NEET Physics Notes - Chapter 21 - Magnetic Effect of Current PDF

1173 60 Magnetic Effect of Current Chapter E3 21 ID Magnetic Effect of Current Oersted found that a magnetic field is established around a current carrying conductor. a vector quantity. Its direction is same as the direction of current. Magnetic lines of forces D YG i Fig. 21.1 Biot-Savart's Law U B...

1173 60 Magnetic Effect of Current Chapter E3 21 ID Magnetic Effect of Current Oersted found that a magnetic field is established around a current carrying conductor. a vector quantity. Its direction is same as the direction of current. Magnetic lines of forces D YG i Fig. 21.1 Biot-Savart's Law U Biot-Savart’s law is used to determine the magnetic field at any point due to a current carrying conductor. ST This law is although for infinitesimally small conductor yet it can be used for long conductors. In order to understand the Biot-Savart’s law, we need to understand the term current- element. According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element i dl is given by the expression, dB k i dl sin  r2  nˆ also B  d B  0 i dl sin ˆ. n 4 r2 In C.G.S. k = 1 and in S.I. : k  where  0  4  10 7  0 4 = Absolute permeability of air or vacuum Wb. It's other units are Amp  metre N Tesla  metre Henry or or Ampere metre Amp 2   i(dl  rˆ )  i(dl  r )  0  Vectorially, d B  0  4 4 r2 r3 P dl r i Fig. 21.2 Direction of Magnetic Field The direction of magnetic field is determined with the help Current element It is the product of current and length of infinitesimal of the following simple laws : (1) Maxwell’s cork screw rule : According segment of current carrying wire. The current element is taken as Current element AB = i dl U Magnetic field exists as long as there is current in the wire. i A B to this rule, if we imagine a right handed dl Fig. 21.3 1174 Magnetic Effect of Current screw placed along the current carrying linear conductor, be is directed perpendicular and out of the plane of the paper it is rotated such that the screw moves in the direction of flow of represented by  (dot) current, then the direction of rotation of the thumb gives the direction of magnetic lines of force. (2) Right hand thumb rule : According conductor is held in the right hand such B direction of current flow, then the direction of folding fingers will represent In : Magnetic field is away from the observer or perpendicular inwards. Fig. 21.4 the E3 that the thumb of the hand represents the 60 to this rule if a straight current carrying Out direction of magnetic lines of force. : Magnetic field is towards the observer or perpendicular outwards. Ampere’s Law this rule if the direction of B current in circular conducting coil is in the direction of then the direction magnetic field due to a given current distribution. Line integral of the magnetic field B around any closed curve is equal to  0 times the B U folding fingers of right hand, Amperes law gives another method to calculate the ID (3) Right hand thumb rule of circular currents : According to i of the area enclosed by the curve Fig. 21.5 D YG magnetic field will be in the  net current i threading through direction of stretched thumb. (4) Right hand palm rule If we stretch our right hand i3 i1 i.e. i5 i2  B  dI   0 i4  i   0 (i1  i3  i2 ) Fig. 21.8 Also using B  0 H (where H = magnetising field)   H.dl   i   H.dl   i 0 0 Total current crossing the above area is (i1  i3  i2 ). Any the point. At which magnetic field current outside the area is not included in net current. (Outward U such that fingers point towards is required while thumb is in the ST direction of current then normal to  +ve, Inward  – ve) B Table 21.1 : Biot-Savart's law v/s Ampere's law Fig. 21.6 the palm will show the direction of magnetic field. Meaning of Cross  and dot  If magnetic field is directed perpendicular and into the plane of the paper it is represented by  (cross) while if magnetic field i B Out i i B In B In CW i B Out Fig. 21.7 In Out ACW Biot-Savart's law Ampere's law this law is valid for all