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)

Question

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) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-liaracyadhs4.png" /> (A) σ 1=(- Ïµ0 vB/2), σ 2=(Ïµ0 vB/2) (B) σ 1=ϵ0vB, σ 2=-ϵ0vB (C) σ 1=(Ïµ0 vB/2), σ 2=(- Ïµ0 vB/2) (D) σ 1=σ 2=ϵ0vB

Tardigrade Admin · Accepted Answer

σ 1=-σ 2 [due to magnetic force] E=(σ /ϵ 0)=vB σ =ϵ 0vB

$σ_{1} = \frac{- ϵ _{0} v B}{2}, σ_{2} = \frac{ϵ _{0} v B}{2}$

$σ_{1} = ϵ_{0} v B, σ_{2} = - ϵ_{0} v B$

$σ_{1} = \frac{ϵ _{0} v B}{2}, σ_{2} = \frac{- ϵ _{0} v B}{2}$

$σ_{1} = σ_{2} = ϵ_{0} v B$

$σ_{1} = - σ_{2}$ [due to magnetic force]
$E = \frac{σ}{ϵ _{0}} = v B$
$σ = ϵ_{0} v B$

σ1​=2−ϵ0​vB​,σ2​=2ϵ0​vB​

σ1​=ϵ0​vB,σ2​=−ϵ0​vB

σ1​=2ϵ0​vB​,σ2​=2−ϵ0​vB​

σ1​=σ2​=ϵ0​vB

σ1​=−σ2​ [due to magnetic force] E=ϵ0​σ​=vB σ=ϵ0​vB

$σ_{1} = \frac{- ϵ _{0} v B}{2}, σ_{2} = \frac{ϵ _{0} v B}{2}$

$σ_{1} = ϵ_{0} v B, σ_{2} = - ϵ_{0} v B$

$σ_{1} = \frac{ϵ _{0} v B}{2}, σ_{2} = \frac{- ϵ _{0} v B}{2}$

$σ_{1} = σ_{2} = ϵ_{0} v B$

$σ_{1} = - σ_{2}$ [due to magnetic force]
$E = \frac{σ}{ϵ _{0}} = v B$
$σ = ϵ_{0} v B$