Question

A current 'I' flows through a thin wire as shown in the figure. If there exists an external magnetic field B in the same plane of the wire. The torque acting on the coil is:

• A. $I(\frac{\pi a^2}{2}+8a^2)B$
• B. $I(\frac{\pi a^2}{2}+4a^2)B$
• C. $I(\pi a^2+8a^2)B$
• D. Zero

Answer 1

Answer: (A)

$I(\frac{\pi a^2}{2}+8a^2)B$

$\overline{t}=\overline{M}\times \overline{B}$

$=I[8a^2+\frac{\pi a^2}{2}]Bsin{90}^0$

$=I(\frac{\pi a^2}{2}+8a^2)B$