steady current in the circuit, the source emf has to U do work against of self inductance of coil and whatever energy consumed for this work stored in magnetic field U Parabolic ST of coil this energy called as magnetic potential energy ( U) of coil U  i 0 Lidi  Ni 1 1 2 Li ; Also U  (Li)i  2 2 2 Note :  Energy density is given as U  1 B 2 2 0 i. (v) Calculation of self inductance for a current carrying coil : If a coil of any shape having N turns, carries a current i, then total flux linked with coil N   Li Also  = BA cos ; where B = magnetic field produced at the centre of coil due to it’s current; A = Area of each turn;  = Angle between normal to the plane of coil and direction of magnetic field. Electromagnetic Induction 31 L N NBA cos  ;  i i If  = 0o,  max  BA So L NBA i Circular coil calculated as follows as field at the centre of coil due to   2Ni  2 N 0.  (r )  N 2 r 4 r   L  0 i 2  Note :  own current B 2 (For constant r)  0 2Ni. r 4  ID LN L1  N 1    L 2  N 2  If radius is doubled so self inductance will also doubles. (L  r) (If N = constant) U  2 its E3 Magnetic 60 If a circular coil of N turns carrying current i and its each turn is of radius r then its self inductance can be  If area across section is doubled (N = constant) i.e. So A '  2 A  r' 2  2  r 2  r'  2r 2 L i.e. increase in self induction is 41.4%. D YG L'   If a current carrying wire of constant length is bend into circular coil of N-turns then N (2r)  l ; l N r Now as If N - given L  N 2r If r - given  l  L  N 2   L  N N L l r   L  l 2 r r U e.g. If a wire of length l first bent in single turn circular coil then in double turn (concentric ST coplanar) coil so by using L  N we can say that L in second case twice that in first case. Other important cases Square coil Triangular coil l Toroid Winding Core l l O Solenoid O i l i r 32 Electromagnetic Induction B L 2 20 N a  L N. B   0 ni  l   18 Ni   3 2  N 0 l    4  l   4  L i   8 2 Ni  2 a N 0.  4 a   L i 2  0 18 Ni 4 L 2 9 3 0 N 2l 8  L  N2 l   Ni  N  0 . A  l  L ; i  N2A L  N2 L 0 l B  0 Ni 2r   Ni  N  0  r 2  N 2r 2r   0 L  i 2 For iron cored solenoid L 0 r N 2 A N 2 A  (  0 r ) l l Inductance at the ends of a solenoid is half of it's the inductance at the centre. ID Note :   0 Ni 60 0 8 2 i. N 4 a E3 B (2) Mutual Induction U 1    Lend  Lcentre . 2   Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a D YG neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’. The coil or circuit in which the current changes is called ‘primary’ i i2 1 while the other in which emf is set up is called ‘secondary’. Variable current P S Load R ST U M In case of mutual inductance for two coils situated close to each other, total flux linked with the secondary due to current in the primary is N2 2 and N2 2  i 1  N 2  2  Mi 1 where N1 - Number of turns in primary; N2 - Number of turns in secondary; 2 - Flux linked with each turn of secondary; i 1 - Current flowing through primary; M-Coefficient of mutual induction or mutual inductance. Electromagnetic Induction 33 According to Faraday’s second law emf induces in secondary e 2   N 2 di d 2 di 1 Amp ; e 2   M 1 ; If 1  dt dt sec dt of change of current in primary coil is unity. Units and dimensional formula of M are similar to self-inductance (L) (a) Number of turns (N1, N2) of both coils (b) Coefficient of self inductances (L1, L2) of both the coils ID (c) Area of cross-section of coils E3 (i) Dependence of mutual inductance 60 then |e2| = M. Hence coefficient of mutual induction is equal to the emf induced in the secondary coil when rate (d) Magnetic permeability of medium between the coils ( r) or nature of material on which two coils are U wound (e) Distance between two coils (As d  = M ) D YG (f) Orientation between primary and secondary coil (for 90 o orientation no flux relation M = 0) (g) Coupling factor ‘K’ between