Electrostatic field at the centre of a disk

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Homework Statement


A thin conducting disc has radius a thickness b and electrical resistivity ρ. It is
placed in a uniform time-dependent magnetic induction ##B = B_0 sin ωt## directed parallel
to the axis of the disc. Assuming that ρ is large, find E at a distance r < a from the axis
of the disc, in the plane of the disc, and obtain an expression for the induced current density as a function of r

Homework Equations


Maxwells equations

The Attempt at a Solution


So there are two ways I can think of doing this question. One is to consider the emf induced in the disk using Faradays law ##emf= -{d\phi}/{dt}## and then calculate the associated current using the resistivity, followed by calculating the associated magnetic field. From this I can then calculate the electric field which is produced as a result of this current using ##integral E.ds= -integral {dB}/{dt} .ds## this can then be subtracted from the initial magnetic field to give the resulting.

However, since the resistivity is large, I'm going to assume that this current is essentially negligible. So can I just skip to the point where I can calculate E using ##\integral E.ds= -\integral {dB}/{dt} ds##? My only concern with this, is this electric field is the same regardless of whether the disk is there or not.

Either way, once I have the electric field I can easily calculate the current density. I just can decide which of these methods is more appropriate..

Many thanks

Edit: Really sorry- Don't know how to get the integral sign, nor can I remember where to find how to do it. If you can help me with either of these, that would be much appreciated! :)
 

Answers and Replies

  • #2
The integral you wrote down is valid whatever the shape is however when you are trying to find the electric field fom that integral equation the shape matters. The disk allows you to conclude that the electric field will be dependent only on radius and the total length element 2*pi*r, so it allaws you to extract E from that equation. After that you can find the current density
 
  • #3
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The integral you wrote down is valid whatever the shape is however when you are trying to find the electric field fom that integral equation the shape matters. The disk allows you to conclude that the electric field will be dependent only on radius and the total length element 2*pi*r, so it allaws you to extract E from that equation. After that you can find the current density
Should I take into account the induced current though? Many thanks :)
 
  • #4
While calculating the electric field, the integral you wrote down is enough, after you calculate the electric field you can find the induced current density by J=δE (Ohm's Law).
 
  • #5
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While calculating the electric field, the integral you wrote down is enough, after you calculate the electric field you can find the induced current density by J=δE (Ohm's Law).
Ah okay. Thank you so much!

Sorry, I should've included this above, but the next part asks for the power dissipated, for which I will need the resistance.


So far I have done
considering ## R=rho * l/A## where L and A are the lengths and areas respectively. The area that is perpendicular to the current is given by ##bdr## where dr is the infinitesimal change in radial distance, and ##l=2*pi*r##. so ##R=rho * 2*pi*r/bdr ## ? I don't feel as though this is right?

Many thanks :)
 
  • #6
Since you can calculate the potential difference(emf), you can calculate the power with VI instead of going into resistance.
 
  • #7
rude man
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It is correct for the OP to consider including the mag field introduced by the current, but it is also correct for the OP to assume that this mag field is negligible since ρ is large.

So Faraday works nicely: contour∫E(r)⋅ds = -∂Φ(r)/∂t = emf.

If ρ were not large the problem would be prohibitively difficult for an introductory course, since the inductance of the disk would have to be considered which is next to impossible to compute even with advanced math & special functions (elliptic integrals).

To get current density think of a differentially thin annulus within the disc, of width dr and thickness b, and compute the differential current di. I suggest using conductivity and conductance rather than resistivity and resistance in the calculations. If you post your result we can compare.
 
  • #8
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It is correct for the OP to consider including the mag field introduced by the current, but it is also correct for the OP to assume that this mag field is negligible since ρ is large.

So Faraday works nicely: contour∫E(r)⋅ds = -∂Φ(r)/∂t = emf.

If ρ were not large the problem would be prohibitively difficult for an introductory course, since the inductance of the disk would have to be considered which is next to impossible to compute even with advanced math & special functions (elliptic integrals).

To get current density think of a differentially thin annulus within the disc, of width dr and thickness b, and compute the differential current di. I suggest using conductivity and conductance rather than resistivity and resistance in the calculations. If you post your result we can compare.
I got the electric field to be ##E=\frac{B_0 \omega}{2 \pi r} cos( \omega t)##
and the Current density to be ##E=\frac{B_0 \omega}{2 \pi r \rho} cos( \omega t)##

However, I'm struggling to find the power dissipated. Many thanks
 
  • #9
rude man
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I got the electric field to be ##E=\frac{B_0 \omega}{2 \pi r} cos( \omega t)##
Uh oh. The E field goes to infinity at r = 0???
I think you forgot to include the area in Φ?
BTW your dimensions are also wrong. Always do a dimension check of all terms in an equation. It's the most powerful checking method there is!
 
Last edited:
  • #10
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Uh oh. The E field goes to infinity at r = 0???
I think you forgot to include the area in Φ?
BTW your dimensions are also wrong. Always do a dimension check of all terms in an equation. It's the most powerful checking method there is!
My mistake!

I should have got
##E=\frac{- B_0 \omega r}{2 } cos( \omega t)## (I think this is dimensionally correct)
##J=\frac{- B_0 \omega r}{2 \rho} cos( \omega t)## ?
 
  • #11
rude man
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My mistake!

I should have got
##E=\frac{- B_0 \omega r}{2 } cos( \omega t)## (I think this is dimensionally correct)
##J=\frac{- B_0 \omega r}{2 \rho} cos( \omega t)## ?
All correct! Good going.
 
  • #12
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All correct! Good going.
Thank you for your help! Is there any chance you can also help me find the power dissipated?
 
  • #13
BvU
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Don't know how to get the integral sign, nor can I remember where to find how to do it. If you can help me with either of these, that would be much appreciated!
\int is the magic incantation. See e.g. here
 
  • #14
rude man
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Thank you for your help! Is there any chance you can also help me find the power dissipated?
Well, you now know the current density so you can compute the differential power in a thin annulus of radius r and cross-section b dr, then you can sum the differential power over the radius of the disc.
 
  • #15
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Well, you now know the current density so you can compute the differential power in a thin annulus of radius r and cross-section b dr, then you can sum the differential power over the radius of the disc.
I'm afraid I don't know how to compute the differential power in a thin annulus
 
  • #16
rude man
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I'm afraid I don't know how to compute the differential power in a thin annulus
The length is 2 pi r, the thickness is b, the width is dr, the current density you know. Surely you can compute power from V^2/R.
 

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