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WBJEE · Physics · Magnetic Effects of Current

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of \(60^{\circ}\) and the work done is \(W\). The torque on the magnetic needle at this position is

  1. A \(2 \sqrt{3} w\)
  2. B \(\sqrt{3} w\)
  3. C \(\frac{\sqrt{3}}{2} w\)
  4. D \(\frac{\sqrt{3}}{4} \mathrm{W}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{3} w\)

Step-by-step Solution

Detailed explanation

Given, work done \(=W\) and \(\theta=60^{\circ}\) We know that Hence, \[ \begin{array}{l} W=M B(1-\cos \theta) \\ W=M B\left(1-\cos 60^{\circ}\right) \\ W=\frac{M B}{2} \\ |\tau|=M B \sin 60^{\circ}=\sqrt{3} W \end{array} \]