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 1 and σ 2 are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects): (a) σ 1 = ε 0 v B, σ 2 = -ε 0 v B (b) σ 1 = (ε₀ v B)/(2), σ 2 = (-ε₀ vB)/(2) (c) σ 1 = σ 2 = ε₀ v B (d) σ 1 = (-ε₀ vB)/(2), σ 2 = (ε₀ vB)/(2),

Answer: (a). Force on + charge: qv × B with v up and B into the page points to the left. So the left surface becomes positive and the right surface negative. Equilibrium inside: qE = qvB ⇒ E = vB. Field between two opposite charged surfaces: E = (σ)/(ε₀) ⇒ σ = ε₀ vB So the answer is (a) σ₁ = ε₀ vB, σ₂ = -ε₀ vB.

Consider a thin metallic sheet perpendicular to the plane of the… | JEE Main 2016 (10 Apr, Shift 1)

Q50JEE Main 2016 · 10 APR · SHIFT 1Difficult

Motional EMF

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):

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Answer: (a)

Force on + charge: $q\vec{v} \times \vec{B}$ with $\vec{v}$ up and $\vec{B}$ into the page points to the left.
So the left surface becomes positive and the right surface negative.
Equilibrium inside: $qE = qvB \Rightarrow E = vB$ .
Field between two opposite charged surfaces: $E = \dfrac{\sigma}{\epsilon_0} \Rightarrow \sigma = \epsilon_0 vB$
So the answer is (a) $\sigma_1 = \epsilon_0 vB$ , $\sigma_2 = -\epsilon_0 vB$ .

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Motional EMF question from JEE Main 2016 (10 Apr, Shift 1)

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