The rate of current through a solenoid - clarification needed.

[Kruum]

Homework Statement

http://www.aijaa.com/img/b/00106/3735947.jpg

Homework Equations

$$\epsilon =- \frac {d \Phi}{dt}=- \frac {d BA}{dt}= \oint \vec{E} \cdot d \vec{l}$$
$$B= \mu _0nIA$$, where n is solenoid's turns per meter.

The Attempt at a Solution

I was able to solve this through trial and error, almost. What I got is $$\epsilon =- \frac { \mu _0 nAdI}{dt} \Longrightarrow \frac {dI}{dt}=- \frac {E2 \pi r_1}{ \mu _0n \pi r_{2}^2}=-9.21 \frac{A}{s}$$, where r₁=0.035m and r₂=0.011m. Our assistant teacher wasn't able to tell me why we to take the line integral over a circle in $$\epsilon = \oint \vec{E} \cdot d \vec{l}$$ and why my sign is wrong. Can somebody help?

 

[alphysicist]

Hi Kruum,

Homework Statement

http://www.aijaa.com/img/b/00106/3735947.jpg

Homework Equations

$$\epsilon =- \frac {d \Phi}{dt}=- \frac {d BA}{dt}= \oint \vec{E} \cdot d \vec{l}$$
$$B= \mu _0nIA$$, where n is solenoid's turns per meter.

The Attempt at a Solution

I was able to solve this through trial and error, almost. What I got is $$\epsilon =- \frac { \mu _0 nAdI}{dt} \Longrightarrow \frac {dI}{dt}=- \frac {E2 \pi r_1}{ \mu _0n \pi r_{2}^2}=-9.21 \frac{A}{s}$$, where r₁=0.035m and r₂=0.011m. Our assistant teacher wasn't able to tell me why we to take the line integral over a circle in $$\epsilon = \oint \vec{E} \cdot d \vec{l}$$ and why my sign is wrong. Can somebody help?

The point in taking the line integral over a circle is because the E field is constant all along that circular path (from symmetry). (By constant I mean that it has the same magnitude and has the orientation (tangent) relative to each point along the circle.) That allows us to take E out of the integral and then the integral is just the circumference of the circle.

 

[Kruum]

Thanks for the hasty reply, alphysicist!

The point in taking the line integral over a circle is because the E field is constant all along that circular path (from symmetry).

So the electric field is circular around the solenoid? If so, I misinterpreted the text and thought the E field went through the solenoid.

 

[alphysicist]

Thanks for the hasty reply, alphysicist!

Yes, I almost missed your post completely. The first time I saw it had already gone off the front page.

So the electric field is circular around the solenoid? If so, I misinterpreted the text and thought the E field went through the solenoid.

The magnetic field goes straight through the solenoid, and the changing magnetic field will induce an electric field around it.

I think it's useful if you compare Faraday's law for this problem with Ampere's law applied to a long straight wire. That way you can see the relationship between E and the changing B field from this problem, is similar to the relationship between an infinite straight current and it's magnetic field.

 

[Kruum]

I think it's useful if you compare Faraday's law for this problem with Ampere's law applied to a long straight wire. That way you can see the relationship between E and the changing B field from this problem, is similar to the relationship between an infinite straight current and it's magnetic field.

So is this the case with all solenoids? Their magnetic field will create an electric field around them?

Oh, and hopefully my thank yous didn't come out negative. I really appreciate your replies.

 

[alphysicist]

So is this the case with all solenoids? Their magnetic field will create an electric field around them?

That sounds right, as long as the magnetic field is changing.

Oh, and hopefully my thank yous didn't come out negative. I really appreciate your replies.

I'm glad to help! (And for a moment I was wondering if your thank you was just a sarcastic comment. But I know how easy it is to misunderstand and be misunderstood with just text, so I always just take things at face value.)