question_answer

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed 'v' in a uniform magnetic field B going into the plane of the paper (see figure) . If charge densities \[{{\sigma }^{1}}\]and \[{{\sigma }^{2}}\] are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects)   JEE Main Online Paper (Held On 10 April 2016)

A)              \[{{\sigma }_{1}}=\frac{-{{\varepsilon }_{0}}vB}{2},\,{{\sigma }_{2}}=\frac{-2{{\varepsilon }_{0}}vB}{2}\]

B)              \[{{\sigma }_{1}}={{\sigma }_{2}}={{\varepsilon }_{0}}vB\]              

C) \[{{\sigma }_{1}}=\frac{{{\varepsilon }_{0}}vB}{2},\,{{\sigma }_{2}}=\frac{-{{\varepsilon }_{0}}vB}{2}\]

D)              \[{{\sigma }_{1}}={{\varepsilon }_{0}}vB,\,{{\sigma }_{2}}=-{{\varepsilon }_{0}}vB\]

Correct Answer: D

Solution :

                                      direction of \[\overrightarrow{V}\times \overrightarrow{B}\]is towards left therefore induced change dentition will be electric field \[=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}\] \[\frac{\sigma }{{{\varepsilon }_{0}}}\times W=(B)(V)\,(\omega )\]              \[\sigma =BV{{\varepsilon }_{0}}\]