Question

A circular loop carrying a current' I' is made of a wire of length 'L'. A uniform magnetic field 'B' exists parallel to the plane of the loop. The torque experienced by the circular loop will be

• A. \(ILB\)
• B. Zero
• C. \(\frac{IL^{2}B}{2\pi }\)
• D. \(\frac{IL^{2}B}{4\pi }\)

Hint:

Find the radius from the wire length, then use torque equals magnetic moment times field with the field in the plane of the loop.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The wire forms the circumference: \(2\pi r=L\), so \(r=\frac{L}{2\pi}\). The area is \(A=\pi r^{2}=\frac{L^{2}}{4\pi}\).

Step 2: Magnetic moment
\(m=IA=\frac{IL^{2}}{4\pi}\). The field lies in the plane of the loop, so it is perpendicular to the normal of the loop.

Step 3: Torque
\(\tau=mB\sin90^{\circ}=\frac{IL^{2}B}{4\pi}\).

Final Answer:
The torque is I L squared B divided by 4 pi. \[ \boxed{\text{(D) }\dfrac{IL^{2}B}{4\pi}} \]