• # 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) $3\,mg\sin \theta -qB\sqrt{2gR\sin \theta }$

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 }$

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