Question

A coaxial wire consists of a copper core of radius \(1.00 \mathrm{~mm}\) surrounded by a copper sheath of inside radius \(1.50 \mathrm{~mm}\) and outside radius \(2.00 \mathrm{~mm}\). A current, \(i\), flows in one direction in the core and in the opposite direction in the sheath. Graph the magnitude of the magnetic field as a function of the distance from the center of the wire.

Short answer

Answer: In the three different regions, the magnetic field's magnitude behaves as follows: 1. Inside the core (radius less than 1 mm) the magnetic field is given by \(B = \frac{\mu_0 i}{2\pi r}\), where \(i\) is the current density, \(r\) is the distance from the center of the core, and \(\mu_0\) is the vacuum permeability. 2. Between the core and sheath (radius between 1.5 mm and 2 mm), the magnetic field depends on the current in the sheath and is given by \(B = \frac{\mu_0 i_s}{2\pi r}\), where \(r\) is the distance from the center of the coaxial wire, and \(i_s\) is the current in the sheath. 3. Outside the sheath (radius greater than 2 mm), the magnetic field's magnitude is zero as there is no net current enclosed.

Step-by-step solution

Analyze the different regions

In this problem, there are three regions where we need to determine the magnetic field: inside the core (radius less than 1 mm), between the core and the sheath (radius between 1.5 mm and 2 mm), and outside the sheath (radius greater than 2 mm).

Apply Ampere's Law within the core

When we are inside the core (radius less than 1 mm), we apply Ampere's Law. The magnetic field (B) is given by: \(B = \frac{\mu_0 i}{2\pi r}\) where \(i\) is the current density, \(r\) is the distance from the center of the core, and \(\mu_0\) is the vacuum permeability.

Apply Ampere's Law between the core and sheath

In this region (radius between 1.5 mm and 2 mm), the magnetic field depends on the current in the sheath. Let \(i_s\) represent the current in the sheath. Ampere's Law gives: \(B = \frac{\mu_0 i_s}{2\pi r}\) where \(r\) is the distance from the center of the coaxial wire.

Apply Ampere's Law outside the sheath

When we are outside the sheath (radius greater than 2 mm), there is no net current enclosed. The magnetic field (B) is zero.

Calculate the current in the sheath

To calculate the current in the sheath, we use the current density, since the current flows in the opposite direction in the sheath compared to the core. Assuming the volume of the sheath is a perfect cylindrical shell, \(i_s = i \frac{a^2 - b^2}{a^2}\) Where \(a\) is the outer radius of the sheath (2 mm), and \(b\) is the inner radius of the sheath (1.5 mm).

Combine the results and plot the graph

Now that we have determined the magnetic field in different regions, we will combine the results and provide a graph that represents the magnitude of the magnetic field as a function of distance from the center of the wire. The graph will show a drop in the magnetic field between the core and the sheath and then further decrease outside the sheath.