Find induced E field inside a disk in an uniform magnetic field.

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Homework Help Overview

The problem involves finding the induced electric field in a circular region due to a uniform time-varying magnetic field directed along the z-axis. The original poster attempts to solve this using two different methods, leading to differing results.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster describes two methods for calculating the induced electric field, one using the electromotive force and the other using the curl of the electric field. Some participants question the calculations related to the curl and the dependence of the electric field on the radial coordinate.

Discussion Status

Participants are actively discussing the differences in the results obtained from the two methods. Some provide insights into the implications of the differential equation derived from the curl of the electric field and suggest that the first method's result is a valid solution to the differential equation.

Contextual Notes

There is an emphasis on the region of constant flux change being restricted to a circular area, which may influence the interpretation of the solutions provided.

yungman
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Homework Statement



Uniform time varying magnetic field \vec B_{(t)} pointing at z direction, filling up a circular region on xy-plane. Find the induced E field.

I tried two different ways and get two different answers. Please tell me what did I do wrong.

Homework Equations



Emf =-\frac{d\Phi}{dt}

\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}

The Attempt at a Solution



We know \vec E = \hat {\phi} E

1) Using

Emf =-\frac{d\Phi}{dt}

\Phi=\int_{S'} \vec B_{(t)} \cdot d S' = \pi s^2 B_{(t)} \;\hbox { where s is the radius of the circle and B is uniform. }

\frac {d \Phi}{dt} = \pi s^2 \frac{\partial \vec B}{\partial t} \;\rightarrow\; Emf = \int_C \vec E \cdot d\vec l = 2\pi s E = -\pi s^2 \frac{\partial \vec B}{\partial t} \; \Rightarrow \; \vec E = -\hat {\phi} \frac s 2 \frac{\partial \vec B}{\partial t}

The above is the same as in the book.




2) This one I use the fact of uniform B and \vec E = \hat \phi E

\nabla \times \vec E = \frac 1 r \left |\begin {array}{ccc} \hat r &amp; r\hat {\phi} &amp; \hat z \\ <br /> \frac {\partial }{\partial r} &amp; \frac {\partial }{\partial \phi} &amp; \frac {\partial }{\partial z}\\<br /> 0 &amp; rE_{\phi} &amp; 0 \end {array}\right |_{r=s} =<br /> \hat z \frac{E_{\phi_{(t)}}}{s} =-\frac {\partial \vec B}{\partial t} \;\Rightarrow \; \vec E = -\hat {\phi} s \frac{\partial \vec B}{\partial t}




As you can see, the two methods differ by 1/2! What did I do wrong?
 
Last edited:
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Anyone? I don't see how there is a difference of 1/2!
 
E depends on the r - coordinate.
So you calculated the curl wrong.
 
qbert said:
E depends on the r - coordinate.
So you calculated the curl wrong.

Can you show where I did wrong?

Thanks
 
yungman said:
Can you show where I did wrong?

Thanks

Your second method does not directly give you the solution right away. Instead it gives you a differential equation to solve. I won't go through the process of solving the differential equation, but will just point out that the first solution you found clearly is a solution for your differential equation.

So, the first step is to calculate the differential equation from the curl equation. You get the following.

{{1}\over{r}}{{d}\over{dr}}(rE_\phi)=-{{\partial B}\over{\partial t}}

This then leads to the following.

{{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=-{{\partial B}\over{\partial t}}

Now, you can solve this differential equation, but since we already know the answer, we need only verify that solution. Do it out and you will see that it does work out.

yungman said:
Uniform time varying magnetic field pointing at z direction, filling up a circular region on xy-plane.

Now, there is another aspect to the question. Strictly, the problem says that the region of constant flux change is restricted to a circular area. Hence, the solution requires that we center the origin in the middle of the circle and express the full solution as follows.

E_\phi={{-r}\over{2}}{{\partial B}\over{\partial t}} if r&lt;R

and

E_\phi={{-R^2}\over{2r}}{{\partial B}\over{\partial t}} if r&gt;R

where R is the radius of the region of flux.

Notice that even this second solution outside the region of flux (i.e. the region has dB/dt=0) also obeys your differential equation. {{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=0
 
Last edited:
stevenb said:
Your second method does not directly give you the solution right away. Instead it gives you a differential equation to solve. I won't go through the process of solving the differential equation, but will just point out that the first solution you found clearly is a solution for your differential equation.

So, the first step is to calculate the differential equation from the curl equation. You get the following.

{{1}\over{r}}{{d}\over{dr}}(rE_\phi)=-{{\partial B}\over{\partial t}}

This then leads to the following.

{{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=-{{\partial B}\over{\partial t}}

Now, you can solve this differential equation, but since we already know the answer, we need only verify that solution. Do it out and you will see that it does work out.



Now, there is another aspect to the question. Strictly, the problem says that the region of constant flux change is restricted to a circular area. Hence, the solution requires that we center the origin in the middle of the circle and express the full solution as follows.

E_\phi={{-r}\over{2}}{{\partial B}\over{\partial t}} if r&lt;R

and

E_\phi={{-R^2}\over{2r}}{{\partial B}\over{\partial t}} if r&gt;R

where R is the radius of the region of flux.

Notice that even this second solution outside the region of flux (i.e. the region has dB/dt=0) also obeys your differential equation. {{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=0

Thanks

Got it.
 

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