JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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)

- A \({\sigma _1} = \frac{{ - { \in _0}\,vB}}{2}\,,\,{\sigma _2} = \frac{{{ \in _0}\,vB}}{2}\)
- B \({\sigma _1} = { \in _0}\,vB\,,\,{\sigma _2} = - { \in _0}\,vB\)
- C \({\sigma _1} = \frac{{{ \in _0}\,vB}}{2}\,,\,{\sigma _2} = \frac{{ - { \in _0}\,vB}}{2}\)
- D \({\sigma _1} = {\sigma _2} = { \in _0}\,vB\)
Answer & Solution
Correct Answer
(B) \({\sigma _1} = { \in _0}\,vB\,,\,{\sigma _2} = - { \in _0}\,vB\)
Step-by-step Solution
Detailed explanation
\(\because F=q E\) and \(F=q v B\) \(\therefore \mathrm{E}=\mathrm{vB}\) And Gauss's law in Electrostatics \(\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}\) \(\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}=\mathrm{vB} \Rightarrow \sigma=\varepsilon_{0} \mathrm{vB}\)…
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