Question 5.11: A long filament with a current I is oriented along the z-axi......
A long filament with a current I is oriented along the z-axis in free space. It is then enclosed by a long hollow cylinder of permeability \mu_{1}, with an inner radius a and an outer radius b, as shown in Fig. 5.18. Find
(a) H and B everywhere by Ampere’s circuital law and
(b) magnetization current density in the hollow cylinder,
(c) magnetization surface current densities on the inner and outer surfaces.
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
\nabla \times \pmb{M} =0Then, 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).