2
$\begingroup$

The following magnetic field is given (and nothing else is known): $$ \begin{cases} B_0 \frac{t}{T} & \quad\quad\quad r < R \\ 0 & \quad\quad\quad r > R \end{cases} $$

I need to calculate the electric field (everywhere).

Using the differential Faraday's law:
$$ \begin{align} & \vec \nabla \times \vec E = - \frac{\partial B}{\partial t} \\ & RHS = \frac{\partial B}{\partial t} = \frac{B_0}{T} \\ & LHS = \vec \nabla \times \vec E \overset{\text{ cylindrical }}{=} \frac{1}{r} \left( \frac{\partial}{\partial r}(r E_\phi) - \frac{\partial}{\partial \phi} E_r\right) \end{align} $$

Now, it is clear to me that in order to solve this problem, I need to infer that there's a symmetry in the problem, thus $\frac{\partial}{\partial \phi} E_r = 0$ and then, I can easily solve the ODE.

But nothing is known regarding the symmetry. Perhaps there are radial & angular electric fields which add up to make a constant magnetic field?

I don't think that having the magnetic field merely in the $\hat z$ direction, establishes any symmetry in the problem.


I would like to add:

Using Ampere's law, $\vec \nabla \times \vec B = \mu_0 \vec j$:
$\vec \nabla \times \vec B = 0$ therefore $\vec j = 0$ in $r < R$, thus $\vec E = 0$ in $r < R$.
So we conclude that $\vec E$ exist in $r =R$, and we should be able to find it with the known rule $\Delta B = \mu_0 k$ ($k$ is linear current density).
But according to the answers, $\vec E \ne 0$ in $r < R$.


Also here's an example to show the asymmetry:
Setting $E_r = \frac{B_0\, r\, \phi}{T}$ and $E_\phi = \frac{r\, B_0}{T}$, then: $$\vec \nabla \times \vec E \overset{\text{ cylindrical }}{=} \frac{1}{r} \left( \frac{\partial}{\partial r}(r E_\phi) - \frac{\partial}{\partial \phi} E_r\right) = \frac{B_0}{T} \hat z = \vec B$$

$\endgroup$
  • $\begingroup$ related: induced E from changing uniform magnetic field $\endgroup$ – Physiks lover Jun 9 '15 at 18:56
  • $\begingroup$ Dor: I'm unclear about the question, but note that there aren't radial & angular electric fields etc. The field is the electromagnetic field. You combine radial electric lines of force with concentric magnetic lines of force to depict it. See Maxwell's convergence + curl sketch on page 7 of this paper. Here's a slightly better depiction. Electromagnetic field interactions result in linear electric force and/or rotational magnetic force. Think spinor. $\endgroup$ – John Duffield Jun 9 '15 at 19:05
  • $\begingroup$ @Physikslover Please see my edit. $\endgroup$ – Dor Jun 9 '15 at 19:53
  • $\begingroup$ @JohnDuffield Please see my edit. $\endgroup$ – Dor Jun 9 '15 at 20:05
  • $\begingroup$ You're right when you say: "I don't think that having the magnetic field merely in the z^ direction, establishes any symmetry in the problem." for reasons given by Lubos in his answer in the link I gave. You should therefore be able to answer your own question in the answers below ;) $\endgroup$ – Physiks lover Jun 9 '15 at 20:06
1
$\begingroup$

You can't literally solve for $\vec E$ because you could, for instance, add any constant vector field to $\vec E$ and get the same $\vec \nabla \times \vec E.$ So you just don't have enough information unless you know a lot more about the electric field.

So you're right in the sense that you either need symmetry or you need boundary conditions. But while that might seem impossible, consider the similar situation where you have a constant $\vec J$ inside a wire and zero current outside the wire.

If you are comfortable with the magnetic field going in a circle when solving $\vec \nabla \times \vec B = \mu_0 \vec J$ then you should be equally happy with solving $\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}.$ Same techniques work for the same reasons (take a derivative of $\vec B$ slap a minus sign on it and treat it like $\mu_0\vec J$ and call the result an electric field instead of a magnetic field, but nothing changed).

Anything that bothers you in that situation should have bothered you for the equivalent problem with the uniform current through a wire.

