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) \[mg\,\sin \theta +qB\sqrt{2g\,R\,\sin \,\theta }\]

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

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

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

Correct Answer: B

Solution :

[b] \[v=\sqrt{2gR\,\sin \theta }\] and \[F=qvB\] Now \[N-(F+mg\,\sin \theta )=\frac{m{{v}^{2}}}{R}\] \[\therefore \,\,N=F+mg\,\sin \,\theta +\frac{m{{v}^{2}}}{R}\] \[=qB\,\,\sqrt{2gR\,\sin \,\theta }+mg\sin \theta +2mg\sin \theta \] \[\therefore \,\,\,N=3mg\sin \theta +qB\sqrt{2gR\sin \theta }\]