Question

A metal disc of radius R = 25 cm rotates with a constant angular velocity $\omega$ = 130 rad s$^{-1}$ about its axis. Find the potential difference between the center and rim of the disc if the external uniform magnetic field B = 5.0 mT is directed perpendicular to the disc.

• A. 0 V
• B. 0.020 V
• C. 0.040 V
• D. 0.060 V

Answer 1

Answer: (B)

0.020 V

Answer 2

When an external uniform magnetic field B is directed perpendicular to a rotating disc, a motional electromotive force (emf) is induced. For a small element of the disc at a radial distance $r$ from the center, moving with linear velocity $v = \omega r$, the induced electric field is $E = vB = \omega r B$. The potential difference $dV$ across a radial element $dr$ is $dV = E dr = \omega r B dr$. Integrating from the center ($r=0$) to the rim ($r=R$), the total potential difference is $V = \int_{0}^{R} \omega r B dr = \frac{1}{2} \omega B R^2$.

Given: $R = 25 \text{ cm} = 0.25 \text{ m}$ $\omega = 130 \text{ rad s}^{-1}$ $B = 5.0 \text{ mT} = 5.0 \times 10^{-3} \text{ T}$

$V = \frac{1}{2} \times (130 \text{ rad s}^{-1}) \times (5.0 \times 10^{-3} \text{ T}) \times (0.25 \text{ m})^2$ $V = \frac{1}{2} \times 130 \times 5.0 \times 10^{-3} \times 0.0625 \text{ V}$ $V = 0.0203125 \text{ V}$

Rounding to two significant figures (based on the given values of R and B), the potential difference is approximately $0.020 \text{ V}$ or $20 \text{ mV}$.