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

Answer: (D)

Induced emf $e=\pi l^{2} B n=\frac{l^{2} B \omega}{2}$
$=\frac{(1.0)^{2} \times 0.25 \times 2 \times 3.14 \times 12}{2}=9.42\, V$