Can Symmetry Arguments Simplify Ampere's Law for Solenoids?

In summary: No, they don't. 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...
  • #1
latentcorpse
1,444
0
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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  • #2


Show us what you have.
 
  • #3


[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]
 
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  • #4


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.
 
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  • #5


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!
 
  • #6


The question asks you to use Ampere's and Gauss' Laws. Which one (or both?) of these is useful for finding magnetic fields?
 
  • #7


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.
 
  • #8


Brian_C said:
You can't prove either of those assertions without making certain assumptions about the behavior of the magnetic field beforehand.
You don't really need any assumptions that can't be justified by simple symmetry arguments, do you?
 
  • #9


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?
 
  • #10


latentcorpse said:
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?

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?
 
  • #11


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)?
 
  • #12


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?
 
  • #13


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
 
  • #14


latentcorpse said:
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

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]!
 
  • #15


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?
 
  • #16


Gokul43201 said:
You don't really need any assumptions that can't be justified by simple symmetry arguments, do you?

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.
 
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  • #17


latentcorpse said:
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?

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.
 
  • #18


Brian_C said:
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.

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.
 

1. What is Ampere's Law and how does it relate to a solenoid problem?

Ampere's Law is a fundamental law in electromagnetism that describes the relationship between the magnetic field around a closed loop and the electric current passing through that loop. In the context of a solenoid problem, Ampere's Law can be used to calculate the magnetic field created by a solenoid, which is a long, cylindrical coil of wire with multiple loops.

2. How do you use Ampere's Law to solve a solenoid problem?

To use Ampere's Law, you must first determine the closed loop path around the solenoid. Then, you must calculate the electric current passing through that loop. Finally, you can use the equation B = µ₀I/2πr to calculate the magnetic field at any point along the loop, where B is the magnetic field, µ₀ is the permeability of free space, I is the electric current, and r is the distance from the point to the center of the solenoid.

3. What are the assumptions made when solving a solenoid problem using Ampere's Law?

There are a few key assumptions that are made when using Ampere's Law to solve a solenoid problem. These include: the solenoid is infinitely long, the electric current is constant and evenly distributed throughout the solenoid, and the magnetic field is uniform within the solenoid.

4. Can Ampere's Law be used to solve a solenoid problem with a non-uniform current?

Yes, Ampere's Law can be used to solve a solenoid problem with a non-uniform current, but the equation becomes more complex. In this case, the electric current must be broken down into smaller segments, and the magnetic field at each segment must be calculated separately. Then, the magnetic fields from each segment can be combined to find the total magnetic field at any point.

5. Are there any practical applications of solving a solenoid problem using Ampere's Law?

Yes, there are many practical applications of solving a solenoid problem using Ampere's Law. Some examples include designing and optimizing electromagnets, calculating the magnetic fields in MRI machines, and understanding the behavior of particle accelerators.

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