primary and secondary coil (ii) Calculation of mutual inductance between two coils If two coils (1 and 2) also called primary and secondary coils are placed close to each other (maximum U coupling); N1 and N2 = Number of turns in primary and secondary coils respectively, 2 = Flux linked with each turn of secondary, N22 = Total flux linkage with secondary coils; M = Mutual inductance between two coil ST So N22 = Mi1  N2(B1A2) = Mi1  M  B1 N 2 A 2 i1 Two concentric coplaner circular coils Two Solenoids i1 R l P r i1 S Two concentric coplaner square coils (1) (2) i Primary Secondary (N1 turns) (N2 turns) l L 34 Electromagnetic Induction If two air cored solenoid tightly Magnetic field at the centre due to current in outer coil is wound to each other as shown : current in outer coil is  0 2N 1 i1. , 4 R Magnetic field inside the primary N  solenoid B1   0 n1 i1  n1  1 l  From the above formula M  0 N 1 N 2 r 2 2R  M r2 R From the above formula B1  From the above formula M {A = Area of each solenoid) μ0 N 1 N 2 A l 0 2 2 N 1 N 2l 2 l2  M L L ID M  0 8 2 N 1 i1. 4 L E3 B1  60 Magnetic field at the centre due to U (iii) Relation between M, L1 and L2 For two magnetically coupled coils M  k L1 L 2 ; where k – coefficient of coupling or coupling factor magnetic flux linked in secondary ; magnetic flux linked in primary D YG which is defined as k  P S P 0k1 Air S S P gap M  L1 L 2 (Normal U coupling) If coils are tightly coupled (k = 1) No coupling (k = 0) N N N L Specially for Transformer in ideal case M  2 L1 and M  1 L 2 ; 1   1 L2  N 2 N1 N2 ST Note :  If coils are loosely coupled (0 < k < 1)    2 (3) Combination of inductance (i) Series combination Mutual induction is absent (k = 0) Mutual induction is present and Mutual induction is present and favours self inductance of coils opposes self inductance of coils L2 L1 L1 L2 i (CW) M (CW) L1 i (CW) i M L2 (ACW) (ii) Parallel combination Winding nature same Opposite winding nature Their flux assist each other Their flux opposes each other Leq  L1  L2  2 M Leq  L1  L2  2 M E3 Current in opposite direction Mutual induction is present and U Mutual induction is absent (k = 0) Current in same direction ID L eq  L1  L 2 60 Electromagnetic Induction 35 D YG favours self inductance of coils Mutual induction is present and opposes self inductance of coils L1 L2 L1 L 2 L1  L 2 L eq  L1 L 2  M 2 L1  L 2  2 M L eq  L1 L 2  M 2 L1  L 2  2 M ST U L eq  Note :  If nothing is said then it is to be considered that mutual inductance between the coils is absent. Concepts  A thin long wire made up of material of high resistivity behaves predominantly as a resistance. But it has some amount of 36 Electromagnetic Induction inductance as well as capacitance in it. It is thus difficult to obtain pure resistor. Similarly it is difficult to obtain pure capacitor as well as pure inductor. Due to inherent presence of self inductance in all electrical circuits, a resistive circuit with no capacitive or inductive element in it, also has some inductance associated with it. 60  The effect of self-inductance can be eliminated as in the coils of a resistance box by doubling back the coil on itself. The coil is placed in space as shown in figure below It is not possible to have mutual inductance without self inductance but it may or may not be possible self inductance E3  without mutual inductance.  If main current through a coil increases (i ) so di will be positive (+ve), hence induced emf e will be negative (i.e. opposite dt  If main current through a circuit (coil) decreases (i ) so di will be negative (– ve), hence induced emf e will be positive (i.e. dt D YG same directed emf)  Enet  E  e Sometimes at sudden opening of key, because of high inductance of circuit a high momentarily induced emf produced  U  U ID emf)  E net  E  e  In checking balancing of Wheatstone bridge, always firstly pressed cell key and after–wards galvanometer key, so that and a sparking occurs at key position. To avoid sparking a capacitor is connected across the key. ST One can have resistance with or without inductance but one can’t have inductance without having resistance. momentarily induced current produced, because of self inductance of coil of galvanometer becomes almost zero or disappear.  