Question

A circular coil of $30$ turns and radius $8.0 \,cm$ carrying a current of $6.0\, A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \,T$. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

• A. 4 Nm
• B. 6 Nm
• C. 3.1 Nm
• D. 2.8 Nm

Answer 1

Answer: (C)

Magnitude of torque acting on the current-carrying coil due to the magnetic field,

$ \tau = NIAB \sin \theta $
$=30 \times 6 \times \pi(0.08)^{2} \times 1 \times \sin 60^{\circ} $
$=30 \times 6 \times 3.14 \times 0.08 \times 0.08 \times \frac{\sqrt{3}}{2} $
$=3.133 \,N $