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Electric field of a solenoid

  1. Mar 3, 2006 #1
    A long solenoid with radius 'a' and 'n' turns per unit length carries a time-dependent current [itex]I(t)[/itex] in the [itex]\phi[/itex] direction. Find the electric field (magnitude and direction) at a distance 's' from the axis (both inside and outside the solenoid), in quasi-static approximation.


    What's quasi-static approximation? Anyway, without much prior thought I applied the flux rule :rolleyes: :
    [tex]\varepsilon = \int \vec E \cdot d\vec l = -{d\Phi \over dt}[/tex]

    [itex]\vec B = \mu_0 nI \hat z[/itex], [itex]\vec A = \pi a^2 \hat z[/itex]
    [tex]\Phi = \mu_0 nI \pi a^2[/tex]

    [tex]\int \vec E \cdot d\vec l =-{d ( \mu_0 nI \pi a^2)\over dt}[/tex]

    [tex]E2\pi a = -( \mu_0 n \pi a^2){dI\over dt}[/tex]

    Before I proceed to the final step, someone please check my work.
     
  2. jcsd
  3. Mar 3, 2006 #2

    Galileo

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    Since you have a time varying current, the electromagnetic field created by the solenoid varies with time. The electromagnetic 'news' travels at the speed of light though.
    In this case you are working with a time varying currents, but you are using the apparatus of magnetostatics, like Biot-Savart and Ampère's Law (e.g. Your equation for B came from magnetostatics). Your answers are only approximately correct, but the deviation is small if the current is 'static enough', i.e. it doesn't vary quickly. The approximation is called the quasi-static approximation.

    You are looking at a distance 's' from the solenoid axis right? So [itex]\oint \vec E \cdot d\vec l=E2\pi s[/itex]. Also, the flux through the surface bounding your Amperian loop is [itex]\mu_0 n \pi a^2[/itex] only outside the loop, where [itex]s\geq a[/itex].
     
  4. Feb 5, 2011 #3
    By the quasi static appriximation we mean that B=mu_0 nI (s<a) and B= 0 (s>a)
    then why E is not zero outside(s>a)?
     
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