Frozen-in magnetization in a long cylinder

[issacnewton]

"frozen-in" magnetization in a long cylinder

hi

I am doing this problem from Griffiths EM book on page 272 (3ed.) An infinitely long cylinder
of radius R, carries a frozen-in magnetization 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. Find the magnetic field inside and outside the cylinder by two different methods.

a)Locate all the bound currents and calculate the field they produce

b) Use Ampere's law to find \mathbf{H} and then get \mathbf{B} from

\mathbf{H}=\frac{1}{\mu_o}\mathbf{B}-\mathbf{M}

I am attaching the author's solution here. In part b), how does he invoke symmetry to
say that \mathbf{H} points in z direction ? Unlike the solution of a), in the
solution of b), nothing is assumed about the direction of \mathbf{B}, so
I don't understand the symmetry argument used by Griffiths...

 

[QuantumBunnii]

Symmetry, here, is merited through the definition of H:

H = \frac{1}{\mu}B - M

Since both M and B point in the z-direction (M is given; B is always points in the z-direction inside a solenoid), H must also point in the z-direction.

Hope this helped. :3

 

[issacnewton]

So to get the direction of H, we need to use the bound currents to get the direction of
B field right ? I thought we were not supposed to use the bound currents in the part b.

 

[issacnewton]

So to get the direction of H, we need to use the bound currents to get the direction of
B field right ? I thought we were not supposed to use the bound currents in the part b.

 

[QuantumBunnii]

Ah, I see what you're saying.
There are two ways to go about solving for the B-field here:

1) Ampere's Law, which corresponds to part a:

\ointB \bullet dl = \muI(enc)

2) Invoking the H-field, which corresponds to part b:

\ointH \bullet dl = I(free)

and

H = \frac{1}{\mu}B - M

Clearly, method 1 required a knowledge of the bound currents (how else would we solve for I(enc)?). In method 2, though, we know that the auxiliary-field H = 0, since there is no free current in this problem. Ergo, M = \frac{1}{\mu}B, and we have solved for B without any reference to the bound currents.
As it happens in this particular problem, the B and H fields must point in the same direction, as they differ only by a constant multiple. Perhaps this is what is meant by "symmetry" in this problem.

 

[issacnewton]

awesome quantum... makes sense now