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Find induced E field inside a disk in an uniform magnetic field. |
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| Apr7-11, 12:51 AM | #1 |
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Find induced E field inside a disk in an uniform magnetic field.
1. The problem statement, all variables and given/known data
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. 2. Relevant equations [tex]Emf =-\frac{d\Phi}{dt}[/tex] [tex]\nabla\times \vec E = -\frac{\partial \vec B}{\partial t}[/tex] 3. 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? |
| Apr7-11, 11:31 AM | #2 |
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Anyone? I don't see how there is a difference of 1/2!!
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| Apr7-11, 12:14 PM | #3 |
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E depends on the r - coordinate.
So you calculated the curl wrong. |
| Apr7-11, 02:29 PM | #4 |
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Find induced E field inside a disk in an uniform magnetic field.Thanks |
| Apr7-11, 10:16 PM | #5 |
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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. [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] |
| Apr8-11, 05:51 PM | #6 |
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Got it. |
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