Question

A rod of length 1.0 m is rotated in a plane perpendicular to a uniform magnetic field of induction 0.25 T with a frequency of 12 rev/s. The induced emf across the ends of the rod is

• A. 18.89 V
• B. 3 V
• C. 15 V
• D. 9.42 V

Answer 1

The correct option is D 9.42 V

Induced emf $e=\frac{l^{2}B\omega }{2}$

Given : $l=1.0m,$ $B=0.25T,$ and $\nu =12$ rev/s

Thus angular speed of rotation $w=2\pi \nu =2\times 3.14\times 12$
$e=\frac{(1.0{)}^2\times 0.25\times 2\times 3.14\times 12}{2}=9.42V$