current This law is valid for symmetrical distributions current distributions This law is the differential form of Basically this law is the integral B or H from of B or H This law is based only on the This principle of magnetism principle of electromagnetism. law is based on the 1175 Magnetic Effect of Current At these points B varies linearly with x  (i) Magnetic Field Due to Circular Current constant  If a coil of radius r, carrying current i then magnetic field on it's axis at a distance x from its centre given by (Application of r from the centre of the coil 2 4  0 Ni r and the magnetic field at x  is B  O 2 5 5r (ii) These are located at x   r P O x d2B 0. dx 2 B (7) Helmholtz coils i 60 Biot-Savart's law) dB  dx (i) This is the set-up of two coaxial coils of same radius Fig. 21.9 such that distance between their centres is equal to their radius. μ0 2 πNir ; where N = number of turns in coil.. 4 π (x 2  r 2 )3 / 2 (2) At centre x = 0  Bcentre  2Ni  Ni = 0  Bmax  0. 4 r 2r B 0 2 Nir2 0 2 NiA. . 3 4 4 x3 x D YG (4) If x >> r  Baxis   B centre x   1  2  B axis r   0 Ni R  1.432 B , where B  where A = r2 = Area of each turn of the coil. (5) B-x curve : The variation of magnetic field due to a 0 Ni 2R (iii) Current direction is same in both coils otherwise this arrangement is not called Helmholtz’s coil arrangement. (iv) Number of points of inflextion  Three (A, A, A) U and on it's axis is given by 3/2  0.716 5 5R (3) The ratio of magnetic field at the centre of circular coil 2 8 0 Ni ID (1) Baxis  2 E3 (ii) At axial mid point O, magnetic field is given by a Resultant field (Uniform) a O O2 O1 O1 + + – circular coil as the distance x varies as shown in the figure. A A x   – O a A O2 x  x a 2 2 Fig. 21.11 B varies non-linearly with distance x as shown in figure and is maximum when x 2  min  0 , i.e., the point is at the centre of U the coil and it is zero at x =  . B0 ST A x = – r/2 x=0 Magnetic Field at Centre O in Different Conditions of Circular Current A Condition Figure x = r/2 Fig. 21.10 (6) Point of inflection (A and A) : Also known as points of Magnetic field i Arc subtends  angle  at the centre B r O curvature change or points of zero curvature. i Arc subtends O  r 0  i. 4 r 1176 Magnetic Effect of Current angle (2 – ) B at the centre 0 (2   ) i 4. r opposite direction B  B12  B 22 B2 B arc O 0 i 4. r  0 i loops but their 4r planes r are i2  perpendicular Three quarter r O semi-circular  current 8r  0 i 2r i Concentric co-planer circular loops r1 1 1  B1  0 2i    4  r1 r2  U carries r2 i current in the ST direction Concentric co-planer circular loops carries current in the r2 an angle with B1 i2 B  2 B1 B2 cos   each i Distribution of current across O B=0 the diameter Distribution of i current between any O two points on the i circumference i r1 B12  B22 other D YG carrying arc  ID r O current same  U Circular 0 2i 4 r i12  i22  i1 Concentric planes are at B 2r B2 loops but their carrying arc 0 B1 E3 to each other    2   i 0  2 B. 4 r 3 0 i  60 Semi-circular  i1 Concentric i B2  0 4 1 1  2i    r1 r2  Magnetic Field Due to a Straight Wire B=0 Magnetic Effect of Current 1177 Magnetic field due to a current carrying wire at a point P B=0 which lies at a perpendicular distance r from the wire as shown B Y  0 i. (sin1  sin 2 ) 4 r i From figure   (90 o  1 ) and   (90 o   2 ) Hence B  r 2 1 Magnetic Field Due to a Cylindrical Wire P Magnetic field due to a cylindrical wire is obtained by the  application of Ampere's law X (1) Outside the cylinder Fig. 21.12 o i. (cos   cos  ) 4 r P (B) Thin hollow cylinder P ID r Fig. 21.13 R1 r i R2 (C) Thick hollow cylinder Fig. 21.17 U X (A) Solid cylinder P   i i E3 Y  i So B  0. (2 sin ) 4 r r i which lies on perpendicular bisector of finite length wire 1   2   P R R (1) For a wire of finite length : Magnetic field at a point 60 is given as (2) For a wire of infinite length : When the linear conductor D YG XY is of infinite length and the point P lies near the centre of the conductor 1 = 2 = 90o.  i So, B  0 [sin90 o  sin90 o ] Y 4 