A rod of length l rotates with a small but uniform angular velocity `omega` about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is

Options

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

Answer

A

Explanation

If in time `t`. The rod turns by an angle `q`. The area genrated by the rotation of rod will be
`=(1)/(2)lxxltheta=(1)/(20l^(2)theta`
so the flux linked with the area genrated by the ratation of rod
`varphi=B((1)/(2)l^(2)theta)cos0=(1)/(2)Bl^(2)theta=(1)/(2)Bl^(2)omegat`
and so `e=(dvarphi)/(dt)=(d)/(dt)((1)/(2)Bl^(2)omegat)=(1)/(2)Bl^(2)omega`