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 \[=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}\] \[\frac{\sigma }{{{\varepsilon }_{0}}}\times W=(B)(V)\,(\omega )\] \[\sigma =BV{{\varepsilon }_{0}}\]You need to login to perform this action.
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