EMF between an axis and the surface of long wire

[PumpkinCougar95]

A friend mine of gave me a problem :

Calculate the Emf b/w the axis and surface of a long current carrying wire of radius $r$ and current density $J$.

I am not able to understand why there would be a potential difference between the axis and surface but i think that either of these could happen:

1) Due to the motion of electrons in the wire and the magnetic field an emf is created because of motional emf

using this I get : $B =\frac {\mu i Vd }{4 \pi}$ Vd is the drift velocity

But I guess this should be wrong as electrons aren't really moving with $Vd$

2) Maybe there are some surface charges on the wire?(because of which a potential diff. will be created without any field inside )

I don't know the reason for this but in many problems, i have seen surface charges on wires so I thought that there may be some in this case too.

I am Really confused and don't even know whether the problem is correct or not. Any help would be greatly appreciated.

 

[TSny]

Consider a point p in the wire at a distance r from the axis. Is there a magnetic field B at point p? If so, can you describe the direction of the B field at p?

 

[PumpkinCougar95]

Yes , there is, but what does that have to do with anything ?
in my case 1) I considered that only to find the emf. You have something else in mind?

 

[TSny]

Yes , there is, but what does that have to do with anything ?
in my case 1) I considered that only to find the emf. You have something else in mind?

What effect would the B field have on the electrons moving in the wire?

 

[PumpkinCougar95]

The B field would try to push the electrons inwards , but since they are moving straight in a wire , an opposite E field must exist. Thereby creating a potential difference.

BUT in order to calculate the emf, we would need the velocity of the electrons. I don't think it would be okay to use the drift velocity as the electrons aren't really moving with that velocity. Am I wrong ?

1) Due to the motion of electrons in the wire and the magnetic field an emf is created because of motional emf

using this I get : B=μiVd4πB=μiVd4π B =\frac {\mu i Vd }{4 \pi} Vd is the drift velocity

Besides I already did the calculation for the case you are referring to. ^^

 

[TSny]

The B field would try to push the electrons inwards , but since they are moving straight in a wire , an opposite E field must exist. Thereby creating a potential difference.

Yes. I think that's the idea.

BUT in order to calculate the emf, we would need the velocity of the electrons. I don't think it would be okay to use the drift velocity as the electrons aren't really moving with that velocity. Am I wrong ?

Yes, use the drift velocity. It represents the average velocity of the electrons, which should be what you need.

Besides I already did the calculation for the case you are referring to. ^^

What answer did you get for the potential difference?

 

[PumpkinCougar95]

I get $Emf = \frac { \mu I (Vd) } {4\pi}$ . Sorry I think in my first post I mistakenly used the symbol B instead.

 

[TSny]

I get $Emf = \frac { \mu I (Vd) } {4\pi}$

I don't agree with your result. Please show your work.

Edit: Actually I do get your result! I had written my answer in a different form (in terms of J and the number density of the carriers rather than in terms of I and the drift speed). Sorry for the confusion.