A small bar magnet placed with its axis at 30^{\circ} with the external field of 0.06T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:
Option: 1 6.4 \times 10^{-2}J
Option: 2 9.2 \times 10^{-3}J
Option: 3 7.2\times 10^{-2}J
Option: 4 11.2 \times 10^{-3}J

Answers (1)


Here, \theta=30^{\circ}, \mathrm{\tau }=0.018 \mathrm{~N}-\mathrm{m}, \mathrm{B}=0.06 \mathrm{~T}

And Torque on a bar magnet

\tau =M B \sin \theta\\ \Rightarrow 0.018=\mathrm{M} \times 0.06 \times \sin 30^{\circ}\\ \Rightarrow 0.018=\mathrm{M} \times 0.06 \times \frac{1}{2}\\ \Rightarrow \mathrm{M}=0.6 \mathrm{~A}-\mathrm{m}^{2}

Now using U=-M B \cos \theta
Position of stable equilibrium (\theta=0^0)
\mathrm{U}_{\mathrm{i}}=-\mathrm{MB}

Position of unstable equilibrium (\theta=180^0)
\mathrm{U}_{\mathrm{f}}=\mathrm{MB}

\begin{array}{l} \text { Using work done }: \Delta U \\ \Rightarrow W=2 M B \\ \Rightarrow W=2 \times 0.6 \times 0.06 \\ \Rightarrow W=7.2 \times 10^{-2} \mathrm{~J} \end{array}

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