What is the induced electric field in a wire of radius R?

[decerto]

Homework Statement

https://www.writelatex.com/628580dnzrxr#/1303752/

The Attempt at a Solution

As I said in the link above I tried using the surface orthogonal to the cross section of the wire of radius to integrate.

For the line integral I assume it just reduces to E*2(pi)R but for the surface integral I'm not sure what dA becomes, is it r^2 sin\phi dr d\phi ?

 

[tiny-tim]

hi decerto!

(why are you using an image of your latex? )

Homework Statement

'm having some trouble understanding what is probably a simple application of faraday's law:
The problem is this: Given a magnetic field $$B(r,t)=\frac{\mu_0 I_0 e^{-\alpha t}r}{2 \pi R^2} \hat{\theta}$$ inside a wire of radius $R$, what is the induced electric field inside the wire.

So faraday's law is:

$$\oint_C E \cdot dr=\frac{d}{dt}\int_S B \cdot dA$$

Im having trouble understanding what contour/surface to integrate over/through. My guess is that it is a disc of radius $R$ that is orthogonal to cross section disc of the wire but I'm not sure.

|E| is constant for constant r, so yes you use a circle of radius R, and any surface with that circle as boundary …

obviously, the disc is the easiest! ​

… for the surface integral I'm not sure what dA becomes, is it r^2 sin\phi dr d\phi ?

neither (look at the dimensions) …

rsinφ dr dφ ​

 

[decerto]

Thanks for the help

I used latex because I had already typed it up for /r/askphysics and I didn't know if you could used latex on this site so I just copied and pasted it into an online compiler.


Should the area element not be rdrd\phi as you are integrating from 0-2pi and the angle associated with the sin in spherical coordinates only goes from 0-pi?

Is this the easiest way to do this question?, I'm fairly new to EM so my conceptual understanding and my experience of these problems is pretty terrible at the moment.

 

[tiny-tim]

hi decerto!

sorry, i got confused earlier when i said rsinφ dr dφ

for some reason i thought we were on a sphere

for a disc, it should be r dr dφ

Is this the easiest way to do this question?

yes!

 

[decerto]

hi decerto!

sorry, i got confused earlier when i said rsinφ dr dφ

for some reason i thought we were on a sphere

for a disc, it should be r dr dφyes!

Ye i just edited to ask that questions, thanks again for your help.

 

[tiny-tim]

When I do the integration out for
\int B \cdot dA=\int_o^{2\pi} \int_0^R \frac{\mu_0 I_0 e^{-\alpha t}r}{2 \pi R^2}rdrd\theta
I get the answer
\frac{\mu_0 IR}{3}. So when I take the derivative and set it equal to
E2\pi R as per faradays law I get an E field
with no dependence on r it's just constant.

Is this right?

hmm, that can't be right

ohh, i should have drawn a diagram at the start, I've got it completely wrong …

that disc surface only proves that E has no θ component!

a useful surface needs to be perpendicular to B, ie perpendicular to θ, ie a radial slice

so try a rectangle bounded by radii 0 and r

(since B vanishes at r = 0, i think we can safely assume that E does too)

 

[decerto]

I think you were right the first time, I was integrating out r by integrating to the edge of the wire instead of integrating to a random variable r<R.

I don't think you can evaluate E from E.dr when E is radial and dr is along a rectangle, you need a circle that will be every tangential to the E field.

 

[tiny-tim]

i still think my rectangles work (though on second thoughts using r and r+dr would be easier)

they should give you a differential equation for E_z, and then you can find E_r by using ∇·E = 0