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Difficulty: MediumElectromagnetic Induction

When a straight metallic conductor of length LL moves at a constant velocity vv perpendicular to a uniform magnetic field BB, free electrons inside the conductor accumulate at one end, creating an internal electric field that eventually balances the magnetic force acting on them.

Answer: Answer

Answer

True. Moving a conductor through a magnetic field exerts a Lorentz magnetic force on its free electrons, pushing them toward one end. This separation of charge produces an internal electric field that exerts an opposing electrostatic force, reaching equilibrium when qE=qvBqE = qvB and resulting in an induced motional e.m.f. of E=BLv\mathcal{E} = BLv.
The statement is correct because free charge carriers in a conductor moving through a magnetic field experience a magnetic Lorentz force. This force drives electrons to one side of the conductor, leaving positive ions on the other. The resulting charge separation builds an internal electric field EE until the electric force qEqE equals the magnetic force qvBqvB, creating a stable motional e.m.f. E=BLv\mathcal{E} = BLv.

Step-by-Step Solution

1
Analyze the force acting on free electrons due to motion in a magnetic field.
Each free electron carrying charge qq experiences a magnetic force of magnitude Fm=qvBF_m = qvB directed along the length of the conductor.
According to the Lorentz force law, a charge moving with velocity vv perpendicular to a magnetic field BB experiences a magnetic force perpendicular to both motion and magnetic field.
2
Determine the consequence of electron movement within the conductor.
Electrons accumulate at one end, making that end negatively charged and leaving the opposite end positively charged.
The conductor has finite boundaries, so mobile charge carriers migrate until stopped by the physical ends of the rod.
3
Evaluate the electric field and equilibrium condition established by charge separation.
An electric field EE is formed pointing from the positive end to the negative end, creating an opposing electrostatic force Fe=qEF_e = qE. Accumulation stops when Fe=FmF_e = F_m, leading to E=vBE = vB and motional e.m.f. E=EL=BLv\mathcal{E} = EL = BLv.
Steady-state motional e.m.f. requires electrostatic equilibrium between the magnetic force driving charges apart and the electric force pulling them back.

Key Concept

Motional Electromotive Force and Microscopic Charge Separation
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Topics

Electromagnetic Induction

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