Question

A conducting disc of radius r spins about its axis with an angular velocity $\omega$. There is a uniform magnetic field of magnetude B perpendicular to the plane of the disc. C is the centre of the ring.

• A. No emf is induced in the disc.
• B. The potential difference between C and the rim is $\frac{1}{3}B{r}^2\omega$.
• C. C is at a higher potential than the rim.
• D. Current flows between C and the rim.

Answer 1

Answer: (B), (C)

The correct options are
B The potential difference between C and the rim is $\frac{1}{3}B{r}^2\omega$.
C C is at a higher potential than the rim.

emf induced = ${\int }_0^{r}Bdrr\omega =B\omega \frac{r^2}{2}$
positive charges in the disc will experience a force n the direction given by $q\overrightarrow{v}\times \overrightarrow{B}=q\overrightarrow{\left(r\omega \right)}\times \overrightarrow{B}$ = towards the center C.
Therefore C will be at a higher potential wrt to the rim.