Find the magnetic field of a coaxial cable using Ampere's Law.

In this video, we calculate the magnetic field inside and outside an ideal coaxial cable by using Ampere's law. 🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/ We begin with a quick discussion of the construction of a coaxial cable: a thin wire runs down the middle of the cable, then a layer of insulating material separates the two conductors. Finally, a thin conducting foil is wrapped around the insulating layer. During the normal operation of coaxial cable, currents in the two conductors will be equal in magnitude but opposite in direction. Next, we show a more schematic view of the cable, viewing the end of the cable face-on. The current in the central conductor is coming out at us, and the current in the foil layer is going directly away from us. Now to find the magnetic field of a coaxial cable using Ampere's law, we need to define an Amperian loop inside the cable. We use the right hand rule to argue that the magnetic field is circumferential and clockwise in this region, and we set up the direction of the path integral (defining the direction of the dl's) so that B is always parallel to dl. Working on the Ampere's Law path integral, we use the fact that B is parallel to dl to conclude that the dot product reduces to the product of magnitudes of B and dl. Next, B must be constant by symmetry, so it can be factored out of the integral, and finally, the path integral of dl over the closed loop is just equal to the circumference, so the left side of Ampere's Law reduces to simply B*(2pir). On the right side of Ampere's law, we have mu_0*I_enc, but the enclosed current is just I. Solving for B, we get the classic result B=mu_0*I/2pi*r, which is the same result we obtained for a simple wire! Finally, we compute the magnetic field outside the coaxial cable. We set up an Amperian loop enclosing the entire cable, and the left side of Ampere's law simplifies in exactly the same way as before. This time, the NET enclosed current is zero, since the two currents penetrate the Amperian loop in opposite directions. So . . . the magnetic field outside a coaxial cable is zero, and we're done!