# Is This a Case of Motional emf?

1. Apr 7, 2012

### MS La Moreaux

If a straight length of wire moves in a suitable direction through the magnetic field of a magnet, there will be a motional emf in the wire. If the magnet moves toward the wire, is there an emf in the wire, motional or otherwise?

2. Apr 8, 2012

### Philip Wood

Strictly, no. Whether the emf is classed as motional or not depends on your reference frame.

If your frame is one in which the magnet is stationary and the wire moves, you have a motional emf. It can be thought of as arising from the qvxB forces on the charge carriers in the conductor, as the conductor moves through space.

If you are in a frame in which the magnet moves and the wire is stationary, but part of a circuit, there will be an emf in that circuit according to the Faraday/Maxwell equation
$$\oint Edl = -\int\int \frac{\partial B}{\partial t}dA.$$ The line integral is evaluated around the closed loop of the circuit, and the surface integral over any surface bounded by the circuit. So the force on the charge carriers is qE, arising from the electric field produced by the changing magnetic field.

Faraday's description of the phenomenon – the emf arises when the magnetic flux linking a circuit changes – fits in either reference frame.

Last edited: Apr 9, 2012
3. Apr 8, 2012

### MS La Moreaux

Thanks. It is as I suspected. To be more precise I should have asked if there would be a potential difference between the ends of the wire, arising from a non-electrostatic emf induced in the wire. The TOTAL emf would be zero in that case, also.

4. Apr 29, 2012

### MS La Moreaux

It has been pointed out to me elsewhere that a moving magnet is accompanied by an electric field. Therefore, the wire, being in that field, will have an emf.

5. Apr 30, 2012

### Philip Wood

That's what I meant when I said (above) "So the force on the charge carriers is qE, arising from the electric field produced by the changing magnetic field."

6. Apr 30, 2012

### MS La Moreaux

Yes, but you specified that there had to be a circuit, and there is none.

7. Apr 30, 2012

### Philip Wood

I did indeed specify a circuit, as I was trying to deal with the case of an emf in a wire, as posed in your original question. But I agree that the Faraday/Maxwell equation that I quoted does not require the line integral to be taken around a conducting circuit. I'm sorry if I inadvertently misled.

8. Apr 30, 2012

### MS La Moreaux

No problem; I believe that it is all clear, now. Thanks.