# Homework Help: Magnetic field calculation. Method of Images/superimposition.

1. Mar 29, 2014

### binbagsss

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 curent 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, moreso 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 !

-

2. Mar 29, 2014

### rude man

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.