Question

A circular coil of 30 turns and radius 8·0 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of 1·0 T. The field lines make an angle of 30\(^{\circ}\) with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged?

Hint:

Always read carefully whether the given angle is with respect to the normal to the plane (\(\theta\)) or with the plane of the coil itself (\(\alpha\)). If it is with the plane, use \(\tau = N I A B \cos\alpha\) or translate it via \(\theta = 90^{\circ} - \alpha\). Also, torque is shape-independent as long as the boundary lies in a single flat plane and encloses the same area.

Answer 1

Concept: The magnetic torque \(\tau\) experienced by a current-carrying planar coil placed inside a uniform magnetic field is governed by the expression: \[ \tau = N I A B \sin\theta \] Where:
• \(N\) is the total number of turns in the coil.
• \(I\) is the electric current flowing through the loop.
• \(A\) is the cross-sectional area enclosed by the planar loop.
• \(B\) is the magnitude of the uniform external magnetic field.
• \(\theta\) is the angle between the magnetic field vector \(\vec{B}\) and the area vector \(\vec{A}\) (normal to the plane of the coil). If \(\alpha\) is the angle that the magnetic field lines make directly with the plane of the coil, then the angle \(\theta\) with respect to the normal is given by the relation: \[ \theta = 90^{\circ} - \alpha \]

Step 1: Extracting given data and converting units.
From the problem description, we have: Number of turns, N &= 30
Radius of the circular coil, r &= 8.0 cm = 8.0 \times 10^{-2} m = 0.08 m
Current in the coil, I &= 6 A
Magnetic field strength, B &= 1.0 T
Angle with the plane of the coil, \alpha &= 30^{\circ}

Step 2: Finding the area of the circular coil and the correct operational angle.
The formula for the area \(A\) of a circular cross-section is: \[ A = \pi r^2 \] Substituting the value of the radius: \[ A = \pi \times (8.0 \times 10^{-2})^2 = \pi \times 64.0 \times 10^{-4} \text{ m}^2 = 64\pi \times 10^{-4} \text{ m}^2 \] Now, calculating the angle \(\theta\) between the magnetic field and the normal to the plane of the coil: \[ \theta = 90^{\circ} - 30^{\circ} = 60^{\circ} \]

Step 3: Calculating the magnitude of the restoring/external torque.
To hold the coil stationary and prevent it from turning, the applied external torque must precisely equal the magnitude of the magnetic torque generated within the system: \[ \tau = N I A B \sin\theta \] Substituting all our numeric variables into the expression: \[ \tau = 30 \times 6 \times (64\pi \times 10^{-4}) \times 1.0 \times \sin(60^{\circ}) \] We know that \(\sin(60^{\circ}) = \frac{\sqrt{3}}{2}\) and taking \(\pi \approx 3.1416\) and \(\sqrt{3} \approx 1.732\): \[ \tau = 180 \times 64 \times 3.1416 \times 10^{-4} \times 1.0 \times \frac{1.732}{2} \] \[ \tau = 90 \times 64 \times 3.1416 \times 1.732 \times 10^{-4} \] \[ \tau = 5760 \times 5.4412 \times 10^{-4} \] \[ \tau = 3.134 \text{ N}\cdot\text{m} \approx 3.13 \text{ N}\cdot\text{m} \]

Step 4: Analysis of replacing the shape with an irregular planar coil.
The expression for the magnetic torque \(\tau = N I A B \sin\theta\) contains only the area parameter \(A\), and is entirely independent of the geometric shape (circular, rectangular, hexagonal, or irregular) of the planar boundary. Since the problem explicitly states that the replacement irregular planar coil encloses the exact same area \(A\) and keeps all other parameter constants (\(N, I, B, \theta\)) completely unchanged, the magnitude of the torque experienced will remain completely unaltered.