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

Magnetic Dipole Moment and Counter Torque of a Circular Coil
We are given:
- Number of turns: N = 100
- Radius of the coil: r = 10/√π cm = 0.1/√π m
- Current: I = 5.0 A
- Magnetic field: B = 2.0 T
- Angle between the normal to the coil and magnetic field: θ = 30°
We are asked to calculate the magnetic dipole moment and the counter torque.

Step 1: Magnetic Dipole Moment of the Coil
The magnetic dipole moment of a coil is given by:
m = N I A
where A is the area of one turn:
A = π r²
Substitute r = 0.1/√π m:
A = π (0.1/√π)² = 0.01 m²
Therefore, the magnetic dipole moment is:
m = 100 × 5 × 0.01 = 5 A·m²

Step 2: Torque on the Coil
The torque on the coil is given by:
τ = m B sin θ
Substitute the values:
τ = 5 × 2 × sin 30° = 10 × 0.5 = 5 N·m
The counter torque required to prevent rotation is equal in magnitude:
τ_counter = 5 N·m

Step 3: Summary
- Magnetic dipole moment: m = 5 A·m²
- Counter torque to prevent rotation: τ_counter = 5 N·m
The coil experiences a torque that tends to align its magnetic moment with the magnetic field. A counter torque of 5 N·m must be applied to keep it stationary.