How to Find the Magnetic Field on the Axis Between Two Co-axial Current Loops?

Welshy
Messages
20
Reaction score
0

Homework Statement



Two co-axial, parallel, circular wires of radius R each carrying current I are placed a distance 2d apart. Find an expression for magnetic field B on the axis at distance e from the midpoint between the two loops up to and including terms of 0(e3).

Homework Equations


The field for one such loop was calculated to be:

B = (u0IR2)/(2(z2+R2)(3/2))


The Attempt at a Solution


I'm not sure how the superposition works at all. I've seen it in a few past papers and the method isn't in my notes or the course textbook. (Introduction to Electrodynamics, Griffiths)

Any help or hints on the method would be greatly appreciated!
 
Physics news on Phys.org
Superposition is nothing more than the ability to calculate the magnetic field due to either ring and then add them up together (vectorial). So the first thing you should do is put yourself a distance e from the midpoint. Express the distance to said point in terms of z,d,e for both loops. Then check the direction of the magnetic fields, do they amplify each other or weaken or is there an angle between the field lines etc.
 
Last edited:
Cyosis said:
Superposition is nothing more than the ability to calculate the magnetic field due to either ring and then add them up together (vectorial). So the first thing you should do is put yourself a distance e from the midpoint. Express the distance to said point in terms of z,d,e for both loops. Then check the direction of the magnetic fields, do they amplify each other or weaken or is there an angle between the field lines etc.

Hmm. What I get is:

B = (u0IR2cos(O)/((d2+e2+R2)(3/2))

where O is the angle between the (e to centre of coils) vector and the z axis and the B vector is in the z axis. I'm not sure if that's right because I don't see how that can be expanded to powers of e in any way.
 
Welshy said:
Find an expression for magnetic field B on the axis at distance e from the midpoint between the two loops up to and including terms of 0(e3).

From this I understand that the point e lies on the axis that goes through the center of both rings. Assuming this is the case I don't see how you get a cosine in your answer.

This is the configuration we are talking about as far as I can see.
.'''''''''''''''''''''''''''' .
|''''''''''''''''''''''''''''|
|-----|--e---| ---------------------> (positive z-direction)
|''''''''''''''''''''''''''''|
x''''''''''''''''''''''''''''x

the . is where the magnetic field comes out of the "paper" and the x is where the magnetic field goes into the "paper". I have assumed that both currents run in the same direction.

Now try to answer the following questions with regards to point e.:

Left ring
1)In what direction does the magnetic field point in point e?
2)What is the distance z from the center of the ring in point e?

Right ring
Same questions as for the left wing.

3)Do the fields of both rings point in the same direction in point e, are they parallel to each other?

Now enter the data in Bleft and Bright and add them together (vectorial).
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top