r   0 2i 4 r P i Fig. 21.14  (2) Inside the hollow cylinder : Magnetic field inside the hollow cylinder is zero. U X  In all above cases magnetic field outside the wire at P  i B.dl   0 i  B dl  0 i  B  2r  0 i  B out  0 2r i In all the above cases Bsurface  0 2R B0 (3) For a wire of semi-infinite length : When the linear B=0 B=0 (A) Thin hollow cylinder (B) Thick hollow cylinder ST conductor is of infinite length and the point P lies near the end Y or X. 1  90 o and  2  0 o So, B  0 i [sin90 o  sin0 o ] 4 r i  0 4 r Y Fig. 21.18 i X Fig. 21.15 P (3) Inside the solid cylinder : Current enclosed by loop (i) is lesser then the total current (i) (4) For axial position of wire : When point P lies on axial position of current carrying conductor then magnetic field at P i Fig. 21.16 R r Loop r i P Fig. 21.19 Loop 1178 Magnetic Effect of Current The current enclosed by the loop is i = jl. Therefore, according to Ampere’s law 2 Bl  0 ( jl) or B   r2 A'  i  2 A R  0 ir Bin. d l  0 i'  B . 2 R 2 Current density is uniform i.e. J = J  i'  i  Hence at inside point      60 (r 2  R12 ) A'  i 2 A (R 2  R12 ) Q Q Loop E3 R1 2 Solenoid (4) Inside the thick portion of hollow cylinder : Current enclosed by loop is given as i'  i  0 j i R2 Fig. 21.20 A cylinderical coil of many tightly wound turns of insulated  B. d l Hence at point Q   0 i'  B  0 i (r 2  R12 ). 2r (R22  R12 ) ID wire with generally diameter of the coil smaller than its length is called a solenoid. U Magnetic Field Due to an Infinite Sheet Carrying Current S B=0 N The figure shows an infinite sheet of current with linear i D YG current density j (A/m). Due to symmetry the field line pattern above and below the sheet is uniform. Consider a square loop l a ST U Fig. 21.21  B.dl  a  c B.dl  b  d c B.dl   B.dl  0 ; b  a  a d B.dl  0 i (By Ampere’s law) B.dl  0 d Also, B || dl along the path a b and c d, thus  b a B.dl   a d B.dl  2 Bl A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of solenoid. (1) Finite length solenoid : If N = total number of turns, l = length of the solenoid, n = Since B  dl along the path b c and d a, therefore, c Fig. 21.22 d l c b b i B of side l as shown P in the figure. i Solenoid B r  P number of turns per unit length   N l Fig. 21.23 (i) Magnetic field inside the solenoid at point P is given by B  0 4 (2 ni)[sin  sin  ] Magnetic Effect of Current (ii) Infinite length solenoid : If the solenoid is of infinite length and the point is well inside the solenoid i.e.     ( / 2). Bin  μ0 ni i.e. ( B end  (iii) Rest charge i.e. v = 0  F = 0 (iv) Moving charge i.e.  = 0o or  = 180o  F = 0 (2) Direction of force : The force F is always perpendicular (iii) If the solenoid is of infinite length and the point is near one end (ii) Neutral particle i.e. q = 0  F = 0 and  0 so   ( / 2) Bend 1  (  0ni ) 2 to both the velocity v and the field B in accordance with Right Hand Screw Rule, though v and B themselves may or may not 60 So 1179   other. be perpendicular to each 1 B in ) 2 Fm Fm  Toroid A toroid can be considered as a ring shaped closed solenoid. Hence it is like an endless cylindrical solenoid. i O r P ID dl B r B  v v 90° Fig. 21.26 Winding Core B  E3  Direction of force on charged particle in magnetic field can Fig. 21.24 U also be find by Fleming's Left Hand Rule (FLHR). F D YG Consider a toroid having n turns per unit length. Magnetic B field at a point P in the figure is given as B  0 Ni N   o ni where n  2r 2r v Fig. 21.27 Force On a Charged Particle in Magnetic Field If a particle carrying a positive charge q and moving with velocity v enters a magnetic field B then it experiences a force F U which is given by the expression F  q(v  B)  F  qvB sin ST where v  velocity of the particle, B  magnetic field                     q, m  v   B  Fig. 21.25 (1) Zero force : Force on charged particle will be zero ( i.e. F = 0) if (i) No field i.e. B = 0  F = 0 Here, First finger (indicates) Direction of magnetic field Middle finger Direction of motion of positive charge or direction, Opposite to the motion of