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

Options

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.

Answer

A

Explanation

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