A closed coil having 100turns is rotated in a uniform magnetic field B = 4.0 × 10^–4 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm² and its resistance is 4.0 Ω. Find

(a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field,

(b) the average emf in a full turn and

(c) the net charge displaced in part (a).

Given:

No. of turns in the coil

Magnetic field intensity

Angular velocity of rotation

Area of the coil

Resistance of the coil

Magnetic flux through the circular coil ϕ can be given by formula

…. (i)

Where B=magnetic field intensity

A=area of cross section

N=no. of turns in the coil

θ =angle between area vector and magnetic field

(a) Initially, angle between area vector and magnetic field is 0°

Therefore, initial flux through the coil is

When it is rotated by 180° flux passing through the coil is given by

Average induced emf in time interval Δt is given by

…(i)

Where

are flux across the cross section at time intervals respectively.

Average induced emf is then given by

……(ii)

Now,

Angular velocity of coil

Time taken to complete half revolution i.e. rotate by π radian

Putting the values of N, B, A and Δt in eqn.(ii)

Therefore average emf induced in the coil in half a turn is

(b) In a full term coil returns to its original position

And hence emf induce in the coil using eqn.(ii) is

Therefore, average emf induced in full turn in the coil is zero

(c) Emf induced in the coil in part (a) is

Hence the current i flowing through the coil of resistance R is

So the charge displaced in time interval Δt =0.1s is

Therefore net charge displaced in part (a) is