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
- A \(2 \sqrt{3} w\)
- B \(\sqrt{3} w\)
- C \(\frac{\sqrt{3}}{2} w\)
- D \(\frac{\sqrt{3}}{4} \mathrm{W}\)
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, \[ \begin{array}{l} W=M B(1-\cos \theta) \\ W=M B\left(1-\cos 60^{\circ}\right) \\ W=\frac{M B}{2} \\ |\tau|=M B \sin 60^{\circ}=\sqrt{3} W \end{array} \]
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