JEE-Physics Magnetic Effect of Current PDF

JEE-Physics...

JEE-Physics MAGNETIC EFFECT OF CURRENT The branch of physics which deals with the magnetism due to electric current or moving charge (i.e. electric current is equivalent to the charges or electrons in motion) is called electromagnetism. ORESTED'S DISCOVERY The relation between electricity and magnetism was discovered by Orested in 1820. Orested showed that the electric current through the conducting wire deflects the magnetic needle held below the wire. S N When the direction of current in conductor is reversed then deflection of magnetic needle is also reversed On increasing the current in conductor or bringing the needle closer to the North conductor the deflection of magnetic needle increases. Oersted discovered a magnetic field around a conductor carrying electric current. Other related facts are as follows: – + (a) A magnet at rest produces a magnetic field around it while an S I electric charge at rest produce an electric field around it. (b) A current carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving with a uniform velocity has an electric as well as a magnetic field around it. N (c) An electric field cannot be produced without a charge whereas a magnetic field can be produced without a magnet. N (d) No poles are produced in a coil carrying current but such a coil Oersted's experiment. Current in the wire deflects the compass needle shows north and south polarities. (e) All oscillating or an accelerated charge produces E.M. waves also in additions to electric and magnetic fields. Current Element dB P A very small element ab of length d of a thin conductor carrying r current I called current element. Current element is a vector I I quantity whose magnitude is equal to the product of current and  a length of small element having the direction of the flow of current. d b d Biot – Savart's Law With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65  on the following experimental observations for the magnetic field dB at a point P associated with a length  element d  of a wire carrying a steady current I. 1 Id  sin  0 Id  sin  dB I, dB  d, dB  sinand dB   dB   dB = r2 r2 4 r2 Vector form of Biot-Savar t's law   Id  sin    dB  0 nˆ n =unit vector perpendicular to the plane of ( Id  ) and ( r ) 4 r2     0 Id   r   dB  [ Id  × r = (Id) (r)sin n̂ ] 4 r3 E 1 JEE-Physics GOLDEN KEY POINTS     Id   r  According to dB  0 , direction of magnetic field vector (dB) is always perpendicular to the plane of 4 r3     vectors  Id   and ( r ), where plane of  Id   and ( r ) is the plane of wire. Magnetic field on the axis of current carrying conductor is always zero (=0° or  = 180°) Magnetic field on the perimeter of circular loop or coil is always minimum. MAGNETIC FIELD LINES (By Michal Faraday) In order to visualise a magnetic field graphically, Michal faraday introduced the concept of field lines. Field lines of magnetic field are imaginary lines which represents direction of magnetic field continuously. GOLDEN KEY POINTS Magnetic field lines are closed curves. Tangent drawn at any point on field line represents direction of the field at that point. Field lines never intersects to each other. At any place crowded lines represent stronger field while distant lines represents weaker field. In any region, if field lines are equidistant and straight the field is uniform otherwise not. Non-uniform Field Uniform Field Magnitude is Direction is Both magnitude Both magnitude not constant not constant and direction are and direction are not constant constant Magnetic field lines emanate from or enters in the surface of a magnetic material at any angle. Magnetic field lines exist inside every magnetised material. Magnetic field lines can be mapped by using iron dust or using compass needle. RIGHT HAND THUMB RULE This rule gives the pattern of magnetic field lines due to current carrying wire. NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 (i) Straight current (ii) Circular current Thumb In the direction of current Curling fingers  In the direction of current, Curling fingers  Gives