Question

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?
Figure

Answer 1

Explanation:

The conducting wire’s length = l

Conducting wire’s velocity = v

Inward magnetic field = B

The wire moves with velocity v. So, this can be considered as the electrons net motion with velocity v inside the wire.

(a) On a wire’s free electron, the average magnetic force

= e(v $\times$ B) = evB, where e is the electron’s charge.

(b) When there is a balance of the magnetic force, the redistribution halts when the electric force is just composed.

Magnetic force, F = evB and electric force, F = eE  

When two forces are equated, we obtain:

eE = evB

⇒E = vB

(c) Between the wire ends, the potential difference is established: V = lE = lvB, where the potential difference across the wire ends and V is the potential difference.