A circular coil of radius R and a current `I`, which can rotate about a fixed axis passing through its diameter is initially placed such that its plane lies along magnetic field B kinetic energy of loop when it rotates through an angle `90^(@)` is : (Assume that `I` remains constant)

Options

A. `pi R^(2) BI`
B. `(pi R^(2) BI)/(2)`
C. `2 piR^(2)BI`
D. `(3)/(2) pi R^(2) I`

Answer

A

Explanation

Loss in potential energy = gain kinetic energy
`(-MB cos 90^(@))-(-MB cos 0^(@))=KE`