negative charge. Thumb Direction of force Trajectory of a Charged Particle in a Magnetic Field (1) Straight line : If the direction of a v is parallel or antiparallel to B ,  = 0 or  = 180o and therefore F = 0. Hence the trajectory of the particle is a straight line. q  = 0o v B q  = 180o Fig. 21.28 1180 Magnetic Effect of Current (i) The radius of this helical path is r  hence particle will experience a maximum magnetic force Fmax = qvB which act's in a direction perpendicular to the motion of charged particle. the trajectory × Therefore × × × × × of the particle is a × × × × × × × × × × v + × × + × × v × F × × × × × × × × × × + v + v × × × × × × × × × × (iii) The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by p  T (v cos  )  2 m (v cos  ) qB (iv) If pitch value is p, then number of pitches obtained in length l given as E3 circle. (ii) Time period and frequency do not depend on velocity qB 2 m and   and so they are given by T  2 m qB 60 (2) Circular path : If v is perpendicular to B i.e.  = 90o, m ( vsin θ ) qB Number of pitches  l l and time required t  v cos  p Lorentz Force Fig. 21.29 the moving charged particle is subjected ID When simultaneously to both electric field E and magnetic field B , the moving charged particle will experience electric force magnetic force provides the necessary centripetal force i.e. mv 2 qvB  r mv p   qB qB of path D YG r radius  2mK 1  qB B 2 mV q where p = momentum of charged particle and K = kinetic energy of charged particle (gained by charged particle after accelerating through potential difference V) then U p  mv  2mK  2mqV (ii) If T is the time period of the particle then T  2m (i.e., qB time period (or frequency) is independent of speed of particle). ST Fe  q E and magnetic force Fm  q(v  B) ; so the net force on it will be F  q [ E  ( v  B )]. Which is the famous ‘Lorentz-force U (i) In this case path of charged particle is circular and equation’. Depending on the directions of v, E and B following situations are possible (i) When v, E and B all the three are collinear : In this situation the magnetic force on it will be zero and only electric    F qE force will act and so a   m m (ii) The particle will pass through the field following a straight-line path (parallel field) with change in its speed. So in this situation speed, velocity, momentum and kinetic energy all will change without change in direction of motion as shown (3) Helical path : When the charged particle is moving at an angle to the field (other than 0o, 90o, or 180o). v q Particle describes  E  B Fig. 21.31 a path called helix. Y v q, m B v sin  p Fig. 21.30 B r v  Z (iii) v , E and B are mutually perpendicular : In this v cos situation if E and B are such that F  Fe  Fm  0 y a  (F / m)  0 X +q v  B z +q  E Fe Fm Fig. 21.32 x i.e., Magnetic Effect of Current 1181 (2) Maximum energy of particle : Maximum energy gained  q2B2  2 r by the charged particle Emax     2m  as shown in figure, the particle will pass through the field with same velocity, without any deviation in path. where r0 = maximum radius of the circular path followed by the positive ion. Hall Effect This principle is used in ‘velocity-selector’ to get a charged The Phenomenon of producing a transverse emf in a 60 And in this situation, as Fe  Fm i.e., qE  qvB v  E / B beam having a specific velocity. current carrying conductor on applying a magnetic field Cyclotron perpendicular to the direction of the current is called Hall effect. charged particles (like, -particles, deutrons etc.) to acquire enough energy to carry out nuclear disintegration etc. charged a High frequency particle W it revolving in circular orbits of constant frequency. Target B z v x (2) – S (1) + + + + + + y proton beam D YG keeps Energetic D1 and the magnetic field D2 oscillator opposite to the direction of flow of current N U accelerates ID The electrons move with drift velocity v the fact that the field charge carriers in a conductor. Consider a conductor having electrons as current carriers. It is based on electric Hall effect