A magnetic needle suspended parallel to a magnetic field requires 3 ~J of work to turn it through 60^ . The torque needed to maintain the needle in this position will be

Correct answer: B. 3 ~J In this case, work done aligned W & =M B ( θ_1- θ_2) & =M B ( 0^ ^ - 60^ ^ ) & =M B (1-(1)/(2))=(M B)/(2) M B & =2 3 ~J ( given W= 3 ~J) & =M B 60^ ^ =(2 3 ) ( 3 2 )J=3 ~J aligned

NEET 2012

Exam: NEET
Year: 2012
Subject: Physics
Chapter: Magnetic Effects of Current

NEET 2012 · Physics · Magnetic Effects of Current

A magnetic needle suspended parallel to a magnetic field requires 3...

A magnetic needle suspended parallel to a magnetic field requires

\sqrt{3} \mathrm{~J}

of work to turn it through

60^{\circ}

. The torque needed to maintain the needle in this position will be

Options

A $2 \sqrt{3} \mathrm{~J}$
B $3 \mathrm{~J}$
C $\sqrt{3} \mathrm{~J}$
D $\frac{3}{2} \mathrm{~J}$

Correct answer

B. $3 \mathrm{~J}$

Step-by-step solution

In this case, work done

\begin{aligned}W & =M B \left(\cos \theta_1-\cos \theta_2\right) \\ & =M B \left(\cos0^{{}^{\circ}}-\cos 60^{{}^{\circ}}\right) \\ & =M B \left(1-\frac{1}{2}\right)=\frac{M B}{2}\\ M B & =2 \sqrt{3} \mathrm{~J} \quad(\because \text{ given }W=\sqrt{3} \mathrm{~J})\\ \tau & =M B \sin 60^{{}^{\circ}}=(2 \sqrt{3}) \left(\frac{\sqrt{3}}{2}\right)\mathrm{J}=3 \mathrm{~J} \end{aligned}

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