First slide

Magnetism

Question

A small bar magnet placed with its axis at $30 °$ with an external field of 0.06 T experiences a torque of $0.018 N m$ . The minimum work required to rotate it from its stable to unstable equilibrium position is :

Moderate

A

$11.7 \times 10^{- 3} J$
B

$7.2 \times 10^{- 2} J$
C

$9.2 \times 10^{- 3} J$
D

$6.4 \times 10^{- 2} J$

Solution

$\begin{array}{l} T o r q u e, τ = M B Sin 30^{\circ} = 0.018 \\ \Rightarrow M B = 0.036 \\ W o r k d o n e i s c h a n g e i n p o t e n t i a l e n e r g y . \\ \begin{array}{l} W = M B [\cos 0^{\circ} - \cos π] = 2 M B \\ = 2 (0.036) = 0.072 J \end{array} \end{array}$

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