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Electromagnetic Induction question

2016 · 10 Apr · Shift 1 · Q65
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Electromagnetic Induction question

2016 · 10 Apr · Shift 1 · Q65

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
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\sigmaσ 1 and σ\sigmaσ 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 106 English
  1. A
    σ\sigmaσ 1 = ∈\in∈ 0 υ\upsilonυ B, σ\sigmaσ 2 = −∈- \in−∈ 0 υ\upsilonυ B
  2. B
    σ\sigmaσ 1 = ∈0υ B2,σ{{{ \in _0}\upsilon \,B} \over 2},\sigma2∈0​υB​,σ 2 = −∈0 υB2{{ - { \in _0}\,\upsilon B} \over 2}2−∈0​υB​
  3. C
    σ\sigmaσ 1 = σ\sigmaσ 2 = ∈0 υB{ \in _0}\,\upsilon B∈0​υB
  4. D
    σ\sigmaσ 1 = −∈0υB2,σ{{ - { \in _0}\upsilon B} \over 2},\sigma2−∈0​υB​,σ 2 = ∈0υB2,{{ { \in _0}\upsilon B} \over 2},2∈0​υB​,
View written solutionFree

Correct answer: A

  1. Physical situation

A conducting metallic sheet moves with speed vvv in a uniform magnetic field B⃗\vec BB directed into the plane of the paper.

Free charges inside the conductor experience the magnetic Lorentz force:

F⃗=q(v⃗×B⃗)\vec F = q(\vec v \times \vec B)F=q(v×B)

This causes charge separation until an internal electric field E⃗\vec EE is established.

  1. Direction of force on positive charges

Take the sheet to be moving upward as shown in the standard figure. With:

  • v⃗\vec vv upward,
  • B⃗\vec BB into the page,

we get

v⃗×B⃗\vec v \times \vec Bv×B

pointing toward the left.

So positive charges accumulate on the left face, and negative charges accumulate on the right face.

Hence:

  • left surface charge density σ1>0\sigma_1 > 0σ1​>0,
  • right surface charge density σ2<0\sigma_2 < 0σ2​<0.
  1. Equilibrium condition

In steady state, electric force balances magnetic force:

qE=qvBqE = qvBqE=qvB

so

E=vBE = vBE=vB

This electric field is directed from the positive surface to the negative surface, i.e. from left to right.

  1. Electric field due to two oppositely charged large conducting faces

Ignoring fringe effects, the two faces act like two infinite sheets with surface charge densities +σ+\sigma+σ and −σ-\sigma−σ.

Field due to one infinite sheet is:

Esheet=σ2ε0E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}Esheet​=2ε0​σ​

Between the two sheets, these fields add, so total field inside is:

E=σ2ε0+σ2ε0=σε0E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}E=2ε0​σ​+2ε0​σ​=ε0​σ​

Set this equal to the equilibrium field vBvBvB:

σε0=vB\frac{\sigma}{\varepsilon_0} = vBε0​σ​=vB σ=ε0vB\sigma = \varepsilon_0 vBσ=ε0​vB

Therefore,

σ1=+ε0vB,σ2=−ε0vB\sigma_1 = +\varepsilon_0 vB, \qquad \sigma_2 = -\varepsilon_0 vBσ1​=+ε0​vB,σ2​=−ε0​vB
  1. Option check
  • A: σ1=ε0vB,  σ2=−ε0vB\sigma_1 = \varepsilon_0 vB, \; \sigma_2 = -\varepsilon_0 vBσ1​=ε0​vB,σ2​=−ε0​vB ✅
  • B: half values ❌
  • C: both same sign ❌
  • D: opposite signs but half magnitude ❌
  1. Final answer

The correct option is:

A\boxed{A}A​
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