Question

A copper rod of length $l$ is rotated about one end, perpendicular to the uniform magnetic field $B$ with constant angular velocity $\omega$. The induced emf between two ends of the rod is:

• A. $\frac{1}{2}B\omega l^2$
• B. $B\omega l^2$
• C. $\frac{3}{2}B\omega l^2$
• D. $2B\omega l^2$

Answer 1

The correct option is A $\frac{1}{2}B\omega l^2$

A copper rod of length l is rotated about one end, perpendicular to the uniform magnetic field B with constant angular velocity ω. The induced emf between two ends of the rod is

The EMF induced is $dE=\frac{BdA}{dt}=Bvdx$

$v=wx$

$dE=Bwxdx$

$dE={\int }_0^{l}Bwxdx$

$dE=Bw\frac{x^2}{2}=\frac{1}{2}Bw{l}^2$