Electromagnetic Induction-03-Subjective SOLVED

SOLVED EXAMPLES 1. A coil A C D of radius R and number of turns n carries a current i amp. and is placed in the plane of paper. A small conducting ring P of radius r is placed at a distance y from the centre and above the coil A C D. Calculate the induced e.m.f. produced in the ring when the ring is allowed to fall freely. Express induced e.m.f. in terms of speed of the ring. r P y A O R C D i Solution: We know that magnetic induction at some point on the axis of a current carrying coil at a distance y from its centre is given by B  0  4 2n i R 2 (R2  y2 )3 / 2 …(i) Now, the magnetic flux linked with ring P is  2 n i r2 R2   BA  0   r2 4 (R 2  y2 )3 / 2  n i R 2 r2  0  2 (R 2  y2 )3 / 2 …(ii) Let the ring P falls with velocity v. Now y varies with time as y  y0  vt The induced e.m.f. in the ring e  d  0  n i R2 r2 d (R2  y2 )3 / 2 dt 2 dt   3   ni R 2 r2 (R 2  y2 )5 / 2 .2y dy   …(iii) 0   2   3  n i R 2 r2   0 y(v) 2 (R 2  y2 )5 / 2 3  n i R2 r2 y v  0 2 (R 2  y2 )5 / 2 2. A thin wire ring of radius a and resistance r is located in- side a long solenoid so that their axes coincide. The length of the solenoid is equal to l its cross-sectional radius to b. At a certain moment, the solenoid was connected to a source of constant voltage V. The total resistance of the circuit is equal to R. Assuming the inductance of the ring to be negligible, find the maximum value of the radial force acting per unit length of the ring. ring V Solution: The inductance L of the solenoid L   n2b2l The current through the solenoid varies as I(t)  V (1  et R / L ) R B(t) = magnetic induction  0n I(t) …(i) e  d → → & B.A dt M flux through ring   n.a 2 e  d ,  na2 dI dt where i (t) is 0 dt dI(t) dt Force unit per length on the ring  F  BI(t) 𝑙  F(t)  0 n I(t).0 • n I(t) a2 / r 2 .a2 •  0 .n2 I(t) I(t) r  2 a2 V  V Ret R / L  F(t) 0 .n 2  (1  e t R / L ). .  r  R 2 a2 V2    R L   n2   0 .e t / R L (1 et / RL )  r  RL  Initial conditions are F(t) = 0 as t = 0 F (t) = 0 at t   For finding maximum value of force, we differentiate F(t) w.r.t. t and put to zero. This gives the time at which this occurs. et / R L  1 2 2 a2 V2 n 2  a2 v2  Fmax  0  0 4rR L 4Rlb2 on make use of (1). × × × × × × × × × R × × × 3. P and Q are two infinite conducting plates kept parallel to each other and separated by a distance 2r. A conducting ring of radius r falls vertically between the plates such that planes are always tangent to the ring. Both the planes are connected by a resistance R. There exists a uniform mag- netic field of strength B perpendicular to the plane of ring. × × × × × × × × × × × × × × × ×r × × × × × × × × × × × × × × × P× × 2r × ×Q × × × × × × × The arrangement is shown in figure. Plane Q is smooth and friction between the plane P and the ring is enough to prevent slipping. At t = 0, the planes and the ring. Find (a) the current through R as a function of time (b) terminal velocity of the ring (assume g to be constant). Solution: Let v be the velocity of C.M. of the ring at any time t. e.m.f. across the diameter of ring, e = 2 B v r.  Current through R is I = ( 2 B v r/R) …(i) The different force on the ring are shown in figure. F Fm i i Smooth plane mg Q For rotational motion fr  I f  I  Ia r r2 where a = acceleration of C.M. 4B2 r2 (  a / r) …(ii) Now Fm Bi L  R v ( L  2r) …(iii) For translational motion mg  f  Fm  ma Substituting the value of f and Fm Ia 4B2 r2 ...(iv) from eqs. (ii) and (iii) in equation (iv), we get mg   r2 R 4B2 r2 v  ma Ia mg  v  ma  R r2 4B2r2  I  or mg  v  a m  R  r   4B2r2  dv  I  or  mg  B v   dt m  r2    dv  m  I   dt  4B2r2   r2  or mg   v   R  for a ring I  mr2 dv (2m)  dt  mg  4B2 r2  R  …(v)   Integrating equation (iv) with proper limits, we get mR  4B2 r2 v  2B2 r2 loge mg  R v  t  0 mR  mg  4B2 r2  R  or 2B2 r2 loge  mg   t       2B2r2   v  mgR 1  e 4B2 r2  .t mR  …(vi)  For equation (i) and (vi), we get   2B2r2  i  2Bv r  mg 1  e R 2Br  .t mR  .  