Question 1

Inside a solenoid of radius 0.5 m , magnetic field is changing at a rate of 50 10^ -6 ~T / s . Acceleration of an electron placed at a distance of 0.3 m from axis of solenoid will be

Accepted Answer

The changing magnetic field induce an electric field which will accelerate the electron. The acceleration of electron will be a= e E m [ F=q E] To find electric field E, we will use Faraday's electromagnetic induction. emf , = -d d t (B A) where, B is the magnetic field and A is the area of cross-section. =-A d B d t =- r^2 d B d t Now, electric field, E=- emf distance =- d = r^2 d d B d t Now, acceleration, a= e m r^2 d d B d t = 1.6 10^ -19 9.1 10^ -31 22 7 aligned & (0.5)^2 (0.3) 50 10^ -6 \

Question 2

There are two long co-axial solenoids of same length l. The inner and outer coils have radii r_1 and r_2 and number of turns per unit length n_1 and n_2, respectively. The ratio of mutual inductance t

Accepted Answer

The rate of mutual inductance is given by M = _0 n _1 n _2 ~m _1^2 ....(i) The rate of self inductance is given by L = _0 n _1^2 r _1^2 .....(ii) Dividing (i) by (ii) M L = n _2 n _1

Question 3

A rectangular loop of length 2.5 m and width 2 m is placed at 60^ to a magnetic field of 4 T . The loop is removed from the field in 10 sec . The average emf induced in the loop during this time is

Accepted Answer

Averageemf, e = Changein flux Time =- t =- 0- (4 (2.5 2) 60^ ) 10 =+1 ~V

Question 4

A 12 ~V battery connected to a coil of resistance 6 through a switch, drives a constant current in the circuit. The switch is opened in 1 ~ms . The emf induced across the coil is 20 ~V . The inductanc

Accepted Answer

Induced emf, =- L dI dt 20=- L (0-2) 10^ -3 Hence, inductance of the coil. L =10 mH

Question 5

In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency _0. Th

Accepted Answer

The resonance frequency of LC oscillations circuit is = 1 L ^ C ^ L ^ 2 ~L C ^ 8 C = 1 2 ~L 8 C = 1 4 LC 9 _0= 1 LC = _0 4 So, x= 1 4

Question 6

A conducting loop of radius 10 cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop

Accepted Answer

As . |_ t=0.5 sec =- d dt =- A dB dt [ =0^ =1 ] =- ( 10 )^2 10^ -4 0-0.5 0.5 =10^ -2 ~V =10 mV As d B d t = constant Induced emf will not change with time. So, . e |_ 0,5 sec = . e |_ 0.25 sec =10 mV

Question 7

A metal disc of radius 100 ~cm is rotated at a constant angular speed of 60 rad / s in a plane at right angles to an external field of magnetic induction 0.05 ~Wb / m ^ 2 . The emf induced between the

Accepted Answer

Induced emf produced between the centre and a point on the disc is given by array l e= 1 2 BR ^ 2 \ =60 rad / s , B =0.05 ~Wb / m ^ 2 \ and R =100 ~cm =1 ~m \ We get e= 1 2 60 0.05 (1)^ 2 =1.5 ~V array

Electromagnetic Induction — Previous Year Questions & Solutions