Q137 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Physics

A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque required to maintain the needle in this position will be

  • Correct answer
    A
    \(\sqrt{3} W\)
  • B
    \(W\)
  • C
    \(\frac{\sqrt{3}}{2} W\)
  • D
    \(2 \mathrm{~W}\)

Solution

According to the question, work done required to rotate magnetic needle from \(\theta_1=0^{\circ}\) to \(\theta_2=60^{\circ}\). \(W=M B\left(\cos \theta_1-\cos \theta_2\right)\) \(\begin{aligned} & =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\ & =M B\left(1-\frac{1}{2}\right) \\ W & =\frac{M B}{2} \Rightarrow M B=2 W \quad \ldots (i) \\ \therefore \text {Torque, } \tau & =M B \sin 60^{\circ}=M B \frac{\sqrt{3}}{2} \\ & =2 W \times \frac{\sqrt{3}}{2} \quad \text { [from Eq. (i)] } \\ & =W \sqrt{3}=\sqrt{3} W \end{aligned}\)

Hence, the correct answer is option: 'A'.

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