Magnetic field inside a Hollow asymmetrical cylinder

[mekansm]

Homework Statement

A cylinder of radius a has a hole drilled through it of radius b, the hole is parallel to the axis of the cylinder but is displaced from it
by a distance d in the x-direction where d < a - b. The asymmetric hollow cylinder has a uniform current density J = J_ze_z
Show that the magnetic field inside the hole is uniform and show that
B = (μ₀/2)J_zde_y

Homework Equations

Amperes law?

The Attempt at a Solution

I am at a complete loss, I don't know how to use ampere's law for something that is asymmetrical. Any ideas on how I should approach it?

 

[bobred]

Use Ampere's law twice. Get an expression for the magnetic field (in cartesian coordinates) for an intact cylinder and one for the displaced cylinder. Using superposition add both vector fields.

 

[turin]

How is this going to result in B parallel or antiparallel to J? I am stumped.

 

[bobred]

I think information given is only part of the question, that the current densities are antiparallel aligned along the z-axis and of the same magnitude.

 

[marcusl]

B will not be parallel to J, it is perpendicular. (Remember the right hand rule?) In your post, you note that J is along z and you are to show that B is uniform and along y.

 

[bobred]

I'm not saying B is parallel or antiparallel to J, but the original post doesn't give all the facts, rather a current density of $\textbf{J}=J\textbf{e}_z$ along the z-axis of a full cylinder plus the smaller clyinder displaced by d on the x-axis of $\textbf{J}=-J\textbf{e}_z$ in the opposite direction. Adding the field of the full cylinder and the offset one should give the expression for B.

 

[marcusl]

bobred, my comment was directed to turin, who did say that B is aligned with J in post #3.

 

[bobred]

Appolpgies marcusl.

 

[turin]

Sorry for the confusion, y'all. I was squinting when I first read the original post, and all I saw were z's. I tried to delete my post when I realized the e_y in the expression for B, but apparently I'm not allowed to delete it.

(Oh ya, and thanks to marcusl for replying to my question.)

 

[blink_again]

Sorry for the confusion, y'all. I was squinting when I first read the original post, and all I saw were z's. I tried to delete my post when I realized the e_y in the expression for B, but apparently I'm not allowed to delete it.

(Oh ya, and thanks to marcusl for replying to my question.)

I was wondering if you could help me with the magnetic field on the plane of the two axes. would it be the same as the hole?

 

[rude man]

Homework Statement

A cylinder of radius a has a hole drilled through it of radius b, the hole is parallel to the axis of the cylinder but is displaced from it
by a distance d in the x-direction where d < a - b. The asymmetric hollow cylinder has a uniform current density J = J_ze_z
Show that the magnetic field inside the hole is uniform and show that
B = (μ₀/2)J_zde_y

Homework Equations

Amperes law?

The Attempt at a Solution

I am at a complete loss, I don't know how to use ampere's law for something that is asymmetrical. Any ideas on how I should approach it?

The only way I can think of doing this is via Biot-Savart. A bit complex but doable that way.

Maybe solving del x B = 0 could be done but I don't know how.

I don't see how Ampere could be of any use. Any loop you put in the cavity will have a zero circulation integral of B dot dl.

 

[rude man]

I'm not saying B is parallel or antiparallel to J, but the original post doesn't give all the facts, rather a current density of $\textbf{J}=J\textbf{e}_z$ along the z-axis of a full cylinder plus the smaller clyinder displaced by d on the x-axis of $\textbf{J}=-J\textbf{e}_z$ in the opposite direction. Adding the field of the full cylinder and the offset one should give the expression for B.

EDIT: my apologies also, bobred! I didn't get your idea at first. Now I think it's got great potential, so to speak .

Put the large cylinder along z with cross-sectional center at x=y=0. Assume it's solid. Then the field is known everywhere inside the solid cylinder using Ampere, including where the hollow part sits.

Then, remove the solid cylinder and replace it with a solid cylinder with the shape and location of the hole, same current density except the current goes along
-z instead of +z. Again, the field is readily established using Ampere.

Then superpose the two B fields.

Elegant - thanks bobred!