# What is the E-field through the rod?

1. Jul 27, 2011

### aximwolf

1. The problem statement, all variables and given/known data

What is the electric field within the rod? (See attached document for full question)

2. Relevant equations

emf = -d(phi)/dt= integral (E dot dl)

3. The attempt at a solution

The rod is moving within a uniform magnetic field. There is no change in flux through the rod and thus no current is induced because the law of induction says that a change in flux induces an electric current. Without a current no E-field is generated thus the E-field must be zero.

Is this correct?

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2. Jul 28, 2011

### PeterO

Not quite:

when you say "There is no change in flux through the rod and thus no current is induced because the law of induction says that a change in flux induces an electric current" you were referring to the law/rule that applies to loops/solenoids not simply wires

If a positive charge was to move through the magnetic field, in the direction the rod is traveling [up the page in the diagram] it would experience a force to the right.

As such, positive charges in the rod will experience a force towards the right of the rod, and negative charges will experience a force to the left. At first, any charge free to move will move in the appropriate direction. [in actuality only the negative charges (electrons) are likely to have a net flow]
After a short while [very short!!!!] the right hand end of the rod will have a net positive charge, while the left hand end will be negative.
As such there will be an electric field to the left.
From this time on, the electric field will try to drive electric charge in one direction, while the electromagnetic effect will try to drive them in the other direction. Equilibrium will have been reached and there is no further electric current.

Peter

ps: if the electric field is actually in the opposite direction I wouldn't be surprised!!! [as I often get these explanation back to front!!] If only they had defined positive and negative the other way round so that it was the positive bits that move in a wire!!!

3. Jul 28, 2011

### ehild

It does not matter if positive or negative charges move in the rod.
Peter is right, you can take that both the positive ions and the free electrons move together with the road with speed v upward. The force acting in a magnetic field on a charged particle moving with velocity v is Fm=q vxB. (q is the charge and x means vector product.) Applying right hand rule, you can find the direction of the force: it points on the right in case of positive particles and on the left in case of the negative one. In any case, the right-hand side of the rode will became positive and the left-hand side negative. The magnetic force and the electric force are equal in magnitude and opposite in direction: qvxB-qE =0 so E=vxB, the direction of E does not depend on the sign of charge carriers.

ehild

Last edited: Jul 28, 2011