Magnetics-02- OBJECTIVE SOLVED

OBJECTIVE SOLVED PROBLEMS 1. The magnetic field lines due to a bar magnet are correctly shown in N N (a) (b) N N (c) (d) . Ans. (d) Solution :Magnetic lines of forces form closed loops. 2. A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. Magnetic field due to current at teh center of ring is (a) zero, only if θ = 1800 (b) zero for all values of  (c) proportional to 2(1800 - θ ) (d) inversely proportional to r. Ans. (b) Solution : For a current flowing into a circular arc, magnetic induction in the centre B  0I 4 dl r r3  0I 4 r2d r3   0I    The total current is divided into two arcs I  E 1 1  E  E l1  2E (R / 2r)l1 (R / 2r)(r) R Similarly I θ  π E constant 1 R l I  E 2 2  E (R / 2r)l2  E  (R / 2r){r(2  )} 2E R(2  )  cons tan t B  B  B  0  2E  2E   0. 1 2 4r  R R  3. Two very long, straight, parallel wires carry steady currents I and -I respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous magnitude of the force due to the mag- netic field acting on the charge at this instant is : (a) μ0Iqv 2πd (b) μ0Iqv πd (c) 2μ0Iqv πd (d) 0. Ans. (d) Solution: Since the currents are flowing in the opposite directions, the magnetic field at a point equidistant from the two wires will be zero. Hence, the force acting on the charge at this instant will be zero. 4. H+, He+ and O++ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O++ are 1 amu, 4 amu and 16 amu respectively. Then (a) H+ will be deflected most (b) O2+ will be deflected most (c) He+ and O2+ will be deflected equally (d) all will be deflected equally. Ans. (a) and (c) Solution : Bqv  mv2 r  Bqr  mv  momentum  Where E = Kinetic Energy  r  2mE Bq if r and r and r are the radius of circular track of H  , He and O 1 2  r : r 3 : r  2mE : 2(4m)E : 2(16m)E 1 2 3 Bq Bq 1 : 2 : 2 B(2q) 5. A current I flows along the length of an infinitely long, straight, thin-walled pipe. Then (a) the magnetic field at all points inside the pipe is the same, but not zero (b) the magnetic field at any point inside the pipe is zero (c) the magnetic field is zero only on the axis of the pipe (d) the magnetic field is different at different points inside the pipe. Ans. (b) Solution: Fig. shows infinitely, long, straight, thin-walled pipe carrying current I. Let P be any point at a distance r from the axis OO1 of the pipe. Let B be magnetic field at P. Consider a closed circular path passing through point P as shown in figure. From Ampere’s Circuital Theorem, i i  B.dl  0i i = current through the closed path. Obviously, i = 0  2RB  0 or B  0 . 6. A particle of charge +q and mass m moving under the influence of a uniform electric field E iˆ and a uniform magnetic field B kˆ follows a trajectory from P to Q as shown in figure. The velocities at P and Q are v iˆ and -2v iˆ . Which of the following statement(s) is/are x correct? 3  mv 2  (a) E    4  qa  3  mv3  (b) rate of work done by the electric field at P is    a  (c) rate of work done by the electric field at P is zero (d) rate of work done by both the fields at Q is zero. Ans. (a), (b) and (d) Solution: Increase in Kinetic energy of particle  1 m(2v)2  1 mv2  3 mv2 2 2 2 Work done by the uniform electric field, E, in going from P to Q = (qE) × 2a = 2qEa Hence, 2qEa  3 mv2 2 3mv2 or E  4qa Rate of work done by the electric field at Pat, P = F. v = qE . v  qEˆi.vˆi  qEv  q. 3mv2 .v  3 mv2 4qa 4 a Q is Pat ,Q  qEiˆ.(2vˆj)  0 At Q, rate of work done by both the fields is zero. 7. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and desribe circular paths of radii R1 and R2, respectivley. The ratio of the mass of X to that of Y is (a) R / R 1/ 2 (c) R / R 2 Ans. (c) (b) (d) R 2 / R1 R1 / R 2 . Solution: Let the masses be m1 and m2 respectively of X and Y. If E is energy gained by charged particle in electric field. mv2 Bqv  r  Bqr  2mE R1  2m1E ; Bq R 2  2m2E Bq R m  R 2 1   1   1  R 2 m2  R 2  8. A proton moving with a constant velocity passes through region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, this region of space may have (a) E = 0, B = 0 (b) E = 0, B  0 (c) E  0, B = 0 (d) E  0, B  0. Ans. (a) and (d) Solution: As there is no acceleration, either E = 0, B = 0 or E  0, B  0. 9. A regular loop carrying a current i is situated near a long striaght i wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If a steady current I established in the I wire as shown in the figure, the loop will : (a) rotate about an axis parallel to the wire (b) move away from the wire (c) move towards the wire (d) remain stationary. Ans. (c) Solution: F  0 il 2 r Fattraction  Rrepulsion Hence the loop will move towards the wire. 10. Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is (a) μ id EQ πd2  x 2  (b) μ ix πd2  x2  μ ix (c) d2  x2  Ans. (b) Solution: The magnetic field due μ id (d) d2  x2  . B  μ0i 1 2π (d  x) i i B  μ0i 2 2π (d  x) Both the magnetic fields act in opposite direction.  B  B  B  0i  1  1  2 1 2 d  x d  x   0i c  x  d  x  2   0ix d2  x 2  . (d2  x 2 ) 11. An electron with mass m, velocity v and charge e describes half a revolution in a circle of radius r in a magnetic field B, will acquire energy equal to (a) 1/ 2mv 2 (b) 1/ 4mv 2 (c) πrBev (d) zero. Ans. (d) Solution: As energy can neither be created nor destroyed, therefore, its energy will remain constant and will acquire no extra energy. 12. A current (l) carrying circular wire of radius R is placed in a mag- netic field B perpendicular to its plane. The tension T along the circumference of the wire is × I × × × R (a) BIR (b) 2BIR × (c) Ans. (a) BIR (d) 2BIR. × × B × Solution: For small elemental portion 2T sin d  2R d IB 2Td  2RIBd T  IRB . 13. The magnetic moment of an electron orbiting in a circular orbit of radius r with a speed v is equal to: (a) evr/2 (b) evr (c) er/2v (d) none of these. Ans. (a) Solution: Magnetic moment   niA Where n = number of tums of the current loop I = current; Since the orbiting electron behaves as current loop of current i, we can write i  e  T e 2r v  ev 2r A = area of the loop  r2    (1) ev (r2 )      evr . 2 14. A particle of mass m and charge q is released from the origin in a region occupied by electric field E and magnetic field B, B  B0 ˆj; E  E0 iˆ (a) (c) qE0 2m (b) (d) none the speed of the particle is Solution : Since the magnetic field does not perpform any work, therfore, whatever has been gain in kinetic energy it is only becuase of the work done by electric field. Applying work-energy theorem, WE  K qE0  1 mv2  o or v  2 15. A charged particle P leaves the origin with speed v  v0 , at some inclination with the x-axis. There is uniform magnetic field B along the x-axis. P strikes a fixed target T on the x-axis for a minimum value of B = B0. P will also strike T if (a) B = 2B0, v = 2v0 (b) B = 2B0, v = v0 (c) B = B0, v=2v0 (d) B= B0/2, v = 2v0. Ans. (a), (c) Solution: Let d = distance of the tanget T from the point of projection. P will strike T if d an integral multiple of the pitch. Pitch =  2 m vcos  qB    Here, m, q and  are constant  pitch = k v    where k = constant.