Use Ampere's Law to determine the magnetic field (a) inside...

Use Ampere's Law to determine the magnetic field (a) inside and (b) outside the toroid shown in the diagram below. Hint: use Path 1 for (a) and Path 2 for (b). Path 1 r Path 2 (c) Show that the self-inductance $L$ of the toroid, which contains $N$ loops each of diameter $d$, is $L = frac{mu_0 N^2 d^2}{8r}$. Assume that the $B$ field is uniform inside the toroid and that $r >> d$. Is this resultant consistent with the self-inductance for a solenoid?

Transcript

00:02 Hello, in the question we have given that use ampere's law to determine the magnetic field.

00:07 So let us see what is the ampere's law first.

00:11 So ampere's law, so it states that close integral b bar dot dl bar is equal to mu naught i enclosed.

00:25 So this is the formula.

00:28 Now what we have to do is we have to find out the magnetic field inside and outside the toroid.

00:37 So as shown in the diagram below, so they have mentioned one hint that we can use path 1 for a and path 2 for b.

00:48 So let us use the path 1.

00:51 So as the hint suggests, so let us use path 1.

00:56 So the first part we will use path 1.

01:00 So according to the path 1, the close integral.

01:04 So this ampere's law will be this b bar dot dl bar is equal to mu naught i.

01:16 I enclosed is i because this current in this loop is i.

01:20 So it will be mu naught i.

01:26 So but now there are n number of turns.

01:30 So i should multiply this by n because there are n number of turns in this coil given.

01:38 So from here this close integral over dl.

01:44 So this b times this will be 2 pi r because that radial distance, the distance between from the center to that path 1 is given to be r.

01:59 So it will be the circumference.

02:01 So dl is the length element.

02:03 So the length of that path 1 will be 2 pi r that is equal to mu naught n i.

02:09 So from here this b will be equal to mu naught n i divided by 2 pi r.

02:17 So this is what we get.

02:19 So this is the magnetic field which we get inside the toroid.

02:31 Now outside the toroid.

02:32 So outside the toroid.

02:35 So this is the magnetic field inside toroid.

02:41 Now outside the toroid, that is if i consider the path 2, so this magnetic field.

02:47 So again if we use b dot dl close integral b dot dl is equal to mu naught i enclosed.

02:59 So this mu naught, this i enclosed is 0.

03:03 So if the i enclosed, there is no current enclosed in this loop 2, in this path 2.

03:09 So if the enclosed current is 0, so therefore this magnetic field will be 0 in this case.

03:16 So for the path 2.

03:19 Now in the third path, they are saying that show that the self -inductance l of a toroid which contains n loop of diameter d.

03:28 So the diameter is given.

03:35 So the diameter is d...

Question

Please give Ace some feedback