Question

In a coaxial cable, the solid core carries a current \(i .\) The sheath also carries a current \(i\) but in the opposite direction, and it has an inner radius \(a\) and an outer radius \(b\). The current density is equally distributed over each conductor. Find an expression for the magnetic field at a distance \(a

Short answer

Answer: The magnetic field at a distance \(a < r < b\) from the center of the core in a coaxial cable is given by the expression \(B = \frac{\mu_{0} \times i \times r}{2 \pi (b^2 - a^2)}\).

Step-by-step solution

Understanding and Defining Parameters

We are given a coaxial cable with the following parameters: - Solid core carrying a current \(i\). - Sheath carrying a current \(i\) but in the opposite direction. - The sheath has an inner radius \(a\) and an outer radius \(b\). - The current density \(J\) is equally distributed over each conductor. - We need to find the magnetic field at a distance \(a < r < b\) from the center of the core.

Determine the Current Density for the Sheath

For the sheath, since the current density is equally distributed over the sheath, the total current \(i\) is distributed over the area between the inner radius \(a\) and the outer radius \(b\). The area of the annular region can be calculated as: Area \( = \pi (b^2 - a^2)\) Now we can determine the current density \(J\) for the sheath as follows: \(J = \frac{i}{\pi (b^2 - a^2)}\)

Determine the Enclosed Current at a Distance r

Since we are interested in the magnetic field at a distance \(a < r < b\) from the center of the core, we will consider a circular loop of radius r and determine the total enclosed current within that loop. The enclosed area within the loop at radius r is: Area \( = \pi r^2\) The enclosed current \(i_{enc}\) within the loop at radius r can be calculated as: \(i_{enc} = J \times (\pi r^2)\) Plugging J from step 2, we get: \(i_{enc} = \frac{i}{\pi (b^2 - a^2)} \times (\pi r^2)\)

Apply Ampere's Law to Find the Magnetic Field

Next, we will apply Ampere's law to determine the magnetic field \(B\). Ampere's law states that the closed integral of \(\vec{B} \cdot d\vec{l}\) around the path is equal to \(\mu_{0} \times i_{enc}\). Since we are considering a circular path, the path length is equal to the circumference of the circle: \(d l = 2 \pi r\) Now applying Ampere's law: \(B \times (2 \pi r) = \mu_{0} \times i_{enc}\) We can now solve for the magnetic field \(B\): \(B = \frac{\mu_{0} \times i_{enc}}{2 \pi r}\) From step 3, we know the enclosed current \(i_{enc}\): \(B = \frac{\mu_{0} \times i \times r^2}{2 \pi r(b^2 - a^2)}\) Finally, the magnetic field at a distance \(a < r < b\) from the center of the core is: \(B = \frac{\mu_{0} \times i \times r}{2 \pi (b^2 - a^2)}\)