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A charged sphere of mass $m$ and charge $- q$ starts sliding along the surface of a smooth hemispherical bowl, at position $P$ . The region has a transverse uniform magnetic field $B$ . Normal force by the surface of bowl on the sphere at position $Q$ is

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$m g \sin θ - q B \sqrt{2 g R \sin θ}$

$m g \sin θ + q B \sqrt{2 g R \sin θ}$

$3 m g \sin θ + q B \sqrt{2 g R \sin θ}$

$3 m g \sin θ - q B \sqrt{2 g R \sin θ}$

answer is B.

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Detailed Solution

$v = \sqrt{2 g R \sin θ}$

and $F = q v B$

Now $N - (F + m g \sin θ) = \frac{m v^{2}}{R} N = (F + m g \sin θ) + \frac{m v^{2}}{R} = q B \sqrt{2 g R \sin θ} + m g \sin θ + 2 m g \sin θ N = 3 m g \sin θ + q B \sqrt{2 g R \sin θ}$

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