Question

A metallic rod of length $l$ is tied to a string of length $2l$ and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field $B$ in the region, the e.m.f. induced across the ends of the rod is:

• A. $\frac{5B\omega l^2}{2}$
• B. $\frac{2B\omega l^2}{2}$
• C. $\frac{3B\omega l^2}{2}$
• D. $\frac{4B\omega l^2}{2}$

Answer 1

The correct option is A $\frac{5B\omega l^2}{2}$

Induced emf is rate of change of magnetic flux.


$E=\int Bvdx[\text{where,}v=\omega x]$

$={\int }_{2l}^{3l}\left(\omega x\right)Bdx$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

$\Rightarrow E=B\omega {\left[\frac{x^2}{2}\right]}_{2l}^{3l}$

$=B\omega \frac{\left[\right(3l{)}^2-\left(2l{)}^2\right]}{2}=\frac{5B\omega l^2\omega }{2}$

Hence, option $\left(D\right)$ is correct.