field line pattern Thumb  Gives field line pattern Case I : wire in the plane of the paper Case I : wire in the plane of the paper I Magnetic CW ACW field lines B B Towards observer or Away from the observer perpendicular or perpendicular I out-wards inwards 2 E  JEE-Physics Case II : Wire is  to the plane of the paper. Case II : Wire is  to the plane of the paper ACW CW S N N S B I ACW concentric & CW concentric & Towards observer Away from the observer circular field lines circular field lines GOLDEN KEY POINTS When current is straight, field is circular When current is circular, field is straight (along axis) When wire is in the plane of paper, the field is perpendicular to the plane of the paper. When wire is perpendicular to the plane of paper, the field is in the plane of the paper. APPLICATION OF BIOT-SAVART LAW : A Magneti c fi eld surroundi ng a thin straight current carrying conductor AB is a straight conductor carrying current i from B to A. At a point P, whose id perpendicular distance from AB is OP =a, the direction of field is perpendicular  r to the plane of paper, inwards (represented by a cross)  a  O  P =a tan dl=a sec 2 d...(i)  =90°– & r=asec B By Biot-Savart’s law   id  sin  dB  0  (due to a current element id at point P) 4 r2  0 id  sin   i  B=  dB  2 (due to wire AB)  B= 0  cos d  4 r 4 Taking limits of integration as – 2 2 to  1 NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65  0 i 1  i  0 i B  4 a 2 cos d   0  sin  1 4 a 2 4 a  sin 1  sin 2  (inwards) Example Magnetic field due to infinite length wire at point 'P' Solution  I 2 = 90° 90° M P 0 I 0 I BP = [sin90° + sin90°] 1 = 90° BP = 4 d 2 d d  E 3 JEE-Physics Example Magnetic field due to semi infinite length wire at point 'P' I 0 I 90° 2=90° 0 I S o l. BP = [sin + sin90°] M P BP = [sin + 1] 4 d 1= 4 d L d Example Magnetic field due to special semi infinite length wire at point 'P' Solution 0 I 0 I BP = [sin0° + sin90°] BP = 4 d 4 d I 90° 2=90° M 1=° P d Example N Magnetic field due to special finite length wire at point 'P' 90° 2=° M P d 1=0° Solution 0 I 0 I BP = [sin0° + sin] ; BP = sin NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 4 d 4 d Example If point ‘P’ lies out side the line of wire then magnetic field at point ‘P’ : d   P 2   I 1 Solution 0 I  I BP  sin  90  1   sin  90   2    0 (cos 1  cos  2 ) 4 d  4 d 4 E  JEE-Physics Magnetic fi eld due to a loop of current Magnetic field lines due to a loop of wire are shown in the figure B i i i The direction of magnetic field on the axis of current loop can be determined by right hand thumb rule. If fingers of right hand are curled in the direction of current, the stretched thumb is in the direction of magnetic field. Calculati on of magnetic field z-axis Consider a current loop placed in y-z plane carrying current i in id anticlockwise sense as seen from positive x-axis. Due to a small i y-axis  R current element id  shown in the figure, the magnetic field at P 2 2 r= R +x  0 id  sin 90 0 O is given by dB . 4 r2   dB x    i P x-axis The angle between id  and r is 90 0 because id  is along y-  axis, while r lies in x-z plane. The direction of dB is perpendicular  to r as shown. The vector dB can be resolved into two components, dB cos along z-axis and dB sin along x-axis. For any two diametrically opposite current elements, the components along x-axis add up, while the other two components cancel out. Therefore, the field at P is due to x-component of field only. Hence, we have  0 id   id  R  0 iR B=  dB sin    sin  =  0 2   B=  d   d   2 R 4 r2 4 r r 4 r 3 z-axis dBcos 2  dB  0 i  2 R 2 =  0 i  2 R   r  R 2  x2  NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65  B = P 4 r3 4 R 2  x2 / 2 3   x-axis dBsin 0 i (a) At the centre, x=0, B centre = 2R 3 / 2 0 i  x2  0 i  3x 2  (b) At points very close to centre, xR  B= 4 x3 0 2M (d) The result in point (c) is also expressed as B = 4 x3 where M=   R 2 , is called magnetic dipole moment. E 5 JEE-Physics Example Find the magnetic field at the centre of a current carrying conductor bent in the form of an arc subtending angle  at its centre. Radius