helps us to know the nature and number of E3 Cyclotron is a device used to accelerated positively F – – – – – – VH – Fig. 21.34 Fig. 21.33 Force acting on electron Fm  e(v  B). This force acts and D2 called dees. The two dees are placed horizontally with a along x-axis and hence electrons will move towards face (2) and small gap separating them. The dees are connected to the it becomes negatively charged. U It consists of two hollow D-shaped metallic chambers D1 source of high frequency electric field. The dees are enclosed in 10–3 mm mercury. The whole apparatus is placed between the In case of current carrying conductor in a magnetic field two poles of a strong electromagnet NS as shown in fig. The force experienced by its small length element is d F  idl  B ; ST a metal box containing a gas at a low pressure of the order of Force On a Current Carrying Conductor In Magnetic Field magnetic field acts perpendicular to the plane of the dees. (1) Cyclotron frequency : Time taken by ion to describe a r m semicircular path is given by t   v qB If T = time period of oscillating electric field then 2 m 1 Bq T  2t  the cyclotron frequency    qB T 2m idl = current element d F  i(d l  B) × × × × × × × × × × × × × × × × × × × × × × × × × × × dl × × × × × × × × × × i × dF Fig. 21.35 × B × 1182 Magnetic Effect of Current Right-hand palm rule : Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point  in the direction of field B and thumb in the direction of current i, then normal to the palm will point in Current the direction of force  is uniform i.e., B = constant F  i[ dl]  B  i(L  B)  dl  L'  vector sum of all the length elements from initial Magnetic to final point. Which is in accordance with the law of vector field 60  Total magnetic force F  d F  i(d l  B). If magnetic field addition is equal to length vector L joining initial to final point. Force (For a straight conductor F  Bil sin ) Fig. 21.38 perpendicular to the plane containing idl and B and is same as that of cross-product of two vectors ( A  B) with A  i dl. Force Between Two Parallel Current Carrying Conductors dF i dl  P The force on a length l of each of two long, straight, parallel B P  ID E3 Direction of force : The direction of force is always i dl wires carrying currents i1 and i2 and separated by a distance a is B F  U dF 0 2i1i2  l 4 a Fig. 21.36 D YG Hence The direction of force when current element i dl  and B are force per unit length F  0 2i1 i2  N  F 2i1 i2  dyne      or    l a  cm  l 4 a m  i1 i2 a Fig. 21.39 perpendicular to each other can also be determined by applying either of the following rules Fleming’s left-hand rule : Stretch the fore-finger, central finger and thumb of left hand mutually perpendicular. Then if the U fore-finger points in the direction of field B and the central in the direction of current i, the thumb will point in the direction of ST force. F Direction of force : If conductors carries current in same direction, then force between them will be attractive. If conductor carries current in opposite direction, then force 1 2   × between them will be× repulsive. i 1        × i2 × × v Fig. 21.37 2 × × × × × × × × × × × × × i1 i2 × × × × × ×  × × × × × ×  × × × × × × F1 F2 F1 Fig. 21.40 B 1 F2 Magnetic Effect of Current Force Between Two Moving Charges X If two charges q1 and q2 are moving with velocities v1 and Fixed l Movable Fixed h q1 q2 v1 F m Fe Fe q1 r Stationary charges Movable i1 Y l q2 r Moving charges In both the situations for equilibrium of XY it's downward Fig. 21.41 4 and Electric force between them is Fe . r2 1 4 0. q1 q 2 r2.... (ii) Fm 1   0  0 v 2 but  0  0  2 ; Fe c F v  where c is the velocity of light in vacuum. So m    Fe c 2 through a spring, then it will contract because current will flow through all the turns in the same direction. Spring U As v < c so Fm < Fe Case 3 : Current carrying spring : If current is passed.... (i) ID  0 q1 q 2 v1v 2 Magnetic force between them is Fm  D YG Standard Cases For Force on Current Carrying Conductors Case 1 : When an