question_answer

When a charged particle, travelling with unifrom speed enters a uniform magnetic field perpendicularly then. Its kinetic energy: 

A)  remains constant

B)  increases

C)  decreases

D)  becomes zero

Correct Answer: A

Solution :

 Key Idea: Force due to magnetic field provides the required centripetal force. The force due to magnetic field B is \[F=qvB\,\sin \theta \] When panicle with charge q enters perpendicularly then\[\theta ={{90}^{o}},\] so \[F=qvB\]and required centripetal force is \[F=\frac{m{{v}^{2}}}{r}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{m{{v}^{2}}}{r}=qvB\]              \[\Rightarrow \] \[v=\frac{qrB}{m}\]              Kinetic energy \[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{qrB}{m} \right)}^{2}}\]              \[=\frac{{{q}^{2}}{{r}^{2}}{{B}^{2}}}{2m}=\text{constant}\] Hence, kinetic energy remains constant.