(III) A toroid has a rectangular cross section as shown in Fig...

Question

(III) A toroid has a rectangular cross section as shown in Fig. 30–26. Show that the self-inductance is

$L = \frac{μ_{0} N^{2} h}{2 π} \ln \frac{r_{2}}{r_{1}}$

where N is the total number of turns and r₁, r₂ and h are the dimensions shown in Fig. 30–26. [Hint: Use Ampère’s law to get B as a function of r inside the toroid, and integrate.]

Accepted Answer

Welcome back. Everyone. In a cutting edge engineering lab, a four sided toroidal coil with a rectangular cross section is being tested for its magnetic properties. This specialized coil is designed with an inner radius A, an outer radius B and a vertical height D for the cross section with N total turns of wire wound around the toroid. Calculate the self inductance L of the coil. A hint is for us to use Ampere's law to determine the magnetic field B as a function of the radial distance are within the toro and integrate to find the magnetic flux allowing you to derive the self inductance. L for our answer choices. After a diagram, it says the self inductance equals Muno NIH divided by two pi multiplied by the natural log of B divided by A B says it's Muno N squared H divided by two pi multiplied by the natural log of B divided by AC and H divided by two pi multiplied by the natural log of B divided by A and the D Muno I divided by two pi multiplied by the natural log of B divided by A. Now let's get back into our problem. OK. We, we essentially want to find a self inductance L of the coil. The question is, what do we know about self inductance? Well, recall that the South Ind's ill, OK, is defined as the number of turns multiplied by the magnetic flux due to the magnetic field B divided by the current. OK. And now, in this case, we already know that we have the current in our expression, we have the number of turns in our expression, but we will need to find an expression to represent our magnetic flux. So if we can figure that out, then we'll be able to find a self inductance. L. So let's give it a try. What do we know that can help us? Well, when we go back to our hint, the first thing it tells us is that we can use Ampere's Law to determine the magnetic field B as a function of the radial distance. R. What does that mean? Well, according to Ampere's Law, OK, the line integral of our magnetic field as a function of the distance is equal to the permeability of free space Muno multiplied by the current enclosed in the loop. OK. And the current enclosed by the path of radius Rie and C is equal to the number of turns. N multiplied by I the current in each turn. So what this tells us then that from Empire's law, we can tell that B multiplied by two pi R is going to be equal to Muno multiplied by N I. Now, if we go ahead and divide through by two pi R, then this tells us that our magnetic field B as a function of the radial distance R equals mu node and I divided by two pi R know that we have an expression for our magnetic field. OK. As a function of the radial distance again, we can go back to our hint. It tells us that we can use this to integrate to find the magnetic flux. What does that mean? Well, no, yeah, we can find the magnetic flux through a single turn of the choir because the flux is the integral of the magnetic field B over the cross sectional area of the to ride. Now, the cross sectional area OK is a rectangle if we go back to our problem with height D and the thickness from R equals A to R equals B OK. So this is our thickness from R equals A to R equals B. OK. Now the differential area element in a cylindrical, in cylindrical coordinates is D A equals HDR. OK. So we let's, let's make a note here where D is equal to HDR. OK. Then the flux, the flux through one turn is going to be equal to the integral between A and B of BR with respect to air, which is going to be the integral between B and A of Muno N I divided by two pi R multiplied by HDR. Now, if we try to get our constant or of our term, OK, then we're going to get the flux to be equal to Muno NIH divided by two pi multiplied by the integral between A and B of one divided by RDR. This is how we can solve or flux. No, that means we'll have to integrate the integral of, of one divided by R between the bones of A and B. Now, when we do that, OK, we get the flux to be equal to Muno NIH divided by two pi multiplied by the natural log of B minus the natural log of A or the natural log of B divided by A based on what we know about natural logs. Now that we have an expression for the flux, that means we can get the self inductance because now that tells us then that L is going to be equal to N divided by I multiplied by or flux of mu not NIH divided by two pi multiplied by the natural log of B divided by A. Let's cancel I from both terms. OK? And now that is gonna give us Muno N squared H divided by two pi multiplied by the natural log of B divided by A. Thus, we can use this expression to describe the self inductance L of the coil. If we look back at our answer choices that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Self-Inductance

Ampère’s Law

Magnetic Field in a Toroid

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