Question

A copper rod of length l is mid-point of rod . perpendicular to the magnetic field B with constant angular velocity $\omega$. The induced emf between the two ends is

• A. $\frac{1}{2}B\omega l^2$
• B. $B\omega l^2$
• C. $2B\omega l^2$
• D. Zero

Answer 1

The correct option is D Zero

Given: A rod of length L rotating in magnetic field B with angular velocity $\omega$.

Solution: We will take mid point as the reference as the rod is rotating about the mid point. So, consider the differential length of the rod $dx$ from its center.

Therefore, in a magnetic field B, the angular velocity of the differential area will be,

$\omega \times x$

So, emf induced in the coil is given by,

$E=Blv$

$E=B\omega xdx$

Integrating within the limits we have,

$E={\int }_{\frac{-L}{2}}^{\frac{L}{2}}B\omega xdx$

$E=B\omega {\int }_{\frac{-L}{2}}^{\frac{L}{2}}xdx$

$E=B\omega {\left[\frac{x^2}{2}\right]}_{\frac{-L}{2}}^{\frac{L}{2}}$

On solving, we get

$E=0$

Hence, the correct option is (D).