# E field induced by toroidal solenoid with changing current

1. Dec 1, 2012

### Eric Wright

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

Suppose we have a toroidal solenoid with square cross section and whose axis of symmetry lies along the z axis. Suppose a current I(t)=kt runs through the solenoid. Find the induced electric field at an arbitrary point on the z axis.

2. Relevant equations

$$\oint \vec E \cdot d \vec l = \frac{d \Phi}{dt}$$

3. The attempt at a solution

I am wanting to use faradays law in integral form, but there doesn't seem to be a good path to integrate around such that the E field will be constant on the path. I have already found the B field generated by a steady current I. Is it applicable to use faradays law just replacing I in the formula for the B field with I(t)?

Any suggestions about how to start this problem correctly would be much appreciated!

Thanks,

Eric

Last edited by a moderator: Dec 2, 2012
2. Dec 1, 2012

### rude man

Seems to me since the B field is confined to within the toroid and the z axis is everywhere outside the toroid, that the answer is zero.

I assume "the axis of symmetry" is perpendicular to the plane of the toroid and so does not pass inside any part of it.

3. Dec 2, 2012

### Eric Wright

That is what I was thinking at first.. but am not sure it is right. Not sure how to actually show that the field vanishes there.

Yes, that is correct. sorry.. i should have been more clear

4. Dec 2, 2012

### rude man

Here is an excerpt of a googled post:

<< If the integration route is chosen outside the torus, it is seen that the net current inside the route is zero, i.e. the magnetic field outside the toroid is zero. The toroid is the only known way to create a magnetic field to a limited part of space. >>

I see this as correct. For a concentric circular path outside the outer toroid radius the net current piercing this area is indeed zero, so ∫B*ds around this path is for sure zero. And by virtue of symmetry we can say that therefore the B field is zero everywhere around this outer closed path.

The same argument applies for any other symmetrical path around the torus if no net current pierces the associated area, for example, a circular path inside the torus' inner radius.

5. Dec 2, 2012

### Eric Wright

We are looking for the induced electric field.. not the B field. We can certainly find that there is some induced E field. We know the flux through a given cross section of the toroid is changing (since the current is changing) and thus there will be a circulation of electric field around a given cross section. if we imagine a loop going through the origin and enclosing a cross section of the toroid, we will get a non zero circulation

$$\oint \vec E \cdot d \vec l \neq 0$$

since

$$\frac{d}{dt} \Phi \neq 0$$

6. Dec 2, 2012

### rude man

Eric, you're right, I'm sorry, I was thinking B instead of E field.

However, to get time-varying flux we need flux, and there is no flux along the z axis as I see it. If flux = 0, so is d(flux)/dt, so is E.

Yes, but that does not mean E is finite everywhere along that loop. It is in fact finite only where a changing B field exists, i.e. within the cross-section only, and not at the toroid's center, or anywhere else along the z axis. ∫E*ds along your loop = d(phi)/dt but that means E on average = d(phi)/dt}/(length of loop) only.

I don't even see how the symmetry argument is compromised due to the torus having a rectangular cross-section instead of the more usual circular one. I was thinking maybe that made a difference, maybe I mised something there but damned if I know what. I hope you'll eventually post what your lecturer gave as the answer and reasoning, either here or privately.

7. Dec 2, 2012

### Eric Wright

we have
$$\vec E = - \nabla V - \frac{\partial \vec A}{\partial t}$$
and the vector potential A can change due to the changing current thus the E field can be non zero

Last edited: Dec 2, 2012