Question

3. A cylindrical bar magnet has radius a and finite length 2, centered at the origin; it has uniform axial magnetization M. (a) Detail the correspondence with the solenoid in HW 5, #3: Evaluate the magnetzation surface-current density i and check the value found for the total magnetic dipole moment m against the definition of M as its volume-density. (b) Express B(o and H) on the axis, inside and outside the magnet, in terms of M, I , a, and乙Draw curves for both functions on the same graph: specify the values at:-0 (get the signs right!) Evaluate each function in the limit (i) 1-0, m constant, and (ii)-00, M constant (#3 above). (c) Detail the analogy between this magnet and the cylindrical bar-electret treated in HW 3, #4.

Answer 1

(a) Since magnetisation is constant, the bound current density is 0, but the surface current is

$\vec{K}_b=\vec{M}\times\hat{n}=M\hat{z}\times\hat{r}=M\hat{\varphi}$

(b) For an ideal linear magnet, inside the magnet

$\vec{H}=M\hat{z}\textup{ and }\vec{B}=2\mu_0M\hat{z}$

For outside we can consider an equivalent dipole moment

$\vec{m}=2\pi a^2lM\hat{z}$

so that for $r\gg1$

4 7T7.

outside there is no medium so magnetisation is 0, hence

$\vec{H}=\frac{m}{4\pi r^3}\left ( 2\cos\theta\hat{r}+\sin\theta\hat{\theta} \right )$