Question

A wire of length '(L)' carries a current '(I)'. If the wire is turned into a square coil of single turn, the maximum magnitude of the torque in a given magnetic field ((\vec{B})) is

• A. (\frac{IBL}{16})
• B. (\frac{IBL}{8})
• C. (\frac{IBL^2}{8})
• D. [suspicious link removed]

Hint:

For a fixed length, a circular coil provides the maximum area and thus the maximum torque compared to a square coil.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A straight wire is reshaped into a square coil. We need to find the maximum torque acting on it in a magnetic field.
Step 2: Key Formula or Approach:
Maximum torque on a current-carrying loop: \(\tau_{\text{max}} = NIAB\).
For a single turn, \(N = 1\).
Step 3: Detailed Explanation:
Let the wire of length \(L\) form a square of side \(s\).
Then, the perimeter is \(4s = L \implies s = \frac{L}{4}\).
The area \(A\) of the square is:
\[ A = s^2 = \left( \frac{L}{4} \right)^2 = \frac{L^2}{16} \]
The maximum torque is:
\[ \tau_{\text{max}} = I \times A \times B \]
\[ \tau_{\text{max}} = I \times \frac{L^2}{16} \times B = \frac{IBL^2}{16} \]
Step 4: Final Answer:
The maximum magnitude of the torque is \(\frac{IBL^2}{16}\).