Question

A circular coil of 50 turns, each of radius 0.1 m, carries a current of 2 A. If the coil is placed in a uniform magnetic field of 0.5 T perpendicular to its plane, what is the magnitude of the torque acting on the coil?

• A. \( 0.157 \, \text{Nm} \)
• B. \( 0.785 \, \text{Nm} \)
• C. \( 1.57 \, \text{Nm} \)
• D. \( 3.14 \, \text{Nm} \)

Hint:

For torque on a current-carrying coil, ensure the angle \( \theta \) is between the magnetic field and the normal to the coil’s plane. When the field is perpendicular to the coil’s plane, \( \sin\theta = 1 \), maximizing the torque.

Answer 1

The Correct Option is B

To find the torque acting on the coil, we use the formula for the torque \( \tau \) on a current-carrying loop in a magnetic field: \(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\). Here,

• \(n\) is the number of turns (\(n=50\)).
• \(I\) is the current (\(I=2 \, \text{A}\)).
• \(A\) is the area of the coil (\(A = \pi r^2\) with radius \(r=0.1 \, \text{m}\)).
• \(B\) is the magnetic field (\(B=0.5 \, \text{T}\)).
• \(\theta\) is the angle between the plane of the coil and the magnetic field. Since the coil is perpendicular to the field, \(\theta=90^\circ\), and \(\sin(90^\circ)=1\).

First, calculate the area \(A\):

\(A = \pi (0.1)^2 = 0.01\pi \, \text{m}^2\).

Now substitute the values into the torque formula:

\(\tau = 50 \cdot 2 \cdot 0.01\pi \cdot 0.5 \cdot 1 = 0.5\pi \, \text{N}\cdot\text{m}\).

Finally, calculate the numeric value:

\(\tau = 0.5 \cdot 3.14 = 1.57/2 = 0.785 \, \text{N}\cdot\text{m}\).

The magnitude of the torque acting on the coil is 0.785 N·m.