Magnetic field calculation. Method of Images/superimposition.

In summary: The symmetry of the problem implies that the magnitude of the B field can only be a function of distance r from the cable axis. (The magnitude of the B field cannot depend upon the angle between r and the cable axis due to the symmetry of the problem.))In summary, the problem can be solved using superposition, considering the B field as if the hole were not present and then adding the contribution of a current equal and opposite to the current in the region of the hole alone. The symmetry of the problem allows for the B field to only depend on the distance from the cable axis.
  • #1
binbagsss
1,254
11
The question is : A cable of circular cross-section and diameter 2 cm has a long cylindrical hole of
diameter 1 mm drilled in it parallel to the cable axis. The distance between the
axis of the hole and the cable is 5 mm. If the cable has a uniform steady current density of J=10
5 Am-2 flowing in it, calculate the magnetic field:


i) centre of cable
ii) centre of the hole.

So I begin by modelling the hole as 2 wires of diamter 1mm , situated on top of each other with opposite and equal current density, J. Let the postive wire be wire A and the negative wire B.


- First of all, I am not entirely sure my understanding of current density as a vector is correct, I am not enirely sure I interpret the term ''uniform steady current denisty' correctly, and this might show through in my questions below.

Here are my thoughts : Current density initally has it's orgin at he centre of the cable. Drilling this hole, in affect, re-defines the origin, such that prior to the hole, the answer to i would be zero. But now, the origin of the distribution is elsewhere. (I'm unsure though, how to picture a origin when the current density has the same magnitude everywhere, more so in terms of Ienclosed, so when applying to Amp's law.)Anyway with my understanding here are my questions:


Questions:

- So for part i, I consider wire A and the remaining body now without the hole, let this be C.Superimposing these two, the system is as it was before, the origin is at the centre, so body A and C contribute 0 to the net field at the centre of the cable.

So field at the centre of the cable= Field at the centre of the cable due to wire B.

However, I am not 100% sure on this interpretation as it doesn't seem correct as I apply it to part ii*, (the origin comments) :

-The correct interpretation is that wire B contributes nothing - my reasoning would be that the origin of it's current density is at the centre of the hole*, and so no charge is enclosed. But that, again modelling wire A and body C as equivalent to the system before drilling the hole, with its origin at the centre of the cable as before, we get a contribution as it would be at this location with no hole present: Field at centre of hole = field at the centre of the hole due to the cable
(where when I refer to the 'cable' i mean before the hole, and body C for the cable with the hole

- So I think my arguements may be flawed as by * couldn't we aregue the same about wire A, that this does not contribute at the centre of the whole, and so only body C does.
OH, OR is this correct as this interpretation takes us back to were we were before modelling body C; we are now considering body C only, whose origin of the current desnity is now not known, but once you solve for this and calculate the field contribution due to C at the centre of the hole, you would attain the same answer?

Are my thoughts and interprations of the concepts okay?

Many thanks to anyone who can help shed some light !

-
 
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  • #2
Whew! Long writeup! Hard to respond to, blow-by-blow.

You seem to have the right idea though: deal with this by superposition. (a) calculate the B field as if the hole were not there. Then, (b) calculate the B field assuming a current equal and opposite to the current that would obtain in part (a) in the region of the hole alone. It should be obvious that the superposition of these two currents is entirely equivalent to your problem.

You can exploit symmetry in both (a) and (b) separately to get the net B field anywhere you want.

(The current flows along the direction of the two axes and is uniform in cross-section except of course in the cavity. It's a vector since it has direction (axial) as well as magnitude.
 

1. What is the Method of Images/superimposition and how does it relate to magnetic field calculation?

The Method of Images/superimposition is a mathematical technique used to calculate the magnetic field around a set of objects. It involves creating a virtual image of the original objects and adding their fields to the existing field, allowing for a more accurate calculation of the total magnetic field.

2. What are the steps involved in using the Method of Images/superimposition for magnetic field calculation?

The first step is to identify the objects in the field and determine their individual magnetic fields. Then, using the principle of superposition, the virtual images are created and their fields are added to the original field. Finally, the total magnetic field is calculated by adding all the individual fields and the fields of the virtual images together.

3. How accurate is the Method of Images/superimposition in calculating magnetic fields?

The Method of Images/superimposition is a highly accurate method for calculating magnetic fields. It takes into account the effects of all objects in the field and allows for a more precise calculation compared to other methods.

4. Are there any limitations to using the Method of Images/superimposition for magnetic field calculation?

One limitation is that the method assumes the objects in the field are stationary and do not interact with each other. Additionally, the objects must be simple and have a uniform magnetic field distribution for the method to be effective.

5. Can the Method of Images/superimposition be used for any type of magnetic field calculation?

The Method of Images/superimposition is most commonly used for calculating the magnetic field of a single point charge or a uniform line/surface charge. It can also be applied to more complex systems, but the accuracy may decrease for these cases.

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