Q: AB is a part of an electrical circuit (see figure). The potential difference " V _ A - V _ B ", at the instant when current i =2 ~A and is increasing at a rate of 1 amp / second is :

aligned & Apply K.V.L. \\ & aligned & V _ A - LdI dt -5- IR = V _ B \\ & ~V _ A - V _ B = LdI dt +5+ IR \\ & =1 1+5+2 2 \\ & =1+5+4 \\ & =10 ~V aligned aligned

Question 1

Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and _0 is the permeability of free space, the inductance of the solenoid is :

Accepted Answer

The magnetic field inside a long solenoid is given by B = _0 n I, where n is the number of turns per unit length and I is the current. The magnetic flux through a single turn of the solenoid is = B A = ( _0 n I)( r^2). The total number of turns in the solenoid of length l is N = n l. The total magnetic flux linkage is = N = (n l)( _0 n I r^2) = _0 n^2 r^2 l I. The self-inductance L is defined by the relation = L I. Substituting the value of , we get L = _0 n^2 r^2 l. Answer: _0 n^2r^2l

Question 2

Two identical inductors are connected in two different configurations P and Q, where a time varying current I(t) is flowing, as shown in the figure. The induced emf between points a and b for configur

Accepted Answer

Let the inductance of each identical inductor be L. For configuration (P): The two inductors are connected in series. The points a and b are located across the first inductor only. The current flowing through this inductor is I(t). The induced emf between points a and b is given by: E_P = L dI dt For configuration (Q): The two inductors are connected in parallel. The points a and b are the common nodes of this parallel combination. The equivalent inductance L_ eq between a and b is: L_ eq = L L

Question 3

A conducting loop of finite resistance lies on the x-y plane. There is a constant magnetic field in the z direction. The area of the loop varies with time t, as A=A_0(1+ t) in appropriate units. The f

Accepted Answer

The magnetic flux passing through the loop is given by: = B A Given that the area varies with time as A = A_0(1 + t) and the magnetic field B is constant, the flux is: = B A_0(1 + t) According to Faraday's law of electromagnetic induction, the induced emf is: = - d dt = -B A_0 d dt (1 + t) = -B A_0 t The power P dissipated in the loop of resistance R is: P = ^2 R = (-B A_0 t)^2 R = B^2 A_0^2 R ^2 t From this expression, we can see that the power P varies as ^2 t. At t = 0, P = B^2 A_0^2 R ^2(0)

Question 4

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed acro

Accepted Answer

Given: Magnetic field, B = 0.3 T Velocity of the loop, v = 2 cm s^ -1 = 2 10^ -2 m s^ -1 Length of the shorter side, l = 3 cm = 3 10^ -2 m Since the loop is moving in a direction normal to the shorter side, the velocity vector is perpendicular to the shorter side. The motional emf is induced across the side that is perpendicular to the direction of motion. The induced emf e is given by the formula: e = B l v Substituting the given values: e = 0.3 (3 10^ -2 ) (2 10^ -2 ) e = 1.8 10^ -4 V Answer:

Question 5

AB is a part of an electrical circuit (see figure). The potential difference " V _ A - V _ B ", at the instant when current i =2 ~A and is increasing at a rate of 1 amp / second is :

Accepted Answer

aligned & Apply K.V.L. \ & aligned & V _ A - LdI dt -5- IR = V _ B \ & ~V _ A - V _ B = LdI dt +5+ IR \ & =1 1+5+2 2 \ & =1+5+4 \ & =10 ~V aligned aligned

Question 6

Let us consider two solenoids A and B, made from same magnetic material of relative permeability _r and equal area of cross-section. Length of A is twice that of B and the number of turns per unit len

Accepted Answer

L= _0 _r n A N L= _0 _r n A N l l L= _0 _r n^2 A I L n^2 I L_A L_B = n_A^2 n_B^2 I_A I_B L_A L_B = 1 4 2= 1 2

Question 7

In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

Accepted Answer

North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South.

Electromagnetic Induction — Previous Year Questions & Solutions