How much energy is stored in the spring when the rod is in equilibrium?

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Solution

In equilibrium,
Clockwise torque of magnetic force $=$ anticlockwise torque of spring force
$∴ 0.0442 = (k x) (l s i n 53^{o}) = (4.8) (x) (0.2) (\frac{4}{5})$
or $x = 0.057 m$
$U = \frac{1}{2} k x^{2} = \frac{1}{2} \times (4.8) {(0.057)}^{2}$
$= 7.8 \times 10^{- 3} J$

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