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The rate of current through a solenoid - clarification needed.

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. The attempt at a solution

    I was able to solve this through trial and error, almost. What I got is [tex] \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}[/tex], 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 [tex] \epsilon = \oint \vec{E} \cdot d \vec{l} [/tex] and why my sign is wrong. Can somebody help?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 6, 2009 #2


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    Hi Kruum,

    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.
    Last edited by a moderator: May 4, 2017
  4. Mar 6, 2009 #3
    Thanks for the hasty reply, alphysicist!

    So the electric field is circular around the solenoid? If so, I misinterpreted the text and thought the E field went through the solenoid.
  5. Mar 6, 2009 #4


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

    The magnetic field goes straight throught 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.
  6. Mar 6, 2009 #5
    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.
  7. Mar 6, 2009 #6


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

    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.)
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