# Faraday's Law for a linearly rising magnetic field

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1. Aug 7, 2015

### Dorian Black

Hi,

Imagine a conductive wire bent to the shape of a loop without its ends meeting. A magnet is moved with respect to the loop such that the magnetic field crossing it (perpendicularly) is linearly increasing with time (Φ=kt) where k is a constant. The induced emf is the rate of change of magnetic flux; this being a constant (k). But then the emf across an inductor is Ldi/dt. So the current in the loop must be rising linearly with time. This current will consequently place opposite charges at both ends of the wire and build up a potential difference there. This potential should keep on growing as long as the magnetic flux carries on rising linearly in the loop. But here's the problem:

∫E⋅dL (closed path)=-N x dΦ/dt (Faraday's Law)

The left side gives a value that rises continuously with time (voltage across the two ends), while the left doesn't do so (dΦ/dt = k).

Grateful to anyone who points out where I got this fatally wrong.

Many thanks.

2. Aug 7, 2015

### Qwertywerty

So your reasoning is that charges move , so current increases , and so the current causes a ΔV ?

Also , the potential drop , and not the emf , across an inductor is Ldi/dt . I don't see how that fits here though .

3. Aug 7, 2015

### Dorian Black

Using the principle of Continuity of Current (∇⋅J=-dpv/dt), the movement of charges will constitute a current that will have to result in the accumulation of charges somewhere; this somewhere being the ends of the loop. Replacing the open-end with a capacitor might make things clearer, although the open end will effectively act as one anyway.

With regards to your second phrase, are you saying that Ldi/dt is not necessarily equal to -NdΦ/dt (where Φ is the total flux due to both the external magnet and the flux generated by the loop's current)?

My question is how the rising potential across the open-end and the constant emf can be reconciled together. Or is the emf not even constant to start with. Again, any thoughts are greatly appreciated.

4. Aug 7, 2015

### Philip Wood

I'd model the set-up by a CR circuit connected at time t = 0 across a constant emf. Then
$$Q = Q_0\ (1-exp -\frac{t}{CR})$$
The time constant, CR, will be extremely small, but for time t up to a few times CR, the pd, Q/C, across the open end of the loop will rise. But, after that it will be constant and equal to the emf.

I've ignored self inductance, as I'd guess the emf due to this is negligible.

5. Aug 7, 2015

### Staff: Mentor

Here you are using a general theory (Maxwell's equations) and a simplified version of the same theory (circuit theory). When a simplified theory is used there are always some simplifying assumptions which must be met in order for the simplification to be valid.

What are the assumptions for circuit theory and are they met in this scenario?

6. Aug 7, 2015

### Dorian Black

Thanks. The problem is precisely in those last two words you used; "constant emf". This emf is produced across an inductor (the loop). But an inductor cannot maintain an emf across itself without a time-varying current running through it (v=L di/dt). One might reply that the emf here is (v=-N x dΦ/dt), but this should not yield a different result to the previous expression (v=Ldi/dt) since both can be derived directly from each other.

7. Aug 7, 2015

### Dorian Black

Faraday's Law is as general as it gets for a case like this. The only 'circuit theory' expression I reverted to was v=Ldi/dt. I'm hoping someone could show how this can be any different from Faraday's dΦ/dt and how it could ever compromise generality in the current context. Again, the two can be derived from each other quite simply and with no specific conditions that I know of.

(later addition: The only exception I'm aware of is when the magnetic flux is changing in a space where no conductive material exists. But even then, the emf is impossible to measure without the insertion of some conductive material. The present case has a clearly defined conductive path and hence generality for both expressions should hold.).

8. Aug 7, 2015

### Staff: Mentor

Yes, that is the general law.

Yes, that is the circuit theory law. What are the assumptions, and do they apply here? If so, how would you analyze this in circuit theory?

EDIT: Often books introduce circuit theory without explicitly stating the fundamental assumptions. So you may not be aware of them. I will just go ahead and list them here:

1) All EM effects propagate instantaneously across the circuit. This is sometimes called the "small circuit" approximation.

2) The net charge on any component is 0, although there may be charge separation within a component.

3) The magnetic flux outside of any component or between any components is 0, although there may be magnetic flux within a component.

Last edited: Aug 7, 2015
9. Aug 7, 2015

### Philip Wood

But you said originally that the emf is induced by moving a magnet wrt the loop. This gives a rate of change of flux, which can be regarded as separate from that due to the changing current, given by L dI/dt. It's the latter which I'm suggesting might be negligible compared with the former.

10. Aug 7, 2015

### Dorian Black

You're right Philip. The inductor has two 'emfs'; one due to the external source and one due to its own current. -N x dΦ/dt is not equal to Ldi/dt as I had blasphemously believed. Thanks a lot for helping resolve this.