1) An infinitely long cylinder, of radius R, carries...

1) An infinitely long cylinder, of radius R, carries a "frozen-in" magnetization (M), parallel to the axis $\mathbf{M} = ks \hat{z}$. where k is a constant and s is the distance from the axis; there is no free current anywhere. a. Find the volume current density $\mathbf{J}_b$. b. Find the surface current density $\mathbf{K}_b$. c. Calculate the auxiliary field H and B inside the cylinder. d. Calculate the auxiliary field H and B outside the cylinder. e. Sketch $\mathbf{J}_b$ and $\mathbf{K}_b$, Amperian's loop on the cylinder.

Transcript

00:01 Consider a long solid cylinder which has a circular cross section of radius a and oriented with its axis in the z direction.

00:12 The current density is not constant across the cross section of the wire and it is given by j is equal to 2i0 upon pi a square in bracket 1 minus r by a square.

00:37 In the direction of k or z let's say this equation number one where r is the distance from the center of the wire and i0 is constant in this part we need to show that i0 is the current passing through the entire cross section of the wire consider a small segment a ring with a width of d r that you can see over here and at distance r from the center of the circle the current in this element equals the area of the element multiplied by the current density j that is the i is equal to j d a the area of the element equals the area of the rectangle with a width of d r and length of 2 pi r which is the circumference of a circle with a radius r so we can write d .a is equal to 2 pi r d r d r let's say it equation number 2.

01:52 Thus we can write d i is equal to 4 i 0 upon a square into 1 minus r by a whole square r dr dr.

02:21 The total current passing through the wire equals the integration over the whole wire.

02:28 So here we are taking the integration that is from r is equal to 0 to a.

02:33 So we can write i is equal to in the integration of 0 to a, di is equal to, let's take the constant term outside the integration.

02:48 So which are 4 i0 upon a square and in the integration 0 to a 1 minus r by a square r d r and the solution of this integration will give the answer is equivalent to i 0 so we can say i is equal to i 0 now in the second part we need to use amper law to calculate a magnetic field b for the case r is greater than or equal a.

03:35 The current enclosed by amper's law path is i0.

03:40 Therefore amper's law gives us the equation bdl is equal to mu 0 i enclosed.

03:55 But we know bdl is equal to b2 pi r is equal to mu zero and i enclosed is i0 and from this we can write b is equal to mu zero i 0 upon 2 pi r.

04:20 In the third case we need to find the enclosed current for r is less than or equal a...

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