Question

Find :
(a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field,
(b) the net charge displaced in part (a).

Answer 1

(a)

The average emf is given as,

$e=\frac{2NBA}{t}$

$e=\frac{2\times 50\times 2\times {10}^{-4}\times 100\times {10}^{-4}}{\frac{1}{600}\times 60}$

$e=2\times {10}^{-3}V$

(b)

Net charge is given as,

$q=\frac{2NBA}{R}$

$q=\frac{2\times 50\times 2\times {10}^{-4}\times 100\times {10}^{-4}}{4}$

$q=5\times {10}^{-5}C$