Question

A rod of length $10\text{cm}$ made up of conducting and non-conducting material (shaded part is non-conducting). The rod is revolving with constant angular velocity of $10\text{rad/s}$ about point $O$, in a constant magnetic field of $2\text{T}$ perpendicular to the plane of rotation of the rod as shown in the figure. The induced emf between the point $A$ and $B$ is -

• A. $0.02\text{V}$
• B. $0.05\text{V}$
• C. $0.03\text{V}$
• D. $0.04\text{V}$

Answer 1

The correct option is B $0.05\text{V}$


From above figure,
$OB=2\text{cm}+5\text{cm}=7\text{cm}$ and $OA=10\text{cm}$

Consider an element of thickness $dr$ at a distance $r$ from the point $O$ lying in the region $BA.$


EMF developed across this element is,

$dE=Bvdr=B\left(\omega r\right)dr[\because v=\omega r]$

So, EMF developed between the point $A$ and $B$ of the rod is,

$E=\int dE={\int }_{r_1}^{r_2}B\omega rdr$

$=B\omega {\left[\frac{r^2}{2}\right]}_{r_1}^{r_2}=B\omega \left(\frac{r_2^2-r_1^2}{2}\right)$

Here,
$B=2\text{T}$
$\omega =10\text{rad/s}$
$r_1=7\text{cm}=0.07\text{m}$
$r_2=10\text{cm}=0.1\text{m}$

Substituting all values, we get,

$E=0.05\text{V}$

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