A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

D

Explanation

`vectau = MB sin theta`
`tau_(1)/tau_(2) = (sin theta_(1))/(theta_(2))` (couple considered as torque)
`tau_(1)/tau_(2) = (sin 90^(@))/(sin theta_(2))`
`sin theta_(2) = tau_(2)/tau_(1)`
`sin theta_(2) =1/2 rArr theta_(1) = 30^(@)`
angle of rotation `=90^(@) - 30^(@) = 60^(@)`