QUESTION 1 a) As shown in Fig. 1, a coaxial cable has...

QUESTION 1 a) As shown in Fig. 1, a coaxial cable has conductor radii a = 0.4 cm and b = 0.8 cm. Material of relative permeability $mu_r$ = 2 exists in the region between the two conductors. DERIVE the inductance per unit length. You need to include your assumptions and steps. (50%) p=b p=a Fig. 1 Conducting cylinders b) As shown in Fig. 2. The plane z = 0 defines the boundary between two magnetic materials with different relative permeability. At Point $P_1$ on the surface the magnetic field $H_1$ in Material 1 exhibits an angle $alpha_1$ = 30° with respect to the normal direction of the boundary surface. Find the direction angle $alpha_2$ of the magnetic field $H_2$ in Material 2 at Point $P_1$. Assume that there is no surface current. (50%) Material 1 $H_1$ $mu_{r1} = 1$ $P_1$ $z = 0$ $H_2$ $mu_{r2} = 2$ Material 2 Fig. 2

Transcript

00:01 Hello, in the question we have given that as shown in figure a coaxial cable has a conductor radii.

00:07 So, a is given that is 0 .4 centimeters.

00:11 So, the inner conductor has radius of 0 .4 centimeters and b is 0 .8 centimeters.

00:17 So, the material of relative permeability that is mu r is equal to 2 exists in the region between the two conductors.

00:25 So, in between these two conductors, this material is present.

00:28 So, what we have to do is we have to derive inductance per unit length formula.

00:33 So, for this, the magnetic field, the magnetic field due to coaxial cable is given by.

00:49 So, it is b is equals to mu times i divided by 2 pi rho, i will write.

00:56 So, now over here mu naught.

00:58 So, here mu is equals to mu naught mu r i divided by 2 pi rho.

01:06 Now, i know that the flux magnetic flux is equal to integral b dot ds.

01:15 So, this is the flux the magnetic flux.

01:19 So, this will be phi will be equal to integral mu naught mu r i divided by 2 pi rho into l times d rho.

01:34 So, from here i can write this as.

01:41 So, this constants we can take out.

01:43 So, mu naught mu r i divided by 2 pi and l will also come out from a to b.

01:50 So, this integration limit is from a to b d rho by rho.

01:55 So, from here phi will be equal to mu naught mu r i divided by 2 pi l into ln of rho from a to b.

02:08 The limits of integration is from a to b.

02:11 Now, so, but now this can be written as phi is equals to mu naught mu r divided by 2 pi i l times ln b minus ln of a.

02:27 So, now this becomes simple.

02:31 So, we know that phi is equals to l i.

02:37 So, if i compare if i compare these two equations.

02:42 So, if i remove this i.

02:44 So, this thing mu naught mu r 2 pi and this l and ln b minus ln a will be nothing but l.

02:52 So, will mu naught mu r i sorry i will not be there l will be there divided by 2 pi ln of here it will be b by a.

03:09 Now, so, this is the required formula.

03:13 So, l by l will be equal to mu naught mu r divided by 2 pi ln of b by a.

03:22 So, now if we plug the values we will get this thing.

03:26 So, l by capital l by l will be equal to mu naught is 4 pi into 10 raised to minus 7 into mu r is 2 divided by 2 pi ln of b is 0 .8 divided by 0 .4.

03:39 So, this is ln 2 ln 2 is 0 .693...

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