Question

A circular copper disc, of radius $5\text{cm}$, rotates with $10\text{rad/s}$ about an axis through its center and perpendicular to the disc. A uniform magnetic field of $2\text{T}$ acts perpendicular to the plane of the disc. What is the induced current in the disc if a resistor of resistance $1\Omega$ is connected between the midpoint of a radius and the center of the disc?

• A. $6.25\times {10}^{-1}\text{A}$
• B. $6.25\times {10}^{-2}\text{A}$
• C. $6.25\times {10}^{-3}\text{A}$
• D. $6.25\times {10}^{-4}\text{A}$

Answer 1

The correct option is C $6.25\times {10}^{-3}\text{A}$

Given:
Radius of disc, $r=0.05\text{m}$
Angular speed, $\omega =10\text{rad/s}$
Magnetic field, $B=2\text{T}$
Resistance, $R=1\Omega$

Emf induced between the midpoint of a radius and the center of the disc is,

$E=\frac{B\omega l^2}{2}=\frac{B\omega r^2}{8}(\because l=r/2)$

$=\frac{2\times 10\times {0.05}^2}{8}=6.25\times {10}^{-3}\text{V}$

Now, the current flowing through the resistor is,

$i=\frac{E}{R}=\frac{6.25\times {10}^{-3}}{1}=6.25\times {10}^{-3}\text{A}$