Question

A closed coil having 100 turns is rotated in a uniform magnetic field $B=4.0\times {10}^{-4}T$ about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolution per minute. The area of the coil is $25{cm}^2$ and its resistance is $4.0\Omega$. 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 full turn and (c) the net charge displaced in part (a).

Answer 1

Given, $N=100$ turns

$B=4\times {10}^{-4}T$

$A=25{cm}^2=25\times {10}^{-4}{m}^2$

(a) When the coil is perpendicular to field.

$\varphi =NBA$

Average emf=$\frac{2NBA}{t}$

=$\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{\frac{1}{600}\times 60}=2\times {10}^{-3}V$

(b)In full turn average emf will be zero.

(c)Charged induced=$\frac{2NBA}{R}=\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{4}\approx 5\times {10}^{-5}C$