A conducting wire of length l, lying normal to a magnetic field B, moves with a velocity v, as shown in the figure. (a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electrons concentrate at one end, resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?

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Solution

Given:
Length of the conducting wire = l
Inward magnetic field = B
Velocity of the conducting wire = v
As the wire is moving with velocity v, we can consider this as the net motion of electrons inside the wire with velocity v.
(a) The average magnetic force on a free electron of the wire
= e(v × B) = evB, where e is the charge of an electron.
(b) The redistribution stops when the electric force is just balanced by the magnetic force.
Electric force, F = eE and also magnetic force, F = evB
On equatinging the two forces, we get:
eE = evB
$\Rightarrow$ E = vB
(c) The potential difference is developed between the ends of the wire:
V = lE = lvB, where V is the potential difference across the ends of the wire.

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