Electromagnetic Induction of a Disk

[Mr_Allod]

Homework Statement

A thin conducting disk of thickness h, diameter D, and conductivity σ is placed in a uniform magnetic field B = B0sinωt parallel to the axis of the disk.
(a) Find the induced current density as a function of distance from the axis of the disk.
(b) What is the direction of this current?

Relevant Equations

Flux through a surface: $\phi = \vec A \cdot \vec B$
EMF: $\varepsilon = - \frac {d \phi} {dt}$
Faraday's Law: $\oint \vec E \cdot d \vec l = - \frac {d \phi} {dt}$
Current density and E-field relationship: $\vec J = \sigma \vec E$

Hello I'm having trouble finding the right way to apply Faraday's law to this question. I've found the flux through the disc:
$\phi = \vec A \cdot \vec B = B_{0} \sin{\omega t} \left( \frac D 2 \right)^2 \pi$

and the EMF:
$\varepsilon = - \frac {d \phi} {dt} = -B_{0} \omega \cos{\omega t} \left( \frac D 2 \right)^2 \pi$

Now to find the current density $\vec J$ I believe I need to find the electric field $\vec E$ using Faraday's law. I've seen it applied to a circular wire in a magnetic field in where the closed loop is the wire itself. I'm not sure how to apply this to a disk of thickness h though. Do I need to add up the contributions of circular strips with radii between 0 and $\frac D 2$ and thickness da?

I'm also not sure about how to make it a "function of distance from the axis of the disk".

 

[Delta2]

You have to apply faraday's law properly (i mean the law in integral form $\oint \vec{E}\cdot d\vec{l}=-\frac{d\phi}{dt}$, this is actually the maxwell-faraday law, which is slightly different from faraday's law of induction, anyway the two laws coincide when we don't have moving parts but only time varying magnetic field like in this case) to find the electric field at distance $r$ from the center of the disk and at depth $h$ inside the disk.

Hint: The magnitude of the E-field does not depend on the depth but it depends on the distance $r$. You just have to carefully apply faraday's law in integral form, take advantage of the cylindrical symmetry present in this problem, and then solve for the magnitude of the electric field E.

 

[Mr_Allod]

You have to apply faraday's law properly (i mean the law in integral form $\oint \vec{E}\cdot d\vec{l}=-\frac{d\phi}{dt}$, this is actually the maxwell-faraday law, which is slightly different from faraday's law of induction, anyway the two laws coincide when we don't have moving parts but only time varying magnetic field like in this case) to find the electric field at distance $r$ from the center of the disk and at depth $h$ inside the disk.

Hint: The magnitude of the E-field does not depend on the depth but it depends on the distance $r$. You just have to carefully apply faraday's law in integral form, take advantage of the cylindrical symmetry present in this problem, and then solve for the magnitude of the electric field E.

If E does not depend on h but only the distance $r$ does that mean it's possible to just use the same closed loop integral as if it were a circular wire?

ie. $\oint \vec{E}\cdot d\vec{l}= \int_0^{2\pi} Es \, d\phi = E2\pi s$?

 

[Delta2]

If E does not depend on h but only the distance $r$ does that mean it's possible to just use the same closed loop integral as if it were a circular wire?

ie. $\oint \vec{E}\cdot d\vec{l}= \int_0^{2\pi} Es \, d\phi = E2\pi s$?

Yes exactly. I think the problem wants us to ignore the self induction effect of the disk (the current in the disk will create its own magnetic field which in turn will create another electric field) and hence we can solve as we suggest.

But be careful on the right hand side of the equation you have to put the (time derivative of) the flux that passes through the circular disk of radius $r$ and not the flux of the whole conducting disk.

 

[Mr_Allod]

Yes I see the mistake I made. After checking with the differential form of Faraday's law $\nabla \times \vec E = \frac {\partial \vec B} {\partial t}$ I see it's true for $r$ as opposed to $D$. Thanks very much for the help!

 

[Delta2]

Yes I see the mistake I made. After checking with the differential form of Faraday's law $\nabla \times \vec E = \frac {\partial \vec B} {\partial t}$ I see it's true for $r$ as opposed to $D$. Thanks very much for the help!

Both the integral and differential form are true (they hold) for any $(\vec{r},t)$. In the differential form this is more clear as the differential form is in its fully form as follows: $$\nabla \times \vec{E}(\vec{r},t)=-\frac{\partial \vec{B}(\vec{r},t)}{\partial t} $$
In the integral form it depends on the problem situation but in general we can integrate around any closed loop (in this problem situation around and circular loop of radius $r$ and center the center of the conducting disk), for any time instance $t$. And on the right hand side we should consider only the flux through the surface of the closed loop we choose to integrate.