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 e.m.f. between its two ends is

Options

A. `(1)/(2)B omega l^(2)`
B. `(3)/(4)B omega l^(2)`
C. `B omega l^(2)`
D. `2B omega l^(2)`

Answer

A

Explanation

If in time t, the rod turns by an angle `theta`, the area generated by the rotation of rod will be
`=(1)/(2)l xx l theta =(1)/(2)l^(2)theta`
So the flux linked with the area generated by the rotation of rod
`phi=B((1)/(2)l^(2)theta) cos 0 =(1)/(2)Bl^(2)theta = (1)/(2)Bl^(2)omega t`
and so `e=(d phi)/(dt)=(d)/(dt)((1)/(2)Bl^(2)omega t)=(1)/(2)Bl^(2)omega`