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Question

(B) When a metallic sheet moves with velocity $v$ in a magnetic field $B$,the free electrons in the metal experience a magnetic Lorentz force $F_m = q

Vedclass · Accepted Answer

$\sigma_1 = \epsilon_0 vB, \sigma_2 = -\epsilon_0 vB$

Vedclass · Answer

(B) When a metallic sheet moves with velocity $v$ in a magnetic field $B$,the free electrons in the metal experience a magnetic Lorentz force $F_m = q(v 	imes B)$.According to the right-hand rule,for a positive charge moving upwards in a magnetic field directed into the page,the force is directed to the right. Thus,electrons are pushed to the left surface,making it negatively charged,and positive charges accumulate on the right surface.This charge separation creates an internal electric field $E$ directed from right to left.In steady state,the magnetic force is balanced by the electric force: $qE = qvB$,which gives $E = vB$.The electric field between two oppositely charged plates (surfaces of the sheet) is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the magnitude of the surface charge density.Equating the two expressions for $E$: $\frac{\sigma}{\epsilon_0} = vB$,so $\sigma = \epsilon_0 vB$.Since the left surface is negatively charged and the right surface is positively charged,we have $\sigma_1 = -\epsilon_0 vB$ and $\sigma_2 = \epsilon_0 vB$.

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