of the arc is R. Solution Let the arc lie in x-y plane with its centre at the origin. y-axis  Consider a small current element id  as shown. id i The field due to this element at the centre is 0 id  sin 90 0  dB= 4 R2  id  and R are perpendicular  R d  R  iRd  0 i Now d   Rd   dB  0 2  dB = d  4 R 4 R The direction of field is outward perpendicular to plane of paper x-axis O  0 i 0 i  0 i Total magnetic field B=  dB  B  4 R  d  4 R   0 0 B= 4 R  Example Find the magnetic field at the centre of a current carrying conductor bent in the form of an arc subtending angle  1 and  2 at the centre. Solution Magnetic field at the centre of arc abc and adc wire of circuit loop b  I  I  B I arc length 1 1 B abc  01 1 and B adc  0 2 2  abc  1 1  angle =   =  1 4 r 4 r B adc I 2 2 radius 2 2 a I1 I1  2 2 I1 R 2   V = I1R1 = I2R2      R  R ) d I2 R 1 I2 1 A I2 c B abc   2   1  B 1 1  B =         adc  1  2 B 2 1 HELMHOLT'Z COIL S ARR ANGEMENT R (N, I, R) (N, I, R) This arrangement is used to produce uniform magnetic field ACW ACW O1 O2 of short range. It consists :- M Two identical co–axial coils (N, I, R same) NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 (uniform magnetic field of short range BHC) Placed at distance (center to center) equal to radius ('R') of coils S P Q Planes of both coils are parallel to each other. O1 M O Current direction is same in both coils (observed from same side) R/2 R/2 2 otherwise this arrangement is not called "Helmholtz coil arrangement". Example A pair of stationary and infinitely long bent wires are placed in the x-y plane I Q as shown in fig. The wires carry currents of 10 ampere each as shown. The R O S segments L and M are along the x-axis. The segments P and Q are parallel to L M the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of P I the magnetic induction at the origin O. 6 E  JEE-Physics Solution As point O is along the length of segments L and M so the field at O due to these segments will be zero. Further,    0 I ˆ  I as the point O is near one end of a long wire, B R  B P  B Q = (k ) + 0 (kˆ) [as RO = SO = d] 4 d 4 d    2I   2  10 ˆ Wb so, B R  0   (kˆ) Substituting the given data, B R  10–7 × (k ) = 10–4 2 (kˆ) 4  d  0.02 m B = 10–4 T and in (+z) direction. Example Calculate the field at the centre of a semi-circular wire of radius R in situations depicted in figure (i), (ii) and (iii) if the straight wire is of infinite length. I a I a b I R R R a c b I O I O I O I b I I c c R R (i) (ii) (iii) Solution The magnetic field due to a straight current carring wire of infinite length, for a point at a distance R from one of its 0 I ends is zero if the point is along its length and if the point is on a line perpendicular to its length while at the 4 R 0 I centre of a semicircular coil is so net magnetic field at the centre of semicircular wire is B R  B a  B b  B c 4R  0 I  I (i) BR = 0 +  + 0 = 0 ( into the page) 4 R 4R  0 I  I  I  I (ii) BR =  + 0  + 0  = 0 [ + 2]  (out of the page) 4 R 4 R 4 R 4 R  0 I  I  I  I (iii) BR =  + 0  + 0  = 0 [ – 2] (in to the page) 4 R 4 R 4 R 4 R Example Calculate the magnetic induction at the point O, if the current carrying wire is b NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 in the shape shown in figure. The radius of the curved part of the wire is a and c I d linear parts are assumed to be very long and parallel. Solution Magnetic induction at the point O due to circular portion of the wire a e µ0 I  0 i 3 3 B1 = =    (out of the page) (= ) 4 R 4 a 2 2 Magnetic induction at O due to wire cd will be zero since O lies on the line cd itself when extended backward. Magnetic induction at O due to infinitely long straight wire ae is µ0 i  µ0 i     µ0 i  B2 = [sin 1  sin 2 ] where r = a,  = 0,  =  B2 = sin 0   sin    = 4 r 1 2 2 4 a   2  4 a Because both the fields are in same direction i.e. perpendicular to plane of paper and directed upwards, hence 0 i  3   the resultant magnetic induction at O is B = B1 + B2 = 1  4 a  2  E 7 JEE-Physics Example B i In the frame work of wires shown in figure, a current i is allowed to flow. A R2 Calculate the magnetic induction at the centre O. If angle  is equal to 90°, C  D then what will be the value of magnetic induction at O ? O Solution R1 i 0 i E Magnetic induction at O due to the segment BC is B1 =  4 R 2 0 i Similarly, the magnetic induction at O due to circular segment AED is B2 = (2    )  4 R 1 Magnetic field due to segments AB and CD is zero, because point 'O' lies on axis of these parts. 