- #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?