If it helps, electromagnetic fields are real things in their own right, they have their own energy and momentum. They can be created and destroyed like anything else and their energy and momentum and move around.

So in a sense the electric fields are just there as things in their own rights and Maxwell really just tells the fields how to change, so $-\vec \nabla \times \vec E$ tells the magnetic field how to change, so you are just find one electric field of many that can make the magnetic field change the way you've been told it should change. Seems less mysterious that there are many possibilities then.

And for symmetry the same thing happens for electric fields. $\frac{1}{\epsilon_0}\left(\frac{1}{\mu_0} \vec \nabla \times \vec B-\vec J\right)$ tells the electric field how to change and since $\vec B$ and $\vec J$ are real things then the electric field has a boss that tells it what to do.

With particles they can have a charge and a mass and a position and a velocity but then they have a boss called a force that tells it how to accelerate. For electromagnetic fields they can have their own values but have no freedom about how to change just like particles have no freedom how about to accelerate.

So your situation is counter intuitive simply because there are multiple electric fields that can make the magnetic field be forced to change that way.

$\endgroup$
  • $\begingroup$ I'm really sorry, despite of your effort in posting a detailed reply, I still don't understand. I mean that I understand everything that you've written, but I can't relate that to my problem. +1 for now, I'll try to read this again tomorrow, maybe I'll be in a different mode of understanding... $\endgroup$ – Dor Jun 10 '15 at 17:26
0
$\begingroup$

Use the magnetic vector potential and Stokes' theorem.

Thus: $\vec{B}=\nabla \times \vec{A}$ implies $$ \oint \vec{A}\cdot d\vec{l} = \int \vec{B}\cdot d\vec{S}\ .$$ That is, the closed line integral of the magnetic vector potential is equal to the magnetic flux through the closed loop.

If all you are told is the magnitude of the B-field, then there are two orthogonal possibilities - either the field is along the z-axis, or it is toroidal - a constant radial field is not allowed since it would have non-zero divergence at $z=0$.

Taking each of these cylindrically symmetric possibilities in turn.

  1. $\vec{B}= B_0 (t/T) \vec{z}$.

For $r<R$, then $ \vec{A}$ must be toroidal and we can construct a circular loop of radius $r$. $$A (2\pi r) = B \pi r^2$$ $$ \vec{A} = \frac{Br}{2}\vec{\phi} = \frac{B_0r t}{2T}\vec{\phi}.$$

Doing the same thing for a circular path with $r>R$ $$ A (2\pi r) = B \pi R^2$$ $$\vec{A} = \frac{B R^2}{2r}\vec{\phi}= \frac{B_0 R^2 t}{2 r T}\vec{\phi}.$$

You can verify that the curl of these A-fields does give the correct specified magnetic fields.

Then using $$\vec{E}= -\frac{\partial \vec{A}}{\partial t} - \nabla V$$ we get $$\vec{E} = -\frac{B_0 r}{2T}\vec{\phi} + \vec{E_0}\ \ \ r<R$$ $$\vec{E}= -\frac{B_0 R^2}{2rT}\vec{\phi} + \vec{E_0}\ \ \ r>R,$$ where $\vec{E_0}$ is an arbitrary curl-free static E-field, where $E_0= -\nabla V$, the gradient of an arbitrary electrostatic potential.

  1. $\vec{B}= B_0 (t/T) \vec{\phi}$.

The A-field must be along the z-direction. For $r>R$ an arbitrary rectangular loop, aligned with the z-axis, would enclose no magnetic flux, meaning the A-field here must have no $r$ dependence - we can set $A=0$, since the curl and time derivative of any constant is zero and does not change the derived E- and B-fields.

Now defining a rectangular loop of length $l$ along the z-axis with one side at a distance $r<R$ from the z-axis and another at $r>R$, we can say $$A l = B l(R-r)$$ $$\vec{A} = B_0(R-r)\frac{t}{T}\vec{z}\ \ \ \ r<R$$ and thus $$\vec{E} = \frac{B_0(r-R)}{T}\vec{z} +\vec{E_0}\ \ \ \ r<R$$

An arbitrary combination of these two possibilities (helical field) has an associated E-field that is the combination of these two orthogonal results.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.