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

Thank you for your help.In summary, the conversation discusses finding the induced electric field of a uniform time-varying magnetic field pointing in the z direction and filling up a circular region on the xy-plane. Two different methods are used, resulting in different answers, with one solution ultimately being verified as correct. The conversation also raises the question of the region of constant flux change and its effect on the solution.
  • #1
yungman
5,718
240

Homework Statement



Uniform time varying magnetic field [itex]\vec B_{(t)}[/itex] 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



[tex]Emf =-\frac{d\Phi}{dt}[/tex]

[tex]\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}[/tex]

The Attempt at a Solution



We know [tex]\vec E = \hat {\phi} E [/tex]

1) Using

[tex]Emf =-\frac{d\Phi}{dt}[/tex]

[tex]\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. }[/tex]

[tex] \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} [/tex]

The above is the same as in the book.




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

[tex] \nabla \times \vec E = \frac 1 r \left |\begin {array}{ccc} \hat r & r\hat {\phi} & \hat z \\
\frac {\partial }{\partial r} & \frac {\partial }{\partial \phi} & \frac {\partial }{\partial z}\\
0 & rE_{\phi} & 0 \end {array}\right |_{r=s} =
\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} [/tex]




As you can see, the two methods differ by 1/2! What did I do wrong?
 
Last edited:
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  • #2
Anyone? I don't see how there is a difference of 1/2!
 
  • #3
E depends on the r - coordinate.
So you calculated the curl wrong.
 
  • #4
qbert said:
E depends on the r - coordinate.
So you calculated the curl wrong.

Can you show where I did wrong?

Thanks
 
  • #5
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.

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

This then leads to the following.

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

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.

[tex]E_\phi={{-r}\over{2}}{{\partial B}\over{\partial t}}[/tex] if [tex] r<R[/tex]

and

[tex]E_\phi={{-R^2}\over{2r}}{{\partial B}\over{\partial t}}[/tex] if [tex] r>R[/tex]

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. [tex] {{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=0[/tex]
 
Last edited:
  • #6
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.

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

This then leads to the following.

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

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.

[tex]E_\phi={{-r}\over{2}}{{\partial B}\over{\partial t}}[/tex] if [tex] r<R[/tex]

and

[tex]E_\phi={{-R^2}\over{2r}}{{\partial B}\over{\partial t}}[/tex] if [tex] r>R[/tex]

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. [tex] {{dE_\phi}\over{dr}}+{{1}\over{r}}E_\phi=0[/tex]

Thanks

Got it.
 

1. How do you calculate the induced E field inside a disk in a uniform magnetic field?

The induced electric field inside a disk in a uniform magnetic field can be calculated using the formula E = B x v, where B is the magnetic field and v is the velocity of the disk.

2. What factors affect the magnitude of the induced E field inside a disk?

The magnitude of the induced electric field inside a disk is affected by the strength of the magnetic field, the speed and direction of the disk's motion, and the size and orientation of the disk.

3. How does the direction of the induced E field inside a disk relate to the direction of the magnetic field?

The direction of the induced electric field inside a disk is perpendicular to both the direction of the magnetic field and the direction of the disk's motion. This is known as Lenz's Law.

4. Can the induced E field inside a disk be negative?

Yes, the induced electric field inside a disk can be negative. This occurs when the disk's motion is in the opposite direction to the magnetic field, resulting in a negative cross product and a negative induced electric field.

5. How does the thickness of the disk affect the induced E field inside it?

The thickness of the disk does not directly affect the induced electric field inside it. However, a thicker disk may have a larger surface area, which can result in a larger induced electric field due to a greater number of charges moving through the magnetic field.

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