4. A rod of length 2 a is free to rotate in a vertical plane, about i a horizontal axis O passing through its midpoint. A long d straight, horizontal wire is in the same plane and is carrying a constant current i as shown in figure. At initial moment of time the rod is horizontal and starts to rotate with con- stant angular velocity  . Calculate e.m.f. induced in rod as a function of time. Solution: The rotated position of the rod after a time t is shown in figure. Consider a small element of length dx of the rod at a distance x from the centre. The velocity of the element v  x and its distance from the wire is r  (d  xsin t) . Magnetic induction at this position B  0i  0i i 2r 2(d  xsin t) The induced e.m.f in this element  i(x) dx d v O t de  Bv dx  0 2(d  xsin t) In order to obtain the resultant e.m.f., we integrate this expression from (–a) to +a. Hence a e  0  x dx 2 a (d  x sin t)  0i 1 2asin t  d log d  a sin t  2  sin2 t   d  asin t   0i 2asin t  dlog d  asin t  . 2sin2 t   d  asin t  5. A wire frame of area 3.92 104 m2 and resistance 20 is suspended freely from a 0.392 m long thread. There is a uniform magnetic field of 0.784 tesla and the plane of wire-frame is perpendicular to the magnetic field. The frame is made to oscillate under gravity by dis- placing it through 2 102 m from its initial position along the direction of magnetic field. The plane of the frame is always along the direction of thread and does not rotate about it. What is the inducted e.m.f. in wire-frame as a function of time? Also find the maximum current is the frame. Solution: The situation is shown in figure. The instantaneous flux through the frame when displaced through an angle  is given by   BAcos Instantaneous induced e.m.f. e   d  BAsin  d dt dt  BA d dt Applying Newton’s law d2 x ( sin   ) m dt2  mgsin  d2 x    or dt2 gsin From figure, sin     x l or x  l d2 x   g   dt2 l d2   g  or dt2 l Putting   , we get d2  2  0 dt2 …(ii) This is the equation of S.H.M. Solution of equation (ii) is given by   0 sin t Substituting the value of  in equation (i), we get e  BA(0 sin t) d ( dt 0 sin t)  BA0 sin t 0cost  1 BA2 sin 2t 2 0 Here,     5sec1 And x 2 102   0  . 0 l 0.392 Substituting the values, we get 1  2 102 2 e   (0.784)  (3.92 104 )  (5)    sin10t 2  2 106 sin10t or e  2 106 volt e 2 106 volt  0.392  7 and imax  max   10 R 20 amp. 6. A variable magnetic field creates a constant emf E in a conductor ABCDA. The resistances of the portions ABC, CDA and AMC are R1 , R 2 and R3 , respectively. What current will be show by the meter M? The magnetic field is concentrated near the axis of the circular conductor. Solution: Let E1 and E2 be the emfs developed in ABC and CDA, respectively. Then E1  E2 = E. There is no net emf in the loop AMCBA as it does not enclose the magnetic field. If E3 is the emf in AMC then E1  E3  0 . The equivalent circuit and distribution of current is shown in figure. By the loop rule R1 (x  y)  R 2 x  E1  E2  E x R3 R2 R1 M and R3 y  R1 (x  y)  E3  E1  0 E2 x-y E1 y x Solving for y, y  ER1 . R1R 2  R2 R3  R3R1 7. A square loop of side a and a straight, infinite conductor b are placed in the same plane with two sides of the square I parallel to the conductor. The inductance and resistance are equal to L and R respectively. The frame is turned through 180º about the axis OO . Find the electric charge that flows in the square loop. Solution: By circuit equation i    L di  / R where   induced emf and L di  self-induced emf  dt  dt    Ri    L di  dt Ridt   dt  L di dt  Rq   d dt  Lif     ( i  0, i  0)  dt i i f initial final  q  (i  f ) / R Consider a strip at a distance x in the initial position. Then inward normal to the plane. B  (0 / 4)(2I/ x) along the  d  ( I / 2x)a dx cos 0  0 Ia dx 1 0  Ia ab dx  i   0 Ia ln 2 x a  b 2 b x 2 b Similarly   0 Ia ln 2a  b f 2 a  b  |    | 0 Ia ln 2a  b i f 2 b | q |  0 Ia ln 2a  b 2R b 8. A rectangular conducting loop in the vertical x-z plane has length L, width W, mass M and resistance R. It is dropped lengthwise from rest. At t = 0 the bottom of the loop is at a height h above the horizontal x-axis. There is a uniform magnetic field B perpendicular to the x-z plane, below the x-axis. The bottom and top of the loop cross this axis at t = t1, and t2 respectively. Obtain the expression for the velocity of the loop for the time t1  t  t2 . Solution : For time t1, the loop is freely falling under gravity, so velocity attained by loop at t  t1 1  gt1  During the time t1  t  t2 , flux linked with the loop is changing, so induced emf e   d  BW dt and Induced current I   BW clockwise R   B2W2 Magnetic Force F = WIB R So, m d  mg  B2W2 dt R dt   md B2W2   mg  R    mR  B2W2  Integrating, t   B2 W2 loge mg  R   A   At t  t1,   1  gt1 mR  B2 W2   A  t1  loge mg  1  Substituting for A, B2W2   R  B2W2  B2 W2  mg   e mR (tt1 )  log  R  mg  1   R  Gives the velocity V of the loop in the interval t1  t  t2 . 