Solve Ampere's Law Problem for Uniform Current Density

[Crazy Gnome]

Homework Statement

A conducting slab of thickness a is bounded by the planes z= $$\pm$$a/2 and carries a uniform current density J=J (y hat)

Use the integral form of Ampere's law to to find magnetic field everywhere
2. Relevant equations

Integral for of Amperes law: Integral (B \bullet ds) = \mu ₀ J_enclosed

The Attempt at a Solution

All of the examples in class and in the book use cylindrical symmetry so I was a bit perplexed as how to approach taking an integral over a closed path dotted into a B in cylindrical coordinates.

 

[Redbelly98]

Try using a rectangular loop for the integration. Each leg of the loop is either perpendicular or parallel to B.

 

[nlightner]

To solve this problem, we can use the integral form of Ampere's law and choose a closed path that follows the symmetry of the problem. In this case, we can choose a rectangular path that follows the edges of the conducting slab.

Starting at the bottom left corner of the slab, we can label the sides of the rectangle as follows:
- The bottom side: from (x=-a/2, y=-a/2, z=-a/2) to (x=a/2, y=-a/2, z=-a/2)
- The right side: from (x=a/2, y=-a/2, z=-a/2) to (x=a/2, y=a/2, z=-a/2)
- The top side: from (x=a/2, y=a/2, z=-a/2) to (x=-a/2, y=a/2, z=-a/2)
- The left side: from (x=-a/2, y=a/2, z=-a/2) to (x=-a/2, y=-a/2, z=-a/2)

Using the symmetry of the problem, we can see that the magnetic field will only have a non-zero component along the z-direction. Therefore, we can rewrite the integral form of Ampere's law as:

∫B·ds = μ0Jenclosed = μ0Jaz∫ds

Where Jaz is the component of the current density along the z-direction.

Now, we can break down the integral into four parts, corresponding to the four sides of the rectangle:

∫B·ds = μ0Jaz∫ds = μ0Jaz∫(dx,y,z = -a/2)^(x=a/2,y,z=-a/2) + μ0Jaz∫(x=a/2,y,z=-a/2)^(x=a/2,y,z=a/2) + μ0Jaz∫(x=a/2,y,z=a/2)^(x=-a/2,y,z=a/2) + μ0Jaz∫(x=-a/2,y,z=a/2)^(x=-a/2,y,z=-a/2)

Since the current density Jaz is constant throughout the slab, we can pull it out of the integral:

∫B·ds =