0 i   2     Hence resultant magnetic induction at O is B = B1 + B2 =   ,  4  R2 R1   0 i   3  0 i  1 3  If  = 90° = , then B=    =    2 4   2R 2 2R 1  8  R 2 R1  Example Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively lie in the same vertical plane containing the north-south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and in Y clockwise, for an observer looking at the coils facing the west. What is the magnitude and direction of the magnetic field at their common centre (i) Due to coil X alone ? (ii) Due to coil Y alone ? (iii) Due to both the coils ? Solution According to the figure the magnitude of the magnetic field at the centre of coil X is 0  x N x 2   10 7 16  20 N Bx =  × = 4 × 10 –4 T 2 rx 2 0.16 Coil X Ix Coil Y Since the current in coil X is anticlockwise, the direction of B x is towards the east as shown in figure. W E By Bx The magnitude of magnetic field at the centre of the coil Y Iy 0  Y N Y 4   10 7 18  25 is given by B Y = = × =9 × 10 –4 T S 2 rY 2 2  since the current in coil Y is clockwise, the direction of field BY is towards the west (see fig.). Since the two fields are collinear and oppositely directed. The magnitude of the resultant field = difference between the two fields and its direction is that of the bigger field. Hence the net magnetic field at the common centre NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 is 5  × 10–4 T and is directed towards the west. Example Z I A long wire bent as shown in the figure carries current I. If the radius of the a Y semi-circular portion is "a" then find the magnetic induction at the centre C. O C Solution X I I  0 I ˆ Due to semi circular part B 1  4a i    I   0 I ˆ  I due to parallel parts of currents B 2  2  0 ( kˆ ) , Bnet = BC = B 1  B 2 = (  i ) + 0 ( kˆ) 4 a 4a 2 a 0 I magnitude of resultant field B = B 12  B 22 = 2  4 4 a 8 E  JEE-Physics Example A piece of wire carrying a current of 6 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field due to this piece of wire at the centre. Solution µ0 I  2 Magnetic field at centre of arc B = ,  = 120° = rad 4 R 3 I 0 I 2  0 I 4   10 7  6  100 120° B=  = = T = 12.57 µT 4 R 3 6R 6  10 Example Y An infinitely long conductor as shown in fig. carrying a current I with a semicircular loop on X-Y plane and two straight parts, one parallel to x-axis and another coinciding with I Z-axis. What is the magnetic field induction at the centre C of the semi-circular loop. I C r Solution X O The magnetic field induction at C due to current through straight part of the I conductor parallel to X-axis is Z 0 I    0 I 0 I B1 = 4  r / 2 sin 2  sin 0  = acting along + Z direction. i.e. B 1 = k̂    2 r 2 r The magnetic field induction at C due to current through the semi-circular loop in X-Y plane is 0 I  I  0 I B2 = 4   r / 2     = 0 acting along + Z-direction i.e. B 2 = k̂ 2 r 2r The magnetic field induction at C due to current through the straight part of the conductor coinciding with Z-axis is 0 I     I  0 I ˆ B3 =  sin  sin 0  = 0 acting along (–X)-axis 4 r / 2  2  2 r i.e. B 3    2 r i     0 I  I  I  I Total magnetic field induction at C is B  B1  B 2  B 3 = k̂ + 0 k̂ – 0 î = 0 1    kˆ  ˆi  2 r 2 2 r 2 r   Example A conductor carrying a current i is bented as shown in figure. Find the magnitude of magnetic field at the origin Solution   i  1  0 i  ˆ 4 r   Field at O due to part 1 B 1  0 ˆi Field at O due to part 2 B 2   4 2r   k   i y 2  1 i  Wire 3 passes through origin when it is extended backwards B 3  0 r 3 x O i     0 i  ˆi kˆ  z B 0  B1  B 2  B 3 =     NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 4r   2  Example 0.03m A conductor of length 0.04 m is tangentially connected to a circular  4A loop of radius 0.03 m perpendicular