9. A very small circular loop of area 5104 m2 , resistance 2 ohm and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius 0.1 m. A constant current of 1 Amp. is passed in the bigger I A loop and the smaller loop is rotated with angular velocity  rad/s about a diameter. Calculate (a) the flux linked with the smaller loop (b) induced emf and induced current in the smaller loop as a function of time. Solution : (a) The situation is shown in figure. The field at the center of larger loop, B  0 1 4 2I  107 21  2106 Wb R 0.1 m2 is initially along the normal to the area of smaller loop. Now as the smaller loop (and hence normal to its plane) is rotating at angular velocity , so in time t it will turn by an → angle   t w.r. to B and hence the flux linked with the smaller loop at time t,   B S cos  (2106 ) (5104) cost i.e., 2  109 cost Wb (b) The induced emf in the smaller loop, e   d2   d (109 cost) 2 dt dt i.e., e  109 sin t volt (c) The induced current in the smaller loop, I  e2  1 109 sin t ampere. 2 R 2 A R2 C 10. Two parallel vertical metallic rails AB and CD are sepa- rated by 1 m. They are connected at the two ends by resis- FM tances R1 and R2 as shown in figure. A horizontal metallic bar of mass 0.2 kg slides without friction, vertically down – + the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane mg of the rails. It is observed that when the terminal velocity is attained, the power dissipated in R1 and R2 are 0.76 W and 1.2 W respectively. Find the terminal velocity of the bar D and the values of R1 and R2. 1 Solution: The rod will acquire terminal velocity only when magnetic force FM = BIl due to electromag- netic induction balances its weight, i.e., BIl  mg, i.e., I  0.2  9.8  9.8 A 0.6 1 3 Now if e is the emf induced in the rod, e × I = P = P1 + P2 So, e  (0.76  1.20)  0.6V (9.8 / 3) Now as this e is generated due to motion of rod with terminal velocity in the magnetic field, i.e., e  B l so   e  0.6 1ms1 T T Bl 0.61 further, as in case of Joule heating V2 P  R V2 i.e., R  P And as here, V1 = V2 = e So, And, e2 R1  1 e2 R2  2  (0.6)2 0.76  (0.6)2 0.76  9  19  0.3 . 11. A coil of inductance L = 50 × 10-6 henry and resistance = 0.5  is connected to a battery of emf = 5.0 V. A resistance of 10  is connected parallel to the coil. Now at some instant the connection of the battery is switched off. Find the amount of heat generated in the coil after switching off the battery. Solution : Total energy stored in the inductor  1 Li2 2 0 1  V 2 EL  2 L r   Fraction of energy lost across inductor  EL. r (R  r)  LV2  50 106  52   4 2r(R  r) 2  0.5(10  0.5) 1.19 10 J Alternative Solution : After switching off the battery, the current at any instant t during discharging process is given by I  I e(Rr) t / L  Energy dissipated across inductance in time dt, dQ  I2r dt   So, Q   I2r dt  I2 re2(Rr)t / Ldt 0 0    2  e2(Rr) t / L  I2rL I0r  2(R  r)   0 2(R  r) Here,  I  V 0 r L 0 V2L 50 106  52 4  Q  2r(R  r)  2  0.5(10  0.5)  1.19 10 J . y × × × × × × × A× 12. A metal rod OA of mass m and length l is kept rotating S with a constant angular speed  in a vertical plane about a R horizontal axis at the end O. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and con- L (A)  × × O × × × × × × C Solution: stant magnetic induction B is applied perpendicular and into the plane of rotation as shown in figure. An inductor L and an external resistance R are connected through a switch S between the point O and a point C on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open. (a) What is the induced emf across the terminals of the switch ? (b) The switch S is closed at time t = 0 (i) Obtain an expression for the current as a function of time (ii) In the steady state, obtain the time dependence of the torque required to maintain the constant angular speed, given that the rod OA was along the positive x-axis at t = 0. (a) As the terminals of the switch S are connected between the points O and C, so the emf across the switch is same as across the ends of the rod. Now to calculate the emf across the rod, consider an element of the rod of length dr at a distance r from O, then dE  Bdr  Brdr (as  r ) l so E  Br dr  0 1 Bl2 2 ...