Question

A copper disc of radius 0.1 m is rotated about its centre with 10 revolutions per second in a uniform magnetic field of 0.1 Tesla with its plane perpendicular to the field. The e.m.f. induced across the radius of disc is
(a) $\frac{\pi}{10}V$
(b) $\frac{2\pi}{10}V$
(c) π×10⁻² V
(d) 2π×10⁻² V

Answer 1

answer : option (a) $\frac{\pi}{10}V$

explanation : use formula, $\xi_{in}=\frac{1}{2}B\omega R^2$

where $\xi_{in}$ denotes induced emf, B is magnetic field, $\omega$ denotes angular frequency of disc and R denotes the radius of disc.

here, B = 0.1 T , R = 0.1 m and $\omega=10rps=20π rad/s$

hence, $\xi_{in}$ = 1/2 × 0.1 × 20π × (0.1)²

= 1/2 × 20π × 0.01

= π/10 V

hence, option (a) is correct

Answer 2

Answer:

Pi*10-2V

Explanation:

In pic