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MHT CET · Physics · Magnetic Effects of Current

A wire of length \(L\) carries a current \(i\). If the wire is turned into a circular coil and kept in a uniform magnetic field \(B\), the maximum magnitude of torque in the given magnetic field will be

  1. A \(\frac{B L^2}{4 \pi}\)
  2. B \(\frac{B i L^2}{4 \pi}\)
  3. C \(\frac{B^2 L^2}{2}\)
  4. D \(\frac{B i L^2}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{B i L^2}{4 \pi}\)

Step-by-step Solution

Detailed explanation

The torque on the current loop is \(\tau=N I A B \sin \alpha---(1)\) If there are \(N\) turns of the circular coil, each of radius \(r\), then \(L=2 \pi r N\)
\(\Rightarrow r=\frac{L}{2 \pi N}\)
Area of the coil, \(A=\pi r^2=\pi\left(\frac{L}{2 \pi N}\right)^2=\frac{L^2}{4 \pi N^2}\) Putting this value in (1), we get,
\(\tau=N I\left(\frac{L^2}{4 \pi N^2}\right) B \sin \alpha=\frac{L^2 I B \sin \alpha}{4 \pi N}--(2)\)
From (2), it is clear that \(\tau\) is maximum if \(N\) is minimum. Torque is maximum if \(\sin \alpha=1\) and \(N=1\)
\(\therefore \tau_{\max }=L^2 I B / 4 \pi\)
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