Question

If a bar magnet of moment is suspended in a uniform magnetic field $\overline{B}$ it is given an angular deflection, w.r.t equilibrium position. Then the restoring torque on the magnet is

• A. $MB\sin \theta$
• B. $MB\cos \theta$
• C. $MBtan\theta$
• D. $M{B}^2\sin \theta$

Answer 1

The correct option is A $MB\sin \theta$

$T=2(F\times a/2\times \sin \theta$)
$T=IAB\sin \theta$
$m=IA$
$T=mB\sin \theta$