Question

A copper disc of radius$0.1m$ is rotated about its center with $20rev/s$ in a uniform magnetic field of $0.1T$ with its plane perpendicular to the field. The emf induced across the radius of the disc is

• A. $\pi /20V$
• B. $10/\pi V$
• C. $20\pi mV$
• D. $10\pi mV$

Answer 1

The correct option is C

$20\pi mV$

Explanation for the correct answer:

Option (C): $20\pi mV$

Step 1. Given data:

Magnetic field, $B=0.1T$

Speed, $N=20\raisebox{1ex}{$rev$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Radius of disc, $R=0.1m$

Step 2. Calculating induced emf, $\mathit{\epsilon }$:

We know that motional emf induced between center and periphery of disk is given by,

$$\begin{gathered} \epsilon =\frac{B\omega R^2}{2} \\ \Rightarrow \epsilon =\left[\frac{\left(0.1\right)\times \left(2\pi N\right)\times {\left(0.1\right)}^2}{2}\right] \end{gathered}$$ $\left[\omega =2\pi N\right]$

$$\begin{gathered} \Rightarrow \epsilon =\left[\frac{\left(0.1\right)\times \left(2\times \pi \times 20\right)\times {\left(0.1\right)}^2}{2}\right] \\ \Rightarrow \epsilon =20\pi \times {10}^{-3}V \\ \therefore \epsilon =20\pi mV \end{gathered}$$

Hence, correct option is (C).