Question

A rod of length $1\text{m}$ rotates with a constant angular velocity of $8\text{rad/s}$ about its perpendicular bisector. A uniform magnetic field of $0.3\text{T}$ exists parallel to the axis of rotation. The potential difference between the center of the rod and its end is :

• A. $0.1\text{V}$
• B. $0.5\text{V}$
• C. $0.3\text{V}$
• D. Zero

Answer 1

The correct option is C $0.3\text{V}$

Given:

Length of the rod $\left(l\right)=1\text{m}$
Magnetic field $\left(B\right)=0.3\text{T}$
Angular velocity $\left(\omega \right)=8\text{rad/s}$

Now, the induced potential difference between the center of the rod and an end is,

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

Here, $r=\frac{l}{2}$

$\therefore E=\frac{B\omega {\left(\frac{l}{2}\right)}^2}{2}$

$=\frac{0.3\times 8\times {\left(\frac{1}{2}\right)}^2}{2}=0.3\text{V}$

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