Question

A conducting rod of length $l=0.5\text{m}$ is hinged at point $\text{O}$. It is free to rotate in a vertical plane. There exists a uniform magnetic field $B=2\text{T}$ as shown in the figure. The rod is released freely. When the rod makes an angle of $\theta ={37}^{\circ }$ from the released position then the potential difference between the ends of the rod is :

• A. $0.5\text{V}$
• B. $1.5\text{V}$
• C. $3.0\text{V}$
• D. $5.0\text{V}$

Answer 1

The correct option is B $1.5\text{V}$

Let the angular velocity of the rod is $\omega$ at angle $\theta$.


Change in height of centre of mass of the rod,

$h=\frac{l}{2}\sin \theta$

From the work energy theorem,

$\frac{1}{2}I{\omega }^2-0=mg\left(\frac{l}{2}\sin \theta \right)$

$\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2=mg\left(\frac{l}{2}\sin \theta \right)[\because I=\frac{m{l}^2}{3}]$

$\therefore \omega =\sqrt{\frac{3g}{l}\sin \theta }$

The emf induced in the rod is,

$E=\frac{B\omega l^2}{2}=\frac{B}{2}\times \sqrt{\frac{3g}{l}\sin \theta }\times l^2$

$=\frac{B}{2}\sqrt{3g{l}^3\sin \theta }$

Here, $B=2\text{T},l=0.5\text{m},\theta ={37}^{\circ }$

$\therefore E=\frac{2}{2}\times \sqrt{3\times 10\times {0.5}^3\times \sin {37}^{\circ }}$

$=1\times \sqrt{3\times 10\times \frac{3}{5}\times (0.5{)}^3}$

$\therefore E=3\times 0.5=1.5\text{V}$

Hence, $\left(\text{B}\right)$ is the correct answer.