(i) and in accordance with Fleming’s right hand rule the direction of current in the rod will be from A to O and so O will be at a higher potential (as inside a source of emf current flows from lower to higher potential) y × × × × × × × A× y × × × I A S  × × × × O × × R O E R L (A) × × × × C  mg FM O × × × (B) C L (C) (b) (i) Treating the ring and rod rotating in the field as a source of emf E given by equation (i) , the equivalent circuit (when the switch S is closed) is as shown in figure. Applying Kirchhoff’s loop rule to it, keeping in mind that current in the circuit is increasing, we get E  IR  L dI  0 or dI  1 dt dt (E  IR) L which on integration with initial condition I = 0 at t = 0 yields I  I (1 et / ) with I  E and   L 0 0 R R So substituting the value of E from Eqn. (1) we have Bl 2 I  2R [1 e(R / L) t ] ...(ii) (ii) As in steady state I is independent of time, i.e., et /  0 i.e., t , so (I) steadystate  Imax  Bl2 2R ...(iii) Now as the rod is rotating in a vertical plane so for the situation shown in figure it will experience torques in clockwise sense due to its own weight and also due to the mag- netic force on it. So the torque on element dr, d  (mg)  r cos  FM  r i.e., d  M (dr)g  r cos   BI dr    M dr and F  BIdr l So total torque acting on the rod.  l M   M  l Mgl l 2    l g cos   BI  rdr  0 cos   BI 2 2 But as rod is rotating at constant angular velocity ,   t I  (Bl2 / 2R) and from equation (iii) So,   Mgl B2l 4 cost  ...(iv) 2 4R And hence the rod will rotate at constant angular velocity  if a torque having magni- tude equal to that given by equation is applied to it in anticlockwise sense. 13: Two long parallel wires carrying current I are separated by a distance 3a. There exists a square loop of side a with I capacitor of capacity C as shown in the figure. The value I of current varies with time as I = I0 sin t (a) Calculate maximum current in the square loop (b) Draw a graph between charge on plate of capacitor as a function of time. Solution: (a) I  I0 sin t Flux linked with square loop 2 a  I  1 1   d   0   adx I a 2  x 3a  x  I  0 Ia [ln x  ln(3a  x) |2a 2 a = 0 Ia [ln 2a  ln a  ln a  ln 2a] 2  0 Ia 2 ln 2  0Ia ln 2 2  Charge an capacitor  C d  C 0a ln 2 dI dt  dt  0Ca ln 2(I ) cost …(i)  0  0C(I0)a ln 2cost   CI 2a …(ii) Maximum current I  max  0 0 ln 2  (b) The graph for charge and time can be drawn from equation (i) as shown in figure. Q 0CI0a ln 2  t  0CI0a ln 2 .  14. An infinitesimally small bar magnet of dipole moment M is pointing and moving with the speed v in the x-direction. A small closed circular conducting loop of radius a and of negli- gible self-inductance lies in the y-z plane with its center at x = 0, and its axis coinciding with the x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the center of the loop is much greater than a. Solution: Field due to the bar magnet at distance x (near the loop) B  0 2M 4 x2  Flux lined with the loop :   BA  a2. 0 2M 4 x3 d  6Ma2 dx emf induced in the loop : e    0 dt 4 x4 dt  6Ma2  0 v. 4 x4 e  3Ma2 3 Ma2  Induced current : i   0 .v.  0 R 2 Rx4 Rx4 Let F = force opposing the motion of the magnet Power due to the opposing force = Heat dissipated in the coil per second. i2R   2  6Ma2 2 R  Fv  i2R  F    0     v2  v 9 2M2a4v F  0 . 4 Rx4  4   Rx4  v 15. A rectangular frame ABCD made of uniform metal wire A has straight connection between E and F made of the same wire as shown in figure. AEFD is a square of side 1 m and EB = FC = 0.5 m. The entire circuit is placed in a steadily increasing, uniform magnetic field direction into the plane 1m E 0.5m B of the paper and normal to it. The rate of change of mag- D F C netic field is 1 T/s. The resistance per unit length of wire is 1 / m . Find the magnitudes and directions of the currents in the segments AE, BE and EF. Solution: Induced e.m.f.   d   d (BA)  A dB dt dt dt  Induced e.m.f. in AEFD = 1  1 = 1V (area = 1 m2) Induced e.m.f. in EBCF  1 1  0.5V 2 (area  1 m2 ) 2 Total induced e.m.f  1  0.5  1.5V Given that resistance per unit length of the wire is 1/ m . Hence the equivalent circuit take the form as shown in figure. The resistances 3 and 2 are in parallel. Hence their equivalent resistance would (3 2)   6  (3  2)    Current from EF  1.5  5 amp. Current in AE  5  2  1 amp. 4 5 2 Current in BE  5  3  3 4 5 4 amp.