Q. 284.0( 3 Votes )

# A conducting wire of length ℓ, lying normal to a magnetic field B, moves with velocity v as shown in 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 the 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?

Answer :

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 take this as the motion of free electrons present inside the wire with velocity *v.*

(a) The average magnetic force on a free electron of the wire

We know, Lorentz force F is given by -

where,

e = charge on an electron

v = velocity of the electron

B=magnetic field

(b)The redistribution of electrons stops when the electric force is just balanced by the magnetic force.

Electric force coulomb’s law,

where

e = charge

E=electric field of the charge

and also magnetic force, we know, Lorentz force F is given by -

where,

e = charge on an electron

v = velocity of the electron

B=magnetic field

On equating these two forces, we get-

(c) The potential difference developed between the ends of the wire , V is -

where l is length of wire and E is applied electric field

From (1)

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