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Ampere's Law solenoid problem

  1. Feb 10, 2009 #1
    I need to use Ampere's Law and Gauss' Law to show that inside a solenoid

    [itex]B_r=0,B_\theta=0,B_z=0[/itex]

    i tried using cylindrical polars on ampere's law but the expression just got really long and i couldn't see any way out of it.
     
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  3. Feb 10, 2009 #2

    Gokul43201

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    Re: Solenoid

    Show us what you have.
     
  4. Feb 11, 2009 #3
    Re: Solenoid

    [itex]\mu_0 J_r = \frac{1}{r} \frac{\partial{B_z}}{\partial{\theta}} - \frac{{\partial{B_\theta}}{\partial{z}}[/itex]
    [itex]\mu_0 J_\theta=\frac{\partial{B_r}}{\partial{z}} - \frac{\partial{B_z}}{\partial{r}}[/itex]
    [itex]\mu_0 J_z=\frac{1}{r}[\frac{\partial}{\partial{r}}(r B_\theta) - \frac{\partial{B_r}}{\partial{\theta}}][/itex]
     
    Last edited: Feb 11, 2009
  5. Feb 12, 2009 #4

    Gokul43201

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    Re: Solenoid

    Fixing your latex:

    [itex]\mu_0 J_r = \frac{1}{r} \frac{\partial B_z}{\partial \theta} - \frac{\partial B_{\theta}}{\partial z}[/itex]

    [itex]\mu_0 J_\theta=\frac{\partial{B_r}}{\partial{z}} - \frac{\partial{B_z}}{\partial{r}}[/itex]

    [itex]\mu_0 J_z=\frac{1}{r}[\frac{\partial}{\partial{r}}(r B_\theta) - \frac{\partial{B_r}}{\partial{\theta}}][/itex]

    But first:

    Should B_z vanish inside a (presumably current carrying) solenoid? Please double-check your question.
     
    Last edited: Feb 12, 2009
  6. Feb 12, 2009 #5
    Re: Solenoid

    my bad.

    yes it's to show [itex]B_z = 0 [/itex] outside and [itex]B_z=\mu_0 NI[/itex] inside. I have no idea how to proceed!
     
  7. Feb 12, 2009 #6

    Gokul43201

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    Re: Solenoid

    The question asks you to use Ampere's and Gauss' Laws. Which one (or both?) of these is useful for finding magnetic fields?
     
  8. Feb 12, 2009 #7
    Re: Solenoid

    You can't prove either of those assertions without making certain assumptions about the behavior of the magnetic field beforehand. The equations you listed are obtained after assuming that the B-field is uniform inside the solenoid, than applying Ampere's Law in integral form to an appropriately constructed loop.
     
  9. Feb 12, 2009 #8

    Gokul43201

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    Re: Solenoid

    You don't really need any assumptions that can't be justified by simple symmetry arguments, do you?
     
  10. Feb 13, 2009 #9
    Re: Solenoid

    ampere's law is [itex]\nabla \wedge \mathbf{B} = \mu_0 \mathbf{J}[/itex]. But I aalready worked out the components of this equation in cylindrical polars above???
     
  11. Feb 13, 2009 #10

    gabbagabbahey

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    Re: Solenoid

    That's the differential form of Ampere's law----which isn't very useful.

    Try working with the integral form of Ampere's law instead:wink:

    Can you think of any symmetries which allow you to evaluate the integral along some path without knowing exactly what B is?
     
  12. Feb 13, 2009 #11
    Re: Solenoid

    ok so

    [itex]\int_C \mathbf{B} \cdot d \mathbf{l} = \mu_0 I_{enc}[/itex]

    i'm not really sure about the symmetry part your talking about but i'll have a stab at it:

    If we take a rectangular Amperian loop through the wall of the solenoid. then if it is of infinitesimal width and length L.
    although i'm not really sure why this would be of any use (it's just frequently used in my notes)?
     
  13. Feb 13, 2009 #12

    gabbagabbahey

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    Re: Solenoid

    In order to pick a useful Amperian loop, you first need to know, qualitatively, what the field must look like.

    First, I assume you are talking about an infinitely long solenoid?

    Let's say the solenoid runs parallel to the z-axis. Take a look at the solenoid by looking at two current loops on the solenoid at a time, equal distances from the center (z=+d and z=-d). At any point in the x-y plane, what does the Biot-savart law tell you about the radial, angular and z-components of the net field from the two loops? (Just concentrate on the v x r term)

    You should see that the only the z-component of the field can be non-zero for the two loops. Now, just treat the solenoid as a Large number of these loops; clearly only the z-component of the field can be non-zero for the solenoid, correct?
     
  14. Feb 13, 2009 #13
    Re: Solenoid

    ok. i have drawn the picture and have the biot savart law but don't seem to be able to get the radial or azimuthal components of B to vanish by taking components
     
  15. Feb 13, 2009 #14

    gabbagabbahey

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    Re: Solenoid

    Call the vector from the first loop to the field point [tex]\vec{r}_1[/tex], and the vector from the second loop to the field point [tex]\vec{r}_2[/tex]

    What do you get for [tex]\vec{v}\times\vec{r}_1+\vec{v}\times\vec{r}_2[/tex]? Note: you need to realize that all of the current is in the azimuthal direction [tex]\vec{v}=v\hat{\theta}[/tex]!
     
  16. Feb 13, 2009 #15
    Re: Solenoid

    i see that your trying to get the Biot savart to give you some vector only in the [itex]\hat{z}[/itex] direction.

    however[itex]\mathbf{r_1,r_2}[/itex] don't point in the radial direction of the cylindrical polar coordinate system do they? r1 is angled downward to the x-y plane and r2 is angled upward to the x-y plane - so surely they both have components in the radial and z direction?
     
  17. Feb 13, 2009 #16
    Re: Solenoid

    You can use symmetry arguments to show that only the z-component is nonzero on the axis of the solenoid, but you can't derive the expression for the z-component without assuming it is uniform. Also, it is not true that B=0 everywhere outside the solenoid. It just vanishes at sufficiently large distances.
     
    Last edited: Feb 13, 2009
  18. Feb 13, 2009 #17

    gabbagabbahey

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    Re: Solenoid

    Yes, but work the math out real quick....what happens to the radial, angular and z-component of vxr1+vxr2...just because the individual loops have non-zero components doesn't mean that the vector sum of two loops has to.
     
  19. Feb 13, 2009 #18

    gabbagabbahey

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    Re: Solenoid

    You can use symmetry arguments to show that only the z-component is nonzero everywhere, not just along the axis. (for an infinite solenoid anyway)

    And there is no need to "assume" that the z-component is uniform, a simple symmetry argument will suffice.
     
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