A horizontal metal rod with a mass m = 10 g and length L = 0.2 m is suspended by two light wires of length $l = 1 c m$ each in a magnetic field induction B = 1 T which is directed vertically downwards (Figure). A capacitor of capacitance $C = 100 μ F$ charged to a voltage of 100 V is connected as shown. If the cosine of the maximum deflection angle of the rod from the initial position after the switch is closed is Z, then find the value of 10Z. Resistances of wire and rod are not taken into account. (Take $g = 10 m / s^{2}$ )

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answer is 8.

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Detailed Solution

$F = i L B$

$m v = \int F d t = \int (i d t) L B$

$m . \sqrt{2 g h} = Q L B$

$m \sqrt{2 g l (1 - \cos θ)} = Q L B$

$θ = 37 °$

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