Magnetic Field Inside Wire Quick Q - (I've asked 3 times and no answers)

[Fusilli_Jerry89]

Homework Statement

There is a wire of radius r with a current i flowing through it. There is also a hole of radius a in the wire a distance b from the centre of the wire. The question asks, can you show that the magnetic field inside the hole is uniform? (assume that if you impose a current in the opposite direction where the hole is, that current has the same current density as in the actual conductor.)

My question is: how is the field inside the hole uniform? If the magnetic field gets stronger as r increases (the distance from the centre of the wire to anywhere in the wire), then wouldn't the magnetic field be larger at the outside edge of the hole rather than the inside edge? At both edges of the hole, the imposed opposite current would cause the same magnitude of B, would it not?

Homework Equations

ampere's law

The Attempt at a Solution

I know how to calculate B for the middle of the hole (it's just the magnetic field o the wire without the hole) and on the outer edge of the hole (magnetic field of wire w/o hole minus magnet field of hole), but when I calculate B on the inside of the hole, I get something different B2(pi)(b-a) = (mu)i(pi)(b-a)^2/(pi)(R^2 - a^2) ;
B = (mu)i(b-a)/2(pi)(R^2 - a^2) , but when I calculate be for the middle or outside of the hole, I get: (mu)ib/2(pi)(R^2 - a^2)

 

[Doc Al]

I don't understand your calculations. What's R? (The radius of the wire?) Call the current density J. Express your answers for the field at the center of the hole, and the near and far points, in terms of J, a, & b.

 

[kkrizka]

I remember being asked this question on one of my assignments, but in my case it was a field uniform along a radial line going through the center of the hole, so I'm not sure if that's the case here too.

Anyways, you might have made the same mistake I origanally did. So let me as you, did you add or subtract the magnetic field of the "hole" from the magnetic field of the wire?

 

[luben]

the distributions of the two B fluxes are circular, one being shifted. Both of them are known, and the superposition of them is the resultant B flux. But it seems to me the flux inside hole would be uniform only when the hole is at the center position. any success?

 

[Fusilli_Jerry89]

k I will write my calculations: (using ampere's law)

for the centre of the hole:

(same as if there was no hole because the centre of the hole contributes no magnetic field)

j = i/(pi)(R^2 - a^2)

B(2(pi)r) = (mu)j(pi)r^2
B = (mu)jr/2
r = b in this case
B = (mu)jb/2

for the outside edge of the hole:

B(w/o hole) - B(hole) = B

(2(pi)r)B(w/o hole) = (mu)j(pi)r^2
B(w/o) = (mu)jr/2
r = b + a in this case
B(w/o) = (mu)j(b+a)/2

(2(pi)a)B(hole) = (mu)j(pi)a^2
B(hole) = (mu)ja/2

B = (mu)j(b+a)/2 - (mu)ja/2
B = (mu)jb/2 <----------- same answer as centre of hole ( the textbook says prove the field inside the hole is uniform so I can assume I did it right?)but for inside edge of the hole:

B(2(pi)r) = (mu)j(pi)r^2
B = (mu)jr/2
r = b - a in this case
B = (mu)j(b-a)/2 <------- this is not the same as the other 2 answers i got

 

[Doc Al]

but for inside edge of the hole:

B(2(pi)r) = (mu)j(pi)r^2
B = (mu)jr/2
r = b - a in this case
B = (mu)j(b-a)/2 <------- this is not the same as the other 2 answers i got

You forgot to add the contribution from the "hole".

 

[Fusilli_Jerry89]

I thought that when you are beneath the surface of something, the outer current/mass/etc. does not contribute due to symmetry. Like gravity. (force of gravity increases linearly inside of a mass rather than 1/r^2 outside of the mass). So can't you just use ampere's law the same way, but just ignore the hole since if you are trying to find B on the inner edge?

 

[Fusilli_Jerry89]

wait, I think I may have got it. The magnetic field due to the hole is still under the hole, it doesn't stop at the edge of the hole. (Jeez!) But then if I subtract B from the hole I get this for B on inside edge of hole:

B(w/o hole) = (mu)j(b-a)/2
B(hole) = (mu)ja/2
B(w/o) - B(hole) = (mu)j[(b-a)-a]/2 = (mu)j(b-2a)/2 <----- I didn't do this right, did I?Also, if B is uniform within the hole, can someone please explain to me how? As far as I understand, B increases with the distance from the centre of the current. This means that B on the outside of the hole would be larger than B on the inside of the hole because the magnetic field that the hole contributes is the same on both edges. Hence, should B be larger on the outer edge of the hole? Maybe it's uniform because of what I said in my previous post? Does the fact that there's less total current involved when r = b - a rather than when r = b + a make this field uniform in the hole? (j(pi)(b-a)^2 is less that j(pi)(b+a)^2)

 

[Doc Al]

wait, I think I may have got it. The magnetic field due to the hole is still under the hole, it doesn't stop at the edge of the hole. (Jeez!) But then if I subtract B from the hole I get this for B on inside edge of hole:

B(w/o hole) = (mu)j(b-a)/2
B(hole) = (mu)ja/2
B(w/o) - B(hole) = (mu)j[(b-a)-a]/2 = (mu)j(b-2a)/2 <----- I didn't do this right, did I?

Almost. To find the field at any point you must add up both contributions. What is the relative direction of each field contribution?

 

[Doc Al]

Let me add two points:

(1) So far, you've only considered three points within the hole. Even if you succeed in showing that the field is the same at those points (as I'm sure you will once you correct your error), you still need to show that it's the same everywhere in the hole. Hint: Express the field as the vector sum of the two contributions.

(2) I don't know of any way to look at this so that it's "obvious" that the field is uniform within the hole. (Perhaps that's just lack of imagination on my part. )