Question 6.13: Suppose the field inside a large piece of magnetic material ...

Suppose the field inside a large piece of magnetic material is B _{0} , so that H _{0}=\left(1 / \mu_{0}\right) B _{0}- M , where M is a “frozen-in” magnetization. 

(a) Now a small spherical cavity is hollowed out of the material (Fig. 6.21). Find the field at the center of the cavity, in terms of B _{0} and M. Also find H at the center of the cavity, in terms of H _{0} and M

(b) Do the same for a long needle-shaped cavity running parallel to M.

(c) Do the same for a thin wafer-shaped cavity perpendicular to M.

Assume the cavities are small enough so M, B _{0}, \text { and } H _{0} are essentially constant. Compare Prob. 4.16. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite magnetization.] 

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(a) The field of a magnetized sphere is  \frac{2}{3} \mu_{0} M (Eq. 6.16), so B = B _{0}-\frac{2}{3} \mu_{0} M , with the sphere removed.

In the cavity, H =\frac{1}{\mu_{0}} B , \text { so } H =\frac{1}{\mu_{0}}\left( B _{0}-\frac{2}{3} \mu_{0} M \right)= H _{0}+ M -\frac{2}{3} M \Rightarrow H = H _{0}+\frac{1}{3} M .

(b)  The field inside a long solenoid is \mu_{0} K. Here K = M, so the field of the bound current on the inside surface of the cavity is  \mu_{0} K, pointing down. Therefore

B = B _{0}-\mu_{0} M .

H =\frac{1}{\mu_{0}}\left( B _{0}-\mu_{0} M \right)=\frac{1}{\mu_{0}} B _{0}- M \Rightarrow H = H _{0} .

(c)  This time the bound currents are small, and far away from the center, so B = B _{0} , while H =\frac{1}{\mu_{0}} B _{0}= H _{0}+ M \Rightarrow H = H _{0}+ M .

[Comment: In the wafer, B is the field in the medium; in the needle, H is the H in the medium; in the sphere (intermediate case) both B and H are modified.]

6.13

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