Magnetic field at the centre of a cable that has an offset hole

[JNBirDy]

Homework Statement

[From Electromagetism by I. S. Grant & W. R. Phillips, Q4.7]

A cable of circular cross-section & diameter 0.2 m has a long cylindrical hole with a diameter of 0.001 m drilled in it parallel to the cable axis. The distance between the axis of the hole and cable axis is 0.005 m. The cable has a uniform steady current density of 10$^{5}$A/m$^{2}$ flowing in it, determine the magnetic field at the centre of the cable.

Answer: ∏ x 10$^{-6}$ Tesla

Homework Equations

Ampere's Law

The Attempt at a Solution

I know that to find the magnetic field I need to calculate the magnetic field of the cylinder carrying the current density and then add to the field of a current density -J in the opposite direction that would be occupying the hole. I've come up with the equation:

B = μ/2 * J * (r$_{1}$ - r$_{2}$) where r$_{1}$ is the distance from the axis of the main cylinder to the centre of it and r$_{2}$ is the distance from the axis of the hole to the centre of the main cylinder. However this gives me an answer of ∏ x 10$^{-4}$ T when I use r$_{1}$ = 0.01 m and r$_{2}$ = 0.005 m.

What am I doing wrong? Any help is appreciated.

 

[anathema]

Thank you for your question. It seems like you have the right idea, but there may be a small error in your calculation. Let's go through it step by step.

First, we can use Ampere's Law to find the magnetic field of the cylinder carrying the current density. Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the permeability of free space times the enclosed current. In this case, our closed loop is a circle with a radius of 0.2 m, and the enclosed current is the current density multiplied by the cross-sectional area of the cylinder.

So, we have:

∫B•dl = μ_0 * J * A

where B is the magnetic field, μ_0 is the permeability of free space, J is the current density, and A is the cross-sectional area of the cylinder.

Since we are looking for the magnetic field at the center of the cylinder, we can use a circular Amperian loop with a radius of 0.2 m. The line integral of the magnetic field around this loop is equal to the magnetic field at the center multiplied by the circumference of the loop (2πr). So we have:

B * 2πr = μ_0 * J * A

Solving for B, we get:

B = μ_0 * J * A / (2πr)

Now, we need to find the magnetic field of the current density -J in the opposite direction that would be occupying the hole. Since the hole has a diameter of 0.001 m, the radius of this current density is 0.0005 m. Using the same formula as before, we get:

B_hole = μ_0 * (-J) * A_hole / (2πr_hole)

Substituting in the values, we get:

B_hole = μ_0 * (-10^5 A/m^2) * (∏ * (0.0005)^2 m^2) / (2π*0.005 m)

= -μ_0 * 10^5 * (∏/4) / 0.01

= -μ_0 * (∏/4)

Now, we can add the two magnetic fields together to get the total magnetic field at the center of the cable:

B_total = B + B_hole

= μ_0 *