Question

A uniform horizontal magnetic field of $7.5\times {10}^{-2}$T is set up at angle of ${30}^o$ with the axis of an solenoid and the magnetic moment associated with it is $1.28$J $T^{-1}$. Then the torque on it is?

• A. $4.8\times {10}^{-2}$N m
• B. $1.6\times {10}^{-2}$N m
• C. $1.2\times {10}^{-2}$N m
• D. $4.8\times {10}^{-4}$N m

Answer 1

The correct option is A $4.8\times {10}^{-2}$N m

Torque, $\tau =MB\sin \theta$
Here, $M=1.28J$ $T^{-1}$, $B=7.5\times {10}^{-2}T,\theta ={30}^o$
$\therefore \tau =1.28\times 7.5\times {10}^{-2}\sin {30}^o$
$=1.28\times 7.5\times {10}^{-2}\times \frac{1}{2}$
$=0.64\times 7.5\times {10}^{-2}=4.8\times {10}^{-2}$N m.