Question

A horizontal copper disc of diameter $20\text{cm}$, makes $10\text{revolution/s}$ about a vertical axis passing through its center. A uniform magnetic field of $100\text{G}$ acts perpendicular to the plane of the disc. Calculate the potential difference across its center and rim.

Take $1\text{G}={10}^{-4}\text{T}$

• A. $2.25\text{mV}$
• B. $3.14\text{mV}$
• C. $6.28\text{mV}$
• D. $1.57\text{mV}$

Answer 1

The correct option is B $3.14\text{mV}$

Given:

Magnetic field, $B=100\text{G}={10}^{-2}\text{T}$

Radius of disc, $r=\frac{d}{2}=0.1\text{m}$

Frequency of rotation, $f=10\text{revolution/s}$

$\Rightarrow \omega =2\pi f=2\pi \times 10=20\pi \text{rad/s}$

Now, emf induced between center and rim is,

$E=\frac{B\omega l^2}{2}=\frac{B\omega r^2}{2}$

$=\frac{{10}^{-2}\times 20\pi \times {0.1}^2}{2}=3.14\text{mV}$

Hence, option $\left(B\right)$ is the correct answer.