Question

A uniform horizontal magnetic field of $7.5\times {10}^{-2}$T is set up at an angle of ${30}^0$ with the axis of an solenoid and the magnetic moment associated with it is $1.28J{T}^{-1}$. Then torque on it is:

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

Answer 1

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

$\overrightarrow{m}=1.28J{T}^{-1}$

$\overrightarrow{B}=7.5\times {10}^{-2}T$

Torque$=\overrightarrow{m}\times \overrightarrow{B}$

$=mB\sin {30}^{\circ }$

$=1.28\times 7.5\times {10}^{-2}\times \frac{1}{2}$

$=0.048Nm$