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

The torque \( \tau \) on a current-carrying coil in a magnetic field is given by \(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\). The parameters are:

• Number of turns, \(n=50\).
• Current, \(I=2 \, \text{A}\).
• Coil area, \(A = \pi r^2\) with \(r=0.1 \, \text{m}\).
• Magnetic field strength, \(B=0.5 \, \text{T}\).
• Angle \(\theta\) between the coil's plane and the magnetic field. With the coil perpendicular to the field, \(\theta=90^\circ\), so \(\sin(90^\circ)=1\).

The coil's area is first computed:

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

Next, these values are substituted 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, the numerical value is calculated:

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

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