Question

A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
the magnetic dipole moment of the coil, and
the magnitude of the counter torque that must be applied to prevent the coil from turning.

Hint:

For circular coils:

\( A = \pi r^2 \)
If radius has \( \sqrt{\pi} \), area often simplifies nicely
Torque = \( mB\sin\theta \)

Answer 1

Concept: Magnetic dipole moment of a coil: \[ m = NIA \] Torque on current loop: \[ \tau = mB \sin \theta \]
Step 1: Convert radius to metres. \[ r = \frac{10}{\sqrt{\pi}} \, \text{cm} = \frac{10}{\sqrt{\pi}} \times 10^{-2} \, \text{m} = \frac{0.1}{\sqrt{\pi}} \, \text{m} \]
Step 2: Area of circular coil. \[ A = \pi r^2 = \pi \left(\frac{0.1}{\sqrt{\pi}}\right)^2 \] \[ A = \pi \cdot \frac{0.01}{\pi} = 0.01 \, \text{m}^2 \]
Step 3: Magnetic dipole moment. \[ m = NIA = 100 \times 5.0 \times 0.01 \] \[ m = 5 \, \text{A·m}^2 \]
Step 4: Torque on coil. \[ \tau = mB \sin \theta \] \[ m = 5, \quad B = 2.0 \, \text{T}, \quad \theta = 30^\circ \] \[ \tau = 5 \times 2 \times \sin 30^\circ \] \[ \tau = 10 \times 0.5 = 5 \, \text{N·m} \] Final Answers:

[(i)] Magnetic dipole moment: \[ m = 5 \, \text{A·m}^2 \]
[(ii)] Counter torque required: \[ \tau = 5 \, \text{N·m} \]