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A wire is bent to form a semi-circle of ...

A wire is bent to form a semi-circle of radius a. The wire rotates about its one end with angular velocity `omega`. Axis of rotation being perpendicular to plane of the semicircle. In the space, a uniform magnetic field of induction exists along the axis of rotation as shown. The correct statement is

A

potential difference between P and Q is equal to `2Bomegaa^(2)`

B

potential difference between P and Q is equal to `2pi^(2)Bomegaa^(2)`

C

P is at higher potential than Q

D

potential difference between P and Q is zero.

Text Solution

Verified by Experts

The correct Answer is:
A

`E=1/2Bl^(2)omega=1/2B(2a)^(2)omega=2Bomegaa^(2)`
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