Question 5.11: A long filament with a current I is oriented along the z-axi......

As far as H is concerned, the hollow cylinder can be completely ignored, because it carries no free current. From symmetry considerations, H is of the form $\pmb{H} = H_{\phi }(\rho ) \pmb{a}_{\phi }$, and thus we choose a circle centered at the filament as the Amperian path.

(a) In the region 0 < ρ < a , Ampere’s circuital law gives

$\pmb{H} =H_{\phi } \pmb{a}_{\phi } = \frac{I}{2\pi \rho}\pmb{a}_{\phi } ,  and  \pmb{B} = \mu_{o} \pmb{H} =\frac{I\mu_{o}}{2\pi \rho} \pmb{a}_{\phi }$                                               (5- 83a)

In the region a ≤ ρ ≤ b , we have

$\pmb{H} = \frac{I}{2\pi \rho}\pmb{a}_{\phi } ,  and  \pmb{B} = \mu_{1} \pmb{H} =\frac{I\mu_{1}}{2\pi \rho} \pmb{a}_{\phi }$                                               (5- 83b)

In the region ρ > b , we have

$\pmb{H} = \frac{I}{2\pi \rho}\pmb{a}_{\phi } ,  and  \pmb{B} = \mu_{o} \pmb{H} =\frac{I\mu_{o}}{2\pi \rho} \pmb{a}_{\phi }$                                               (5- 83c)

Note that the expressions for H in Eq. (5-83) are the same, independent of the hollow cylinder.

(b) In the region a ≤ ρ ≤ b, inserting Eq. (5-83b) into Eq. (5-77) we obtain

$\boxed{\pmb{M} =\chi_{m } \pmb{H}}$                           [A/m]                                       (5-77)

$\pmb{M} =\chi_{m } \pmb{H} =\left\lgroup\frac{\mu_{1}}{\mu _{o} }- 1\right\rgroup \frac{I}{2\pi \rho}\pmb{a}_{\phi }$                                               (5- 84)

Taking the curl of Eq. (5-84) in cylindrical coordinates, we have

Then, from Eq. (5-70) we obtain

$\boxed{\pmb{J}_{m} = \nabla \times \pmb{M}}$                                [A/m²]                                   (5-70)

$\pmb{J}_{m} =0$.

(c) On the cylindrical surface at ρ = a , from Eq. (5-84) we have

$\pmb{M} = \left\lgroup \frac{\mu_{1}}{\mu _{o} }- 1\right\rgroup \frac{I}{2\pi a}\pmb{a}_{\phi }$                                                (5-85)

The unit normal to the surface at ρ = a is $\pmb{a}_{n }= -\pmb{a}_{\rho }$. From Eqs. (5-67) and (5-85) we obtain

$\boxed{\pmb{J}_{ms} = \pmb{M} \times \pmb{a}_{n}}$                               [A/m]                                    (5-67)

$\pmb{J}_{ms1} = \pmb{M} \times \pmb{a}_{n }= \left\lgroup \frac{\mu_{1}}{\mu _{o} }- 1\right\rgroup \frac{I}{2\pi a}\pmb{a}_{z}$                           (ρ = a)                         (5-86)

Similarly, on the cylindrical surface at ρ = b , from Eq. (5-84) we obtain

$\pmb{M} = \left\lgroup \frac{\mu_{1}}{\mu _{o} }- 1\right\rgroup \frac{I}{2\pi b}\pmb{a}_{\phi }$                                                (5-87)

The unit normal to the surface at ρ = b is $\pmb{a}_{n }= \pmb{a}_{\rho }$ , From Eqs. (5-67) and (5-87) we obtain

$\pmb{J}_{ms2} = \pmb{M} \times \pmb{a}_{n }= \left\lgroup \frac{\mu_{1}}{\mu _{o} }- 1\right\rgroup \frac{I}{2\pi b}\pmb{a}_{z}$                          (ρ = b)                           (5-88)

From Eqs. (5-86) and (5-88), we note that two surfaces of the hollow cylinder carry the magnetization surface currents of an equal amount, but flowing in the opposite directions.

If there is no free current in a magnetized material, Ampere’s circuital law tells us that ∇ × H = 0 in the material. In this case, according to Eq. (5- 77) and Eq. (5-70), we have ∇ × M = 0 and $\pmb{J}_{m} = 0$ in the material. Under these conditions, the total magnetic flux can be obtained by adding the external magnetic flux and the flux produced by $\pmb{J}_{ms}$  (see Example 5-12).