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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)
${\sigma _1} = \frac{{ - { \in _0}\,vB}}{2}\,,\,{\sigma _2} = \frac{{{ \in _0}\,vB}}{2}$
${\sigma _1} = { \in _0}\,vB\,,\,{\sigma _2} = - { \in _0}\,vB$
${\sigma _1} = \frac{{{ \in _0}\,vB}}{2}\,,\,{\sigma _2} = \frac{{ - { \in _0}\,vB}}{2}$
${\sigma _1} = {\sigma _2} = { \in _0}\,vB$
Solution
$\because F=q E$ and $F=q v B$
$\therefore \mathrm{E}=\mathrm{vB}$
And Gauss's law in Electrostatics $\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$
$\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}=\mathrm{vB} \Rightarrow \sigma=\varepsilon_{0} \mathrm{vB}$
$\sigma_{1}=-\sigma_{2}$
