# Magnetism problem

1. Homework Statement

The following problem from Griffiths is irritating me for a long time…
A toroidal coil has a rectangular cross section, with inner radius a, outer radius a+w, and height h. It carries a total of N tightly bound wound turns, and the current is increasing at a constant rate dI/dt=k.If w and h are both much less than a, find the electric field at a point z above the centre of the toroid.
[Griffiths gives hint: exploit the analogy between Faraday fields and magnetostatic fields.]

2. Homework Equations

Maxwell's equations!!!

3. The Attempt at a Solution

Here are the two ways I approached the problem.

1. For each turn of wire, a rectangular loop may be assumed. For dI/dt, the magnetic flux Φ (=∫B∙da ) is increasing. So, induced emf ∫E∙dl = ξ = (-dФ/dt ).
Using standard expression of B (=μNI/2πs) (in the circumferential direction) inside the loop, I got ---
ξ = [(μ(N^2)hk)/2π] ln[(a+w)/a]
I understand that this closed line integral is evaluated around the rectangular loop. But I need to evaluate e field at the top of the toroid axis…Is there any way out?

2. Griffiths approach: E can be evaluated from divE=0 and curl E= (-∂B/∂t) and E→ 0 at ∞. Well, I still do not know how to find E(z) specified. I tried with a trick: taking curl of (curl of E) in LHS and curl of B in RHS. So that the LHS reduces to laplacian of E and from that a Poisson like equation should follow. However, the RHS got zero!!!

So this is the case. Please help and note that I need to understand the mathematics in physical terms.

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Galileo
Homework Helper
Did you use all the information?

It is given that w/a<<1. So use that ln(1+e)=e for small e.

Now look at your expression for the flux. Does the form look familiar?

Well.that will certainly change the expression.But what does it do for the problem?

Galileo
Homework Helper
Have you encountered a similar or familiar problem before?
Consider the direction of the change in the magnetic field. Suppose it was a current density, would you be able to find the magnetic field?
Is the relation between current and magnetic field analogous to the relation between the changing magnetic field and the electric field?
If so, you can use the related problem to find the expression for the electric field and replace the current (I in that problem) with the appropriate analog (something prop. to change in flux) and use your expression to get the final expression.

I understand that you are also lesding that way in which Griffiths told.well,I had omitted that hint from the problem.But still a question:
I suppose they are closely related;but how?B and J are related via Bio-Savart's law and it is what we used the similar problem you are referring to.Though I have a relation of curl E=-∂B/∂t,(analogus to Ampere's law),how can I have relation like bio-Savart's law?

Galileo
Homework Helper
I`ll take the general viewpoint first. Whenever the electric field is determined solely by a changing electric field (a Faraday-induced electric field as Griffith's calls it, so the charge density is zero) we have:
$$\nabla \cdot E =0$$
$$\nabla \times E =-\frac{\partial B}{\partial t}$$

While for magnetostatic fields (or general B-fields in the quasistatic approximation, as in this case):
$$\nabla \cdot B =0$$
$$\nabla \times B =\mu_0 J$$

These relations have the same form as above for the E-field, so the analogy is complete. In such cases, then, the E-field is determined by -dB/dt in the exact same way as the B-field is determined by μJ.

You have probably already solved your problem when there's a ring with a steady current flowing and you had to find B at some point z on the axis of the ring. B would be some function of the current I.
Form the above we can say E has the same functional dependance. Just replace I with its corresponding analog.

We have J in the analog instead of I. Well, just use the integral form the above equations (apply Stokes). What is the appropriate analog for I?

well,I got your point much earlier.But can you show me the exact functional dependence of E on B and r?Can I use it as a standard formula?I believe, the role of I is played by B in this problem as both are circumferential.

Galileo
Homework Helper
The same reasoning applies.

It's not necessary to solve the problem. But here's the idea. If the Biot-Savart law is the solution to:
$$\vec \nabla \cdot \vec B =0$$
$$\vec \nabla \times \vec B = \mu_0\vec J$$
(with the boundary condition that B->0 at infinity),
then by simply replacing μJ with -dB/dt you get:

$$\vec E(\vec r)=-\frac{1}{4\pi}\int \frac{\frac{\partial \vec B(\vec r')}{\partial t}\times d\vec R}{R^2}d\tau'$$

and all the 'tricks' you used from Biot-Savart hold. (Here $\vec R=\vec r-\vec r'$, I don't know how to make a 'script' r like Griffiths does).

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Thank you for your help.Now I can do it.
Can you refer to any proof of the formula you have written?Obviously it is the solution of curl E=-∂B/∂t and div E=0 with the condition that E→0 at ∞; but I want to see it proved somewhere...

Galileo
Homework Helper
It's the exact same proof as given in Griffitsh that shows that div B=0 and curl B=μJ (chapter 5.3.2) starting from the Biot-Savart law.

Galileo
Homework Helper
What do you mean by roundabout? Do you mean roundabout in the sense of showing the Biot-Savart formula IS a solution to divB=0, curlB=μJ by checking (and applying the uniqueness theorem) instead of constructing the solution from scratch?

Yes, you could have prescribed for a proof of a parallel Bio-Savart's law (in E) starting from the divergence and curl equations and uniqueness theorem.
What you did is to assume the solution and then show it satisfies the divergence and curl equations...
Well, it is alright.I do not mind.

Galileo
Homework Helper
Assuming the solution of the Laplace equation is known (with the boundary condition that J->0 at infinity), then since div B=0 you can introduce a vector potential A such that curl A=B and div A=0. Then div B = div (curl A) is automatically zero and curl B = curl (curl A)=-(LAP)A= μJ (by some vector differentiation identity).
The solution of which is:

$$\vec A (\vec r) = \frac{\mu_0}{4\pi}\int \frac{J(\vec r')}{\vec R}d\tau'$$

By taking the curl (using again one of the product rules):

$$\vec B (\vec r) = \frac{\mu_0}{4\pi}\int \frac{J(\vec r') \times \hat R}{R^2}d\tau'$$

There's a chapter on the magnetic vector potential in Griffiths that explains the main idea,

Last edited:
Well,I got it.
We are to find a vector potential for E,say C which will satisfy curl C=E and div C=0.Then, with the boundary condition we will find the
C(r)=-(1/4π)∫[(-∂B/∂t)×r/|r|^2]dτ
taking ▼×C(r) we may get E(r)