question_answer

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

A) \[mg\sin \theta +qB\sqrt{2gR\sin \theta }\]

B) \[3\,mg\sin \theta +qB\sqrt{2gR\sin \theta }\]

C) \[\,mg\sin \theta -qB\sqrt{2gR\sin \theta }\]

D) \[3\,mg\sin \theta -qB\sqrt{2gR\sin \theta }\]

Correct Answer: B

Solution :

\[v=\sqrt{2gR\sin \theta }\]

and                   \[F=qvB\]

Now \[N-(F+\operatorname{mg}\,sin\theta )=\frac{m{{v}^{2}}}{R}\]

\[\therefore \]\[N=F+mg\sin \theta +\frac{m{{v}^{2}}}{R}\]

\[a=qB\sqrt{2gR\sin \theta }+mg\sin \theta +2mg\sin \theta \]

\[\therefore \]\[N=3mg\sin \theta +qB\sqrt{2gR\sin \theta }\]