The circuit behaviour of an inductor is quite different from that of a resistor. while a resistor opposes the current i, an inductor opposes the change di in the circuit dt Electromagnetic Induction 37 Example: 52 60 Examples A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self (a) 25 millihenry E3 induction of the coil will be (b) 25 × 10–3 millihenry  0 N 2 r (d) 50 × 10–3 henry (3.14 )  4  (3.14 )  10 7  (500 ) 2  5  10 2  25  10 3 H  25 mH 2 Solution: (a) By using L  Example: 53 A solenoid has 2000 turns wound over a length of 0.30 metre. The area of its cross-section is 1.2 × 10–3 m2. ID 2 ; L (c) 50 × 10–34 millihenry Around its central section, a coil of 300 turns is wound. If an initial current of 2A in the solenoid is reversed (a) 6 × 10–4 V U in 0.25 sec, then the emf induced in the coil is (b) 4 × 10–3 V 0 N1 N 2 A (d) 48 mV {2  (2)} di ; M  3.01  10 3 H  e  3.01  10  3  ; e  48 mV. 0.25 dt Solution: (d) By using M  Example: 54 The coefficient of self inductance of a solenoid is 0.18 mH. If a crode of soft iron of relative permeability D YG and | e |  M (c) 6 × 10–2 V [NCERT 1982] l 900 is inserted, then the coefficient of self inductance will become nearly (a) 5.4 mH We know for air cored solenoid L  (c) 0.006 mH (d) 0.0002 mH 0 N 2 A l U Solution: (b) (b) 162 mH In case of soft of iron core it’s self inductance L'  0 r N 2 A l ; L'   r L. So here L = 900 × 0.18 = 162 mH ST Note :  The self-inductance of a solenoid may be increased by inserting a soft iron core. The function of the core is to improve the flux linkage between the turns of the coil. Example: 55 The current in an inductor is given by i = 2 + 3t amp where t is in second. The self induced emf in it is 9 mV the energy stored in the inductor at t= 1 second is (a) 10 mJ Solution: (b) (b) 37.5 mJ At t = 1 sec, i = 2 + 3  1 = 5A and | e |  L So energy U  (c) 75 mJ (d) Zero di d  9  10–6  L  (2  3 t)  L = 3  10–3 H dt dt 1 2 1 Li  (3  10 3 )  (5) 2 = 37.5 mJ. 2 2 38 Electromagnetic Induction Example: 56 The number of turns in two coils A and B are 300 and 400 respectively. They are placed close to each other. Co-efficient of mutual induction between them is 24 mH. If the current passing through the coil A is 2 Amp then the flux linkage with coil B will be (b) 12 × 10–5 wb (c) 48 mwb (d) 48 × 10–5 wb 60 (a) 24 mwb Flux linkage  N 2 2  Mi1  24  2  48 mwb Example: 57 A coil of wire of a certain radius has 600 turns and a self-inductance of 108 mH. The self-inductance of another similar coil of 500 turns will be (a) 74 mH (b) 75 mH L1  N 1    L2  N 2  2 E3 Solution: (c) (c) 76 mH 2 [MP PMT 1990] (d) 77 mH 108  600    ; L 2  75 mH L2  500   Example: 58 Two different coils have self-inductance L1 = 8mH, L2 = 2mH. The current in one coil is increased at a L  N2   ID Solution: (b) U constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy D YG stored in the first coil are i1, V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2, V2 and W2 respectively. Then choose incorrect option (a) Solution: (b) i1 1  i2 4 By | e |  L (b) e L di  1  1 e2 L2 dt 1 e (c) W2 4 W1  di    same   dt   V1 8  4 V2 2 P  same   i1 e V L 2 1  2  2  2   i2 e1 V1 L1 8 4 U Power P  ei  i  i1 4 i2 i W L 1 Energy stored W  Li 2 ; 1  1   1 2 W2 L 2  i2 V2 1  V1 4  1 1   4    . 