to its plane. Find the magnetic 0.04m field induction at the centre of the loop if 4 ampere current is passed 5m 0.0 through the conductor as shown in fig. Solution Magnetic field induction at the centre of the loop due to the straight current-carrying conductor, 0 I 4   10 7  4  0.04  B = sin 1  sin 2  =  sin 0   0.05  = 1.07 × 10–5T 4 r 4   0.03   Magnetic fields due to the two halves of the loop are equal in magnitude and opposite in direction. So, the magnetic field induction due to the loop at the centre of the loop is zero. So, the magnetic field induction at the centre of the loop is 1.07 × 10–5T. E 9 JEE-Physics Example P Figure shows a right-angled isosceles triangle PQR having its base equal to a. A current of I ampere is passing downwards along a thin straight wire cutting the 90° r r plane of the paper normally as shown at Q. Likewise a similar wire carries an equal current passing normally upwards at R. Find the magnitude and direction Q 45° R a of the magnetic induction at P. Assume the wires to be infinitely long. Solution a Let r = PQ = PR and a2 = r2 + r2 = 2r2 or r = 2 0 I 2 0 I 0 I Magnetic induction at P due to conductor at Q is B1 = = = (along PR) 2 r 2 a 2 a 0 I Magnetic induction at P due to conductor at R is B2 = (along PQ) 2 a 2 2  0 I   0 I  2 0 I 0 I    2 2 Now, resultant of these two is B = B B =   = = 1 2 2 a 2 a  2 a a  The direction of B is towards the mid-point of the line QR. AMPERE'S CIRCUITAL L AW I4 Ampere's circuital law state that line integral of the magnetic field around circulation I3 ACW any closed path in free space or vacuum is equal to 0 times of net I5 current or total current which crossing through the area bounded by the I1 I2   closed path. Mathematically   B. d   0 I Positive  I=(I1 I2+I3) Negative This law independent of size and shape of the closed path. Any current outside the closed path is not included in writing the right hand side of law Note : This law suitable for infinite long and symmetrical current distribution. Radius of cross section of thick cylinderical conductor and current density must be given to apply this law. MAGNETOMOTIVE FORCE (M.M.F.)          B. d   0 I , where B 0 H ,   H. d    I 0 0   H. d   I The line integral of magnetising field around any closed path is equal to net current crossing through the area   NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No-09\Magnetic Effect of current & Magnetism\Eng\01.Theory.p65 bounded by the closed path, also called 'magnetomotive force'. Magnetomotive force (M.M.F.) =   H. d  APPLICATION OF AMPERE'S CIRCUITAL L AW Magnetic field due to infinite long thin current carrying straight conductor   Consider a circle of radius 'r'. Let XY be the small element of length d. B and d  are in the same direction because direction of along the tangent of the circle. By A.C.L.   I  B. d   0  I ,  B d cos   0 I (where  = 0°) r B O y  B d  cos 0    0 I  B   d  0 I (where  d  2 r ) ACW x d I 0 I B (2r) = 0I  B  2 r 10 E  JEE-Physics I Magnetic field due to infinite long solid cylinderical conductor For a point inside the cylinder r < R, Current from area R2 is = I R I I r2 1 2 3 so current from area r2 is = 2 ( r 2 ) = 2 R R By Ampere circuital law for circular path 1 of radius r axis I r2 0 Ir Bin (2r) = 0I' = 0  Bin = Bin  r Cross-sectional R2 2 R 2 view For a point on the axis of the cylinder (r = 0); B axis  0 I 1 2 3 For a point on the surface of cylinder (r = R) By Ampere circuital law for circular path 2 of radius R R 0 I Bs (2  R) = 0I  Bs = (it is maximum) 2 R B For a point outside the cylinder (r > R) :- Bmax r 1 By Ampere circuital law for circular path 3 of radius r  Bout r B r< R 0 I 1 Bout (2  r) = 0I  B out   Bout  r= 0 r= R r> R 2 r r Magnetic field outside the cylinderical conductor does not depend upon nature (thick/thin or solid/hollow) of the conductor as well as its radius of cross section. Magnetic field due to infinite long hollow cylinderical conductor For a point at a distance r such that r < a < b B1  0 b a For a point at a distance r such that a < r < b 12 3 2 2  r a  B2(2r) =0 I'   B2(2r) = 0 I  2 2  b a 