Magnetic field.

[Legendon]

Homework Statement

The figure above (right) shows a cross-sectional view of a coaxial cable. The center of the conductor is surrounded by a rubber layer, an outer conductor and another rubber layer. In a particular application, the current in the inner conductor is I1= 1.00 A out of the page and the current in the outer conductor is I2 = 3.00 A into the page. Assuming d = 1.00 mm, determine the magnitude and direction of the magnetic field at point a.

ans:200 μT towards top of page;

Homework Equations

The Attempt at a Solution

If we consider the field at a by virtue of I₁, then ∫Bds=I₁/μ
B=200μT which is the answer. But we have not yet considered magnetic field due to I₂. I suspect that B field due to I2 is 0 but can't get it.

 

[Simon Bridge]

You suspect that the field inside a hollow wire carrying a current is zero ... but you want to verify this? Why not compute it?

http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Physics/CircularFields.htm

 

[Legendon]

I still don't get why at a the magnetic field must be0 due to I₂...

 

[Simon Bridge]

The B field goes in a loop about the current right?
The strength of the field is proportional to the current inside the loop right?
How much of the second current is inside the loop that goes through a?
Your question amounts to asking "how come?"

If you have two wires carrying the same current, B exactly between them is zero right?

A ring of wires also has zero field in the middle - off-center, there will be a strong contribution from nearby wires but lots more weak contributions from the more distant ones.

The hollow wire is where those wires in the ring touch each other. Now the B field cancels itself out everywhere inside the wire.

This adding up the contribution of different parts of the situation is what the integral does in the formula you posted.

So if you are still unsure - try calculating the field due to the second current at point a.


But what about the field at point b?

 

[Legendon]

At B its much easier since the b field is the superposition of the respective fields.
If it was an electric field, then we can just draw a gaussian surface and just see what the surface encloses. Is it the same with magnetic fields to draw a surface and see what current it encloses?

I can see at the centre the B field is 0. And off centre there is strong contribution from near wires but lots of weak from the further ones. But in school the prof showed the answer he just took I1 and ignored I2. Didn't even mention about it being 0.

 

[Simon Bridge]

At B its much easier since the b field is the superposition of the respective fields.
If it was an electric field, then we can just draw a gaussian surface and just see what the surface encloses. Is it the same with magnetic fields to draw a surface and see what current it encloses?

It works well.
Did you look at the link?

You can divide the current part into small areas and build your integral for each point by the Biot-Savart law for each area.

Some people prefer the differential form, but it means you need to learn about curls.

I can see at the centre the B field is 0. And off centre there is strong contribution from near wires but lots of weak from the further ones. But in school the prof showed the answer he just took I1 and ignored I2. Didn't even mention about it being 0.

He should have explicitly mentioned that result - of course, you could have asked him :)

Part of these courses is usually to prove that the field is zero everywhere inside a hollow conductor for both B and E. That way you don't have to take the profs word for it - good training for later. It may have been a demonstration or an exercise from earlier in your course.

 

[Legendon]

ok thanks. :)
https://www.physicsforums.com/showthread.php?p=3581473#post3581473
If you don't mind can you help me out with this ?