ElMag: Current-carrying Conductor, B inside a Hole

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SUMMARY

The discussion focuses on the magnetic induction produced by a long straight conductor with a cylindrical hole. The derived formula for the magnetic induction at the center of the hole is B = μ₀ bI / (2π(R² - a²)). It is established that the magnetic field inside the hole remains uniform due to the superposition of magnetic fields from the conductor and the hole, which effectively cancel out variations in the magnetic field strength. The analysis employs Ampere's Law and the principle of superposition to explain the constant magnetic field within the hole.

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Hi there,
My boyfriend and I have been bashing our head against this past exam question for some time now, any physical insight would be appreciated.

Homework Statement


A long straight conductor has a circular cross-section of radius R and carries a current I. Through the conductor, there is a cylindrical hole of radius a whose axis is parallel to the axis of the conductor and at a distance b from it. [From the cross-sectional diagram, a<b]

a) Show that the magnitude of the magnetic induction at the centre of the hole is equal to:
[itex]B = \frac{\mu_0 bI}{2\pi(R^2-a^2)}[/itex]

b) Show that the magnitude of the magnetic induction is uniform inside the hole.

2 The attempt at a solution

(a) If I treat the wire and hole as two super-imposed wires, with the "wire" that is the hole carrying the same current in the opposite direction, then I can find B at the centre of the hole (albeit with some difficulty).

(b) I don't know. Any physical insight into why the B-field inside the hole is the same everywhere?

Thanks,
Laura
 
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Laura,

As far as why the magnitude should be constant inside the hole:
When using ampere's law inside (r<R) a cylindrically symmetric current distribution we have that the magnitude of the magnetic field is proportional to the distance from the axis of the cylinder, r.
When also considering that the direction of the magnetic field is [itex]\hat{r}[/itex][itex]\times[/itex][itex]\hat{I}[/itex] we get that [itex]\vec{B}[/itex] [itex]\propto[/itex] [itex]\vec{r}[/itex][itex]\times[/itex][itex]\vec{I}[/itex].

Considering your problem of two equal currents running in opposite directions, the net magnetic field is the sum of the contributions from each current (i.e. use the superposition principle).
[itex]\vec{B_{net}}[/itex] = [itex]\vec{B_{+}}[/itex] + [itex]\vec{B_{-}}[/itex]
Since we are 'inside' both current distributions (r < R) what was said above about the magnetic field still applies so:
[itex]\vec{B_{net}}[/itex] [itex]\propto[/itex] [itex]\vec{r_{+}}[/itex][itex]\times[/itex][itex]\vec{I}[/itex] + [itex]\vec{r_{-}}[/itex][itex]\times[/itex][itex]\vec{-I}[/itex] = ([itex]\vec{r_{+}}[/itex] - [itex]\vec{r_{-}}[/itex]) [itex]\times[/itex] [itex]\vec{I}[/itex]

Where [itex]\vec{r_{+/-}}[/itex] is the vector pointing from the axis of the positive/negative current to the point where we are interested in the magnetic field.
Note that the fact that the magnitude of the currents being equal is important in order to write the proportionality this way.
Now if you draw a picture you can see that [itex]\vec{r_{+}}[/itex] - [itex]\vec{r_{-}}[/itex] is just the vector pointing from the axis of the positive current to the axis of the negative current, therefore it is a constant.
Thus the magnetic field is proportional to the cross product of two constant vectors, and is therefore constant inside the hole.

There is a similar problem in electrostatics where you find the field of a spherical hole inside a spherical charge distribution and you get a similar result (using Gauss' law instead of Ampere's).
I don't know if this counts as 'physical' insight, but the problem is inherently geometrical (which is pretty much true of all classical E&M). The idea is that as you move around in the hole you get closer or further from parts of the current, but on the whole these effects cancel out the magnetic field remains constant.

Hope that helps :)
 

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