Question

Rate of change of torque $\tau$ with deflection $\theta$ is maximum for magnet suspended freely in a uniform magnetic field of induction $B,$ when

• A. $\theta =90^{\circ }$
• B. $\theta =60^{\circ }$
• C. $\theta =45^{\circ }$
• D. $\theta =0^{\circ }$

Answer 1

Answer: (D)

Torque, $\tau =mBsin\theta$
$\tau$ will be maximum, when
$\theta =90^{\circ }$
Rate of change of torque w.r.t. deflection angle is given by, $\frac{d\tau }{d\theta }$
$\Rightarrow \frac{d\tau }{d\theta }=\frac{d}{d\theta }\left(mBsin\theta \right)=mBcos\theta$
And it will be maximum at $cos\theta =1$
$\Rightarrow \theta =0^{\circ }$