Question

The final torque on a coil having magnetic moment 25 A $m^2$ in a 5 T uniform external magnetic field, if the coil rotates through an angle of 60 under the influence of the magnetic field is:

• A. 216.5 Nm
• B. 108.25 Nm
• C. 102.5 Nm
• D. 58.1 Nm

Answer 1

Answer: (B)

108.25 Nm

Here,

$m=25A{m}^2;\theta ={60}^0;B=5T$

$|\overrightarrow{\tau }=|\overrightarrow{m}\times \overrightarrow{B}|=mB\sin \Theta$

$\therefore =25\times 5\times \sin {60}^0$ or

$\tau =125\times \frac{\sqrt{3}}{2}=108.25Nm$