 #1
Tony Hau
 101
 30
 Homework Statement:
 Please see the picture below
 Relevant Equations:

1) ##\vec M = \chi_m \vec H##
2)##\vec B = \mu \vec H##
Boundary conditions:
1) ##H^{\perp}_{above}  H^{\perp}_{below}= (M^{\perp}_{above}  M^{\perp}_{below})##
2) ##H^{\parallel}_{above}  H^{\parallel}_{below} = K_{free} \times \hat n##
The question is as follows:
Solutions given only contain part a) to c), which is as follows:
So I now try to attemp d), e) and f).
d)
The magnetic field of a uniformly magnetized sphere is:
$$ \vec B =\frac{2}{3}\mu_{o} \vec M = \mu_{o}\vec H$$
$$\frac{2}{3}\vec M = \vec H$$
The perpendicular component of ##\vec H## is: $$ H^{\perp}_{s} = \vec H \cdot \hat r = \frac{2}{3} \vec M \cdot \hat r$$
e) ##H^{\parallel}_{above}  H^{\parallel}_{below} = \vec K_{free} \times \hat n =0##,
Therefore, $$H^{\parallel}_{l}(a) = H^{\parallel}_{s}(a)$$
f)
By $$\oint \vec H \cdot d\vec a=  \oint \vec M \cdot d\vec a$$
$$ H^{\perp}_{above}  H^{\perp}_{below}= (M^{\perp}_{above}  M^{\perp}_{below})$$
Given that:
##H^{\perp}_{below} = \vec H_{s}(a) \cdot \hat r##
##M^{\perp}_{above} = \vec M_{above} \cdot \hat r = \chi_{m} \vec H_{l}(a) \cdot \hat r##
##M^{\perp}_{below} = \vec M \cdot \hat r##
Therefore, $$ H^{\perp}_{above}  \vec H_{s}(a) \cdot \hat r =  (\chi_{m} \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r)$$
$$ H^{\perp}_{l} =  (\chi_{m} \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r)+ \vec H_{s}(a) \cdot \hat r =  [(\frac{\mu}{\mu_{o}}1) \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r] +\frac{2}{3}\vec M \cdot \hat r$$
I cannot proceed further, or I do not even know if I am correct. Can anyone give me any advice?
Solutions given only contain part a) to c), which is as follows:
So I now try to attemp d), e) and f).
d)
The magnetic field of a uniformly magnetized sphere is:
$$ \vec B =\frac{2}{3}\mu_{o} \vec M = \mu_{o}\vec H$$
$$\frac{2}{3}\vec M = \vec H$$
The perpendicular component of ##\vec H## is: $$ H^{\perp}_{s} = \vec H \cdot \hat r = \frac{2}{3} \vec M \cdot \hat r$$
e) ##H^{\parallel}_{above}  H^{\parallel}_{below} = \vec K_{free} \times \hat n =0##,
Therefore, $$H^{\parallel}_{l}(a) = H^{\parallel}_{s}(a)$$
f)
By $$\oint \vec H \cdot d\vec a=  \oint \vec M \cdot d\vec a$$
$$ H^{\perp}_{above}  H^{\perp}_{below}= (M^{\perp}_{above}  M^{\perp}_{below})$$
Given that:
##H^{\perp}_{below} = \vec H_{s}(a) \cdot \hat r##
##M^{\perp}_{above} = \vec M_{above} \cdot \hat r = \chi_{m} \vec H_{l}(a) \cdot \hat r##
##M^{\perp}_{below} = \vec M \cdot \hat r##
Therefore, $$ H^{\perp}_{above}  \vec H_{s}(a) \cdot \hat r =  (\chi_{m} \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r)$$
$$ H^{\perp}_{l} =  (\chi_{m} \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r)+ \vec H_{s}(a) \cdot \hat r =  [(\frac{\mu}{\mu_{o}}1) \vec H_{l}(a) \cdot \hat r  \vec M \cdot \hat r] +\frac{2}{3}\vec M \cdot \hat r$$
I cannot proceed further, or I do not even know if I am correct. Can anyone give me any advice?