4 4  2 A current increases uniformly from zero to one ampere in 0.01 second, in a coil of inductance 10 mH it. The ST Example: 59 2 (d) induced emf will be (a) 1 V (b) 2 V (c) 3 V (d) 4 V Solution: (a) e  L dI 1.0  10  10  3  1 volt  | e |  1 volt dt 0.01 Example: 60 The current in a coil varies w.r.t. to time t as I = 3t2 + 2t. If the inductance of coil be 10 mH, the value of induced emf at t = 2s will be (a) 0.14 V Solution: (a) e  L (b) 0.12 V (c) 0.1 V dI d   [3 t 2  2 t]   L[6 t  2]  10  10 3 [6 t  2] dt dt (d) 0.13 V Electromagnetic Induction 39  10  10 3 (14 )   0.14 volt ; | e |  0.14 volt (e)at t 2  10  10 3 (6  2  2) What inductance would be needed to store 1 KWh of energy in a coil carrying a 200 A current (a) 100 H (b) 180 H (c) 200 H (d) 450 H 60 Example: 61 1 1 2 Li  3.6  106   L  (200 ) 2  L = 180 H 2 2 Solution: (b) U = 1KWH = 3.6  106 J. By using U  Example: 62 The self inductance of a coil is L, keeping the length and area same, the number of turns in the coil is (a) 1 L 4 [MP PMT 1997] E3 increased to four times. The self inductance of the coil will now be (b) L (c) 4 L (d) 16 L L  N2 Example: 63 The mutual inductance between a primary and secondary circuit is 0.5 H. The resistance of the primary and ID Solution: (d) the secondary circuits are 20 ohms and 5 ohms respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of (b) 16.0 A/s (c) 1.6 A/s U (a) 4.0 A/s (d) 8.0 A/s e di1 di1 M di1 0.5 di1  0.4  ;  4 A / sec  ; i2  2  R2 R 2 dt dt dt 5 dt By using | e 2 |  M Example: 64 The average emf induced in a coil in which a current changes from 0 to 2 A in 0.05 s is 8 V. The self D YG Solution: (a) inductance of the coil is (a) 0.1 H [CPMT 1999] (b) 0.2 H di1 (2  0) ; 8  L 0.05 dt (c) 0.4 H (d) 0.8 H By using | e |  L Example: 65 A coil of Cu wire (radius-r, self inductance-L) is bent in two concentric turns each having radius is U Solution: (b)  L  0.2 H r. The 2 self inductance now ST (a) 2L (b) L (c) 4 L 2 L1  N 1  r   1  L 2  N 2  r2 2 L 1  r  1       ; L2 = 2 L L2  2  r/2 2 Solution: (a)  L  N 2r ; Example: 66 In the following circuit, the bulb will become suddenly bright if (a) Contact is made or broken (b) Contact is made (c) Contact is broken (d) L / 2  [CBSE 1989] 40 Electromagnetic Induction (d) Won’t becomes bright at all Solution: (c) When contact is broken induced current flows in the same direction of main current. So bulb suddenly glows more brightly. Three inductances are connected as shown below. Assuming no coupling, the resultant inductance will be 60 Example: 67 L2 = 0.50 H (a) 0.25 H E3 L1 = 0.75 (b) 0.75 H L3 = 0.50 H (c) 0.01 H Solution: (d) ID (d) 1 H L2 and L3 are in parallel. Thus their combination gives L'  L2 L3  0.25 H L2  L3 U The L and L1 are in series, thus the equivalent inductance is L = L1 + L = 0.75 + 0.25 = 1H. D YG Tricky example: 7 The current through a 4.6 H inductor is shown on the following graph. The induced emf during the time interval t = 5 milli-sec to 6 milli-sec will be i (Amp) (a) 103 V (b) – 23 103 V A 7 B 5 C 0 2 5 6 t (milli-sec) (c) 23  10 V 3 U (d) Zero Solution : (c) Rate of decay of current between t = 5 ms to 6 ms  di   (Slope of the line BC ) dt ST 5 di       4.6  ( 5  10 3 )  23  10 3 V.    5  10 3 A / s. Hence induced emf e   L 3 dt  1  10  Growth and Decay of Current in LR-Circuit. If a circuit containing a pure inductor L and a resistor R in series with a battery and a key then on closing the circuit current through the circuit rises exponentially and reaches up to a certain maximum value (steady state). If circuit is opened from it’s steady state condition then current through the circuit decreases exponentially. L L Induced Induced current current Main current Main + B K Growth of current current + B K Decay of current E3 60 Electromagnetic Induction 41 The value of current at any instant of time t after closing the circuit (i.e. during the rising of current) is given ID R   t E by i  i 0 1  e L  ; where i0  imax  = steady state current. R     The