Question

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

• A. $\frac{1}{2}B\omega l^2$
  
• B. $\frac{3}{4}B\omega l^2$
  
• C. $B\omega l^2$
  
• D. $2B\omega l^2$

Answer 1

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


If in time t. the rod turns by an angle $\theta$, the area generated by the rotation of 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
$\varphi =B\left(\frac{1}{2}{l}^2\theta \right)cos\theta =\frac{1}{2}B{l}^2\theta =\frac{1}{2}B{l}^2\omega t$
And so $e=\frac{d\varphi }{dt}=\frac{d}{dt}\left(\frac{1}{2}B{l}^2\omega t\right)=\frac{1}{2}B{l}^2\omega$