arbitrary current carrying loop placed in a μ0 2 i1i 2..l 4π h E3 weight = upward magnetic force i.e. mg  From equation (i) and (ii) X Fig. 21.43 v2 F e Fm h i2 60 Fe i1 Y i2 v2 respectively and at any instant the distance between them is r, then 1183 + – Spring K m Hg If current makes to flow through If switch is closed then current start spring, then spring will contract flowing, spring will execute oscillation in vertical plane Fig. 21.44 and weight lift up magnetic field ( to the plane of loop), each element of loop experiences a magnetic force due to which loop stretches and open into circular loop and tension developed in it’s each part. U T B T B d R Fig. 21.42 Case 2 : Equilibrium of a current carrying conductor : When a finite length current carrying wire is kept parallel to another infinite length current carrying wire, it can suspend freely in air as shown below Case 4 : Tension less strings : In the following figure the value and direction of current through the conductor XY so that strings becomes tensionless? O ST B A Strings becomes tensionless if weight of conductor XY balanced by magnetic force (Fm ). String × × × × × × × × × × × × × × × × × × × × l× m × × T X × × Y × × B T × × × × × Fm  Fig. 21.45 X i Y mg 1184 Magnetic Effect of Current Hence direction of current is from X Y and in balanced condition Fm  mg  B i l  mg  i  mg Bl Fig. 21.48 Case 5 : Sliding of conducting rod on inclined rails : When a X i stand Insulated  B Y v F i  B F cos  R mg sin  i  – + (1) Torque : Consider a rectangular current carrying coil PQRS having N turns and area A, placed in a uniform field B , in  mg Behaviour of Current Loop in a Magnetic Field P 60 conducting rod slides on conducting rails. S  n^ such a way that the normal (nˆ ) to the coil Fig. 21.46 E3 makes an angle  with the direction of B. the coil experiences a torque given by  = B Q R Fig. 21.49 NBiA sin. Vectorially   M  B constant velocity if ID In the following situation conducting rod (X, Y) slides at (i)  is zero when  = 0, i.e., when the plane of the coil is Current Loop as a Magnetic Dipole mg tan  il perpendicular to the field. (ii)  is maximum when   90 o , i.e., the plane of the coil is U F cos  mg sin  B i l cos   mg sin  B  parallel to the field  max  NBiA D YG A current carrying circular coil behaves as a bar magnet whose magnetic moment is M = NiA; Where N = Number of turns in the coil, i = Current through the coil and A = Area of the coil Magnetic moment of a current carrying coil is a vector and it's direction is given by right hand thumb rule (2) Workdone : If coil is rotated through an angle  from it's equilibrium position then required work. W  MB(1  cos ). It is maximum when  = 180o  Wmax = 2 MB (3) Potential energy : U = – MB cos  U   M.B Moving Coil Galvanometer Scale Current N U S M Fig. 21.47 Permanen Restoring Pointer torque t magnet ST Magnetic field (1) For a given perimeter circular shape have maximum Pivoted coil (A) Soft-iron core Fig. 21.50 i i (B) Magnetic field torque area. Hence maximum magnetic moment. (2) For a any loop or coil B at centre due to current in loop, In a moving coil galvanometer the coil is suspended between and M are always parallel. the pole pieces of a strong horse-shoe magnet. The pole pieces are made cylindrical and a soft iron cylindrical core is placed B,M B,M Magnetic Effect of Current 1185 within the coil without touching it. This makes the field radial. In of the coil becomes n2 times the previous field i.e. B (n turn) = n2 such a field the plane of the coil always remains parallel to the B(single turn) field. Therefore   90 o and the deflecting torque always has the  When a current carrying coil is suspended freely in maximum value. earth's magnetic field, it's plane stays in East-West direction.  Magnetic field ( B ) produced by a moving charge q is......(i)  def  NBiA Coil deflects, a restoring torque is set up in the suspension 60 fibre. If  is the angle of twist, the restoring torque is 0 q(v  r ) 0 q(v  rˆ ) ; where v = velocity of  4 4 r3 r2 charge and v