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 σ1 and σ2 are induced on the left and right surfaces, respectively, of the sheet then  (ignore fringe effects) :

  • Option 1)

    \sigma _{1}= \epsilon _{0}\nu \, B,\sigma _{2}= -\epsilon _{0}\nu B

  • Option 2)

    \sigma _{1}=\frac{ \epsilon _{0}\nu \, B}{2},\sigma _{2}=\frac{ -\epsilon _{0}\nu B}{2}

  • Option 3)

    \sigma _{1}= \sigma _{2}=\epsilon _{0}\nu B

  • Option 4)

    \sigma _{1}= \frac{-\epsilon _{0}\nu B}{2},\sigma _{2}=\frac{\epsilon _{0}\nu B}{2}

 

Answers (1)

As we learnt in

Force on a charged particle in magnetic field -

\underset{F}{\rightarrow}=q (\underset{V}{\rightarrow}\times \underset{B}{\rightarrow})

F=qVB\sin \theta

- wherein

\underset{V}{\rightarrow} - velocity of the particle 

\underset{B}{\rightarrow}  magnetic field 

 

 

Magnetic field If V(vector), E (vector) and B (vector) are mutually perpendicular -

Fe=Fm

V=\frac{E}{B}

-

 

 Magnetic force on electron in metal sheet

\vec F_v=-e\left | \vec v \times \vec B \right |

At equilibrium, Fm = Fe

\sigma _2=-\sigma_1

\Rightarrow e\:v\:B= e\frac{\sigma}{\epsilon_0}

\Rightarrow E_0\:v\:B=\sigma_1

\sigma_2=\epsilon_0\:v\:B

 


Option 1)

\sigma _{1}= \epsilon _{0}\nu \, B,\sigma _{2}= -\epsilon _{0}\nu B

This is correct option.

Option 2)

\sigma _{1}=\frac{ \epsilon _{0}\nu \, B}{2},\sigma _{2}=\frac{ -\epsilon _{0}\nu B}{2}

This is incorrect option.

Option 3)

\sigma _{1}= \sigma _{2}=\epsilon _{0}\nu B

This is incorrect option.

Option 4)

\sigma _{1}= \frac{-\epsilon _{0}\nu B}{2},\sigma _{2}=\frac{\epsilon _{0}\nu B}{2}

This is incorrect option.

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