Question

A magnetised needle of magnetic moment $4.8\times {10}^{-2}J/T$ is placed at ${30}^o$ with the direction of uniform magnetic field of magnitude $3\times {10}^{-2}T$. Then the torque acting on the needle is:

• A. $7.2\times {10}^{-3}Nm$
• B. $7.2\times {10}^{-4}Nm$
• C. $3.6\times {10}^{-4}Nm$
• D. $14.4\times {10}^{-4}Nm$

Answer 1

The correct option is B $7.2\times {10}^{-4}Nm$

Given, Magnetic momentum, $m=4.8\times {10}^{-2},\theta ={30}^0,B=3\times {10}^{-2}T$

Therefore, $T=mBsin\theta$

$=4.8\times {10}^{-2}\times 3\times {10}^{-2}\times sin{30}^0$

$=4.8\times {10}^{-2}\times 3\times {10}^{-2}\times \frac{1}{2}$

$=7.2\times {10}^{-4}J$

Where $sin{30}^0=\frac{1}{2}$