In a coaxial cable, the solid core carries a current \(i\) The sheath also carries a current \(i\) but in the opposite direction and 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

The first thing we need to do is to find the current passing through a loop between \(a\) and \(b\). We know that the current density \(J\) is equally distributed over each conductor. Thus, we can find the current passing through a loop of radius \(r\) in the sheath by integrating the current density over the area of the loop: \(I_r = \int J dA\) Since the current density is equally distributed, we know that \(J = \frac{i}{\pi a^2}\) for the core and \(J = \frac{i}{\pi (b^2 - a^2)}\) for the sheath. For a loop with an infinitesimally small thickness \(dr\) and radius \(r\) in the sheath, we have \(dA = 2 \pi r dr\). Hence, \(I_r = 2\pi r dr * \frac{i}{\pi (b^2 - a^2)}\). Integrating over the radius of the sheath, we get: \(I_r = \int_{a}^{r} 2\pi r' dr' * \frac{i}{\pi (b^2 - a^2)}\) \(I_r = \frac{2 \pi i}{\pi (b^2 - a^2)} * \int_{a}^{r} r' dr'\)

Now we will calculate the integral to find the total current within the loop: \(I_r = \frac{2 \pi i}{\pi (b^2 - a^2)} * \int_{a}^{r} r' dr'\) \(I_r = \frac{2 \pi i}{\pi (b^2 - a^2)} * \left[\frac{1}{2}r'^2\right]_{a}^{r}\) \(I_r = \frac{2 \pi i}{\pi (b^2 - a^2)} * \left[\frac{1}{2}(r^2 - a^2)\right]\) \(I_r = \frac{i}{b^2 - a^2} * (r^2 - a^2)\)

Now that we have the current inside the loop, we can use Ampère's law to find the magnetic field \(B\). Ampère's law states that the line integral of the magnetic field around a closed loop is equal to \(\mu_0\) times the total current passing through the loop: \(B * 2\pi r = \mu_0 I_r\)

Finally, we can solve for the magnetic field \(B\): \(B * 2\pi r = \mu_0 * \frac{i}{b^2 - a^2} * (r^2 - a^2)\) \(B = \frac{\mu_0 i}{2\pi (b^2 - a^2)} * (1 - \frac{a^2}{r^2})\) So, the magnetic field at a distance \(a < r < b\) from the center of the core is given by: \(B = \frac{\mu_0 i}{2\pi (b^2 - a^2)} * (1 - \frac{a^2}{r^2})\)