value of current at any instant of time t after opening from the steady state condition (i.e. during the  R t L U decaying of current) is given by i  i 0 e (1) Time constant () L ; It’s unit is second. In other words the time interval, during which the current in an R D YG In this circuit   inductive circuit rises to 63% of its maximum value at make, is defined as time constant or it is the time interval, during which the current after opening an inductive circuit falls to 37% of its maximum value. i i0 i i0 ST U i = 0.63 i0 Note :  i = 0.37i0 t= The dimensions of t t= t L are same as those of time i.e. M0L0T1 R  Half life (T) : In this time current reduces to 50% of its initial max value i.e. if t  T then i  again half life obtained as T = 0.693 L R or T = 70% of time constant. i0 and 2 42 Electromagnetic Induction Now from U  i 1 2 Li so in half life time current changes from i0 0 hence energy changes from 2 2 60 U0 4 U0 (2) Behaviour of inductor  The current in the circuit grows exponentially with time from 0 to the maximum value i   E3  E . Just after R closing the switch as i = 0, inductor act as open circuit i.e. broken wires and long after the switch has been closed as i = i0, the inductor act as a short circuit i.e. a simple connecting wire. L R U E Initially S E i S Long after closing Just after closing S i D YG LC Oscillation. i=E/R i=0 i=0 E R ID R U When a charged capacitor C having an initial charge q0 is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is ST dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant. L Frequency of oscillation is given by + q0 C   – 1 rad LC sec or   1 2 LC Hz Electromagnetic Induction 43 The oscillation of the LC circuit are an electromagnetic analog to the mechanical oscillation of a block-spring system. +q0 C + + + + E t=0 – – – – – q0 L 60 i=0 v=0 At t = 0, capacitor is At t = 0, block is k ready to discharge m A i = imax t  q=0 T At t  , capacitor is fully 4 L vmax ID C T E3 x=0 S k discharged i.e. charge q = 0 B and current through the circuit 4 is maximum m A D YG At t  T , block comes in it's 4 mean position i.e. x = 0and velocity of block becomes x=0 U S ready to move maximum i=0 – t  C – – – E T 2 + + + + At t  – q0 + q0 T 2 , capacitor is again L recharged with reverse polarity and i = 0 v=0 k T , block reaches it's 2 m extreme position other side and v = 0 A x=0 S U At t  i = imax C q=0 ST 3T t  4 L At t  3T vmax , capacitor again 4 discharges completely i = imax At t  k , block again 4 m B 3T reaches it's mean position and A it's velocity becomes maximum x=0 S i = imax C t  3T At t  q=0 4 L 3T , capacitor again 4 discharges completely i = imax v=0 k m B S A x=0 At t  3T , block again 4 reaches it's mean position and it's velocity becomes maximum 44 Electromagnetic Induction  60 Concepts Comparison of oscillation of a mass spring system and an LC circuit Mass spring system v/s LC circuit Charge (q) Velocity (v) Current (i) Acceleration (a) Rate of change of current  Mass (m) [Inertia] Inductance (L) [Inertia of electricity] E3 Displacement (x) ID  di    dt  Momentum (p = mv)  dv   dt   Self induced emf   L Equation of free oscillations : 2   2 x ; where   K m D YG d x U  Retarding force   m Magnetic flux ( = Li) dt 2 Force constant K, Kinetic energy  1 mv 2 2 Elastic potential energy = Equation of free oscillations : d 2q dt 2 1  1  2    .q ; where   LC  LC  Magnetic energy = (a) Zero LC 1 2 Li 2 Electrical potential energy  In the figure magnetic energy stored in the coil is ST Example: 68 1 Capacitance C U Examples 1 Kx 2 2  di   dt  [RPET 2000] 2H 10 V 1 q2 2 C 2 (b) Infinite (c) 25 J (d) None of these 2 Solution: (c) 1 1  10  U  Li 2   2    25 J 2 2  2  Example: 69 An emf of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t =  and at t = 1 second is [MP PMT 1994] Electromagnetic Induction 45 e1/2 e Solution: (b) 1/2 e2 (b) 1 Rt   By using i  i0  1  e L   e 1 10 1  2    ; At t = , i = i and at t = 1 sec i  i 1  e  5  ; i  i (1  e  2 )  i  e  1  ; 0 0 0 0  e2          i0 e2  2 i e 1 Example: 70 (d) e– 1 (c) 1  e 1 2 60 (a) An ideal coil of 10 henry is joined in series with a resistance of 5 ohm and a battery of 5 volt. 2 second after (a) e–1 (b) (1 – e–1) (c) (1 – e)  5  1  e 10 5  [MP PET 1995] E3 joining, the current flowing in ampere in the circuit will be (d) e Solution: (b) Rt   By using i  i0  1  e L   Example: 71 A coil of self inductance 50 henry is joined to the terminals of a battery of emf 2 volts through a resistance 52 i   or i  (1  e 1 )   ID  ;   of 10 ohm and a steady current is flowing through the circuit. If the battery is now disconnected, the time (a) 500 seconds U in which the current will decay to 1/e of its steady value is (b) 50 seconds (c) 5 seconds (d) 0.5 seconds L 1 current becomes times of it’s initial value i.e. i0. So R e 50  5 sec. 10 In decaying if t    Example: 72 A solenoid has an inductance of 50 mH and a resistance of 0.025 . If it is connected to a battery, how D YG Solution: (c) t long will it take for the current to reach one half of its final equilibrium value (a) 1.34 ms Solution: (a) t   i  i0  1  e    (b) 1.2 ms t    where i  1 i and   L. Thus 1 i  i  1  e   0 0 0   2 2 R   U Thus t   log e 2  Example: 73 (c) 6.32 ms (d) 0.23 ms t t   or 1  e   or 2  e    2  50  10 3  0.693  1.34  10 3 s  1.34 milli second. 0.025 A 50 volt potential difference is suddenly applied to a coil with L = 5  10–3 henry and R = 180 ohm. The rate of ST increase of current after 0.001 second is (a) 27.3 amp /sec (b) 27.8 amp/sec (c) 2.73 amp/sec (d) None of these Solution: (d) Example: 74 In which of the following circuit is the current maximum just after the switch S is closed R R E E R L S (i) R L E R S S (ii) (iii) 46 Electromagnetic Induction (a) (i) (c) (iii) At t = 0 current through L is zero so it acts as open circuit. The given figures can be redrawn as follow. R R E S R ID (ii) i1  0 i2  E R i3  (iii) E 2R An oscillator circuit consists of an inductance of 0.5 mH and a capacitor of 20 F. The resonant frequency of the circuit D YG is nearly (a) 15.92 Hz Solution: (b) R U Hence i2 > i3 > i1 E S S (i) R E3 E Example: 75 (d) Both (ii) and (iii) 60 Solution: (b) (b) (ii) (b) 159.2 Hz [Kerala PET 2002] (c) 1592 Hz (d) 15910 Hz Tricky example: 8 The resistance in the following circuit is increased at a particular instant. At this instant the value of 10 mH U resistance is 10. The current in the circuit will be now ST (a) i = 0.5 A (b) i > 0.5 A i 5V RH (c) i < 0.5 A (d) i = 0 Solution : (b) If resistance is constant (10) then steady current in the circuit i  5  0. 5 A. But resistance is 10 increasing it means current through the circuit start decreasing. Hence inductance comes in picture which induces a current in the circuit in the same direction of main current. So i > 0.5 A. Electromagnetic Induction 47 Application of EMI. (1) Eddy current 60 When a changing magnetic flux is applied to a bulk piece of conducting material then circulating currents called eddy currents are induced in the material. Because the resistance of the bulk Non-uniform conductor is usually low, eddy currents often have large magnitudes and heat up E3 the conductor. magnetic field These are circulating currents like eddies in water Eddy currents Metallic block ID Experimental concept given by Focault hence also named as “Focault current” (i) Disadvantages of eddy currents U (a) The production of eddy currents in a metallic block leads to the loss of electric energy in the form of heat. D YG (b) The heat produced due to eddy currents breaks the insulation used in the electrical machine or appliance. (c) Eddy currents may cause unwanted damping effect. (ii) Minimisation of losses due to eddy