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