Question

A circular copper disc $10 \,cm$ in diameter rotates at $1800$ revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction $B$ of $1\, Wb\, m^{-2}$ is perpendicular to disc, What potential difference is developed between the axis of the disc and the rim?

• A. $0.023\,V$
• B. $0.23\,V$
• C. $23\,V$
• D. $230\,V$

Answer 1

The Correct Option is B

Here, $l=r=5\,cm=5\times10^{-2}\,m$, $\omega=2 \pi \frac{1800}{60}rad\,s^{-1}$ $=60\, \pi\, rad\,s^{-1}$, $B=1\,Wb\,m^{-2}$ $\varepsilon=\frac{1}{2}Bl^2\omega =\frac{1}{2}\times1\times(5\times10^{-2})^{2}\times60\pi=0.23\,V$