currents U By Lamination, slotting processes the resistance path for circulation of eddy current increases, resulting in to weakening them and also reducing losses causes by them (slots and lamination intercept the conducting ST paths and decreases the magnitude of eddy currents and reduces possible paths of eddy currents) Plane metal × × × × Slotted metal plate × × × × × × × Strong eddies produced Cause excessive electro magnetic damping B plate × × × × × × × Feeble eddies Gradual damping Solid metallic Strong eddy core currents Laminated core (Feeble eddy currents) 48 Electromagnetic Induction (iii) Application of eddy currents : Though most of the times eddy currents are undesirable but they find 60 some useful applications as enumerated below (a) Dead-beat galvanometer : A dead beat galvanometer means one whose pointer comes to rest in the final equilibrium position immediately without any oscillation about the equilibrium position when a current is E3 passed in its coil. We know that the coil of a moving coil galvanometer is wound over a light aluminium frame. When the coil moves due to the torque produced by the current being measured, the aluminium frame also moves in the field. ID As a result the flux associated with the frame changes and eddy currents are induced in the frame. Eddy currents induced in aluminium frame as per Lenz’s law always oppose the cause that produces them. Hence they damp U the oscillation about the final steady position. (b) Electric-brakes : When the train is running its wheel is moving in air and when the train is to be stopped D YG by electric breaks the wheel is made to move in a field created by electromagnet. Eddy currents induced in the wheels due to the changing flux oppose the cause and stop the train. (c) Induction furnace : Here a large amount of heat is to be generated so as to melt metal in it. To produce such a large amount of heat, a solid core of the furnace is taken (as against laminated core in situations where the heat produced is to be minimized). U (d) Speedometer : In the speedometer of an automobile, a magnet is geared to the main shaft of the vehicle and it rotates according to the speed of the vehicle. The magnet is mounted in an aluminium cylinder ST with the help of hair springs. When the magnet rotates, it produces eddy currents in the drum and drags it through an angle, which indicates the speed of the vehicle on a calibrated scale. (e) Diathermy : Eddy currents have been used for deep heat treatment called diathermy. (f) Energy meter : In energy meters, the armature coil carries a metallic aluminium disc which rotates between the poles of a pair of permanent horse shoe magnets. As the armature rotates, the current induced in the disc tends to oppose the motion of the armature coil. Due to this braking effect, deflection is proportional to the energy consumed. (2) dc motors Electromagnetic Induction 49 It is an electrical machine which converts electrical energy into mechanical energy. (i) Principle : It is based on the fact that a current carrying coil placed in the magnetic field experiences a 60 torque. This torque rotates the coil. (ii) Construction : It consists of the following components figure. B C  F2 S1, S2 = split ring comutators B1, B2 = Carbon brushes S  F1 A S1 S2 B1 B2 D B N D S2 S1 B1 B2 S A D YG U ID N, S = Strong magnetic poles N C E3 ABCD = Armature coil    (iii) Working : Force on any arm of the coil is given by F  i(l  B) in fig., force on AB will be perpendicular to plane of the paper and pointing inwards. Force on CD will be equal and opposite. So coil rotates in clockwise sense when viewed from top in fig. The current in AB reverses due to commutation keeping the force on AB and CD in such a direction that the coil continues to rotate in the same direction. U (iv) Back emf in motor : When the