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(m B sin \theta \right)=mBcos\theta $
And it will be maximum at $cos\theta =1$
$\Rightarrow \theta =0^\circ $