Question

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity. The induced e.m.f. between the two ends is

• A. Bωl²
• B. Bωl²
• C. Bωl²
• D. 2Bωl²

Answer 1

The correct option is A Bωl²

If in time t. the rod turns by an angle $\theta$, the area generated by the rotation of the rod will be $\frac{1}{2}l\times l\theta =\frac{1}{2}{l}^2\theta$

So the flux linked with the area generated by the rotation of rod



$\varphi =B\left(\frac{1}{2}{l}^2\theta \right)cos0=\frac{1}{2}B{l}^2\theta =\frac{1}{2}B{l}^{2}wt$
And So $e=\frac{d\varphi }{dt}=\frac{d}{dt}\left(\frac{1}{2}B{l}^{2}wt\right)=\frac{1}{2}B{l}^{2}w$