Three infinitely long, concentric, conducting, cylindrical...

Three infinitely long, concentric, conducting, cylindrical shells of radii a, b, c, (a < b < c) carry currents $I_1$ and $I_2$ on the inner cylinders which are returned on the outer cylinder: i) Determine the magnetic field B(r) everywhere by using Ampere's law. ii) Find the magnetostatic potential energy per unit length. iii) Using the previous result, give the self inductances $L_{11}$ and $L_{22}$ and the mutual inductance M = $L_{12}$ = $L_{21}$ per unit length of the two circuits.

Transcript

00:01 Hello, in the question we have given an infinitely long cylinder with an inner radius a and outer radius b has a linear magnetic susceptibility of chi m.

00:10 So, a wire carrying a current i runs along the centre line as shown in the figure.

00:15 So, we have the cylinder with inner and outer radius a and b and from the centre line of the cylinder we have this current carrying wire.

00:23 Now, this current carrying wire creates a magnetic field.

00:26 So, it magnetizes.

00:27 So, this magnetizes.

00:29 So, it creates a magnetic field and which magnetizes this cylinder because it is a magnetized cylinder.

00:36 So, and this magnetized cylinder will produce its own magnetic field.

00:41 So, we have to find out the h as a function of s.

00:47 Now, see here they have mentioned, they are asking us to calculate the four quantities.

00:51 The first one is we have to calculate h as a function of h s.

00:56 So, as they are saying we will do the same thing.

00:59 So, we will take a loop of radius s.

01:10 So, we have taken this loop as we do in the ampere's law.

01:13 So, they are asking us to calculate this close integral h bar dot dl bar is equal to i free.

01:20 Now, note over here that this integral is possible.

01:24 Why we are not finding directly b, why we are not using ampere's law that is b dot dl is equal to mu naught i because there i should know the surface current and the volume bound current and for which a priori i don't know.

01:40 So, that is why i cannot calculate b directly.

01:43 So, it is very simple.

01:44 So, that it is very intuitive that i will first find out this and from this i can find out this b which is very easy step to do.

01:52 So, that is why we are starting with this.

01:55 So, and this question is guiding us in the same way.

02:00 So, now let us do this.

02:01 So, this will be h into integration over dl will be 2 pi s that is equal to i.

02:07 So, from here h will be equal to i divided by 2 pi s in the phi cap direction by using the right hand thumb rule.

02:18 So, now you are not using the right hand thumb rule for now.

02:21 So, just use it and then in the second part we have to find out the magnetic field as a function of s.

02:28 Now, we know that magnetic field is given by mu r mu naught h.

02:36 Basically, it is mu h because it is a magnetized cylinder.

02:40 So, it is mu h, but mu is given by mu r times mu naught.

02:44 So, now mu r we have a relation that is b is equal to 1 plus chi m mu naught and h is i divided by 2 pi s.

02:58 So, this is the magnetic field which we get.

03:06 So, now in the third part they are asking us to find out the bound surface currents.

03:21 So, now how to find out the bound surface current? so, we have to find out this kb.

03:27 Now, in order to find out this kb, it is m dot n cap.

03:33 So, this we have to find out.

03:35 So, m is required...

Question

Please give Ace some feedback