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QJEEM16Ph9dda: JEE Main 2016 Physics - Class 12 JEEM16Ph9dda Solution: Consider a thin metallic sheet perpendicular to the plane of…

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 σ\sigma 1 and σ\sigma 2 are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects) :

JEE Main 2016 (Online) 10th April Morning Slot Physics - Electromagnetic Induction Question 105 English

Multiple Choice Options

  • A σ\sigma 1 = ∈\in 0 υ\upsilon B, σ\sigma 2 = −- ∈\in 0 υ\upsilon B
  • B σ\sigma 1 = ∈0υ B2,{{{ \in _0}\upsilon \,B} \over 2}, σ\sigma 2 = −∈0 υB2{{ - { \in _0}\,\upsilon B} \over 2}
  • C σ\sigma 1 = σ\sigma 2 = ∈0 υB{ \in _0}\,\upsilon B
  • D σ\sigma 1 = −∈0υB2,{{ - { \in _0}\upsilon B} \over 2}, σ\sigma 2 = ∈0υB2,{{ { \in _0}\upsilon B} \over 2},

Step-by-Step Verified Solution

Magnetic force on electron

F→\overrightarrow F = −- e (V→×B→)\left( {\overrightarrow V \times \overrightarrow B } \right)

F = eVB   [As  V→{\overrightarrow V } and B→{\overrightarrow B } are perpendicular]

Also,   F = e E

and  E = σε0{\sigma \over {{\varepsilon _0}}}

∴\therefore    eVB = e ×\times σε0{\sigma \over {{\varepsilon _0}}}

⇒\Rightarrow     σ\sigma = ε0{{\varepsilon _0}} VB = σ\sigma 1

as    σ\sigma 1 = −- σ\sigma 2

∴\therefore     σ\sigma 2 = −- ε0{{\varepsilon _0}} VB

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