Question

A metallic rod of length ${}^{\prime }{l}^{\prime }$ 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 $‘B^{\prime }$ in the region, the e.m.f. induced across the ends of the rod is

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

Answer 1

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

Let us try to solve it by dimensional analysis first rather than going for conventional approach.
We know that the emf induced in a rod rotating with angular velocity $\omega$ in a region of magnetic field is given by
Emf = $B\omega l^2$
Now, if we see options (a) and (b) are not dimensionally correct so they must be omitted.

Option (c) can be written as $2B\omega l^2$
$⟹2B\omega l^2=B\left(2\omega l\right)l$ = $B(\omega \times 2l)l$
But this would be the emf across metallic string and not across the metallic rod. Emf across the rod needs to be more than $2B\omega l^2$ because length increases further after string.
So, option (d) is the correct answer.