Question 5.12: Since there is no free current in the hollow cylinder in Exa......
Since there is no free current in the hollow cylinder in Example 5-11, the total magnetic flux is given by the sum of the external magnetic flux and the flux produced by the magnetization surface currents \pmb{J}_{ms1} and \pmb{J}_{ms2} residing in free space as shown in Fig. 5.19. Find B everywhere and compare it with the previous result in Example 5-11.
From symmetry considerations, B is of the form \pmb{B} = B_{\phi }(\rho ) \pmb{a}_{\phi } everywhere.
In the region 0 < ρ < a , Ampere’s circuital law gives
\oint_{C}{(\pmb{B}/ \mu _{o})\pmb{\cdot} d\pmb{l}} = I , or (B_{\phi} / \mu _{o}) 2\pi \rho = IThe magnetic flux density is therefore
\pmb{B} = B_{\phi} \pmb{a}_{\phi } = \frac{ I \mu _{o}}{2\pi \rho} \pmb{a}_{\phi } (0 < ρ < a ) (5-89a)
In the region a < ρ < b, Ampere’s circuital law gives
\oint_{C}{(\pmb{B}/ \mu _{o})\pmb{\cdot} d\pmb{l}} = I + 2\pi a J_{ms1}Inserting J_{ms1} expressed by Eq. (5-86) into the above equation, we have
\left\lgroup\frac{B_{\phi }}{ \mu _{o}}\right\rgroup 2 \pi \rho = I + 2\pi a \left\lgroup \frac{ \mu _{1}}{ \mu _{o}} – 1 \right\rgroup \frac{I}{2 \pi a}The magnetic flux density is therefore
\pmb{B} = B_{\phi} \pmb{a}_{\phi } = \frac{ I \mu _{1}}{2\pi \rho} \pmb{a}_{\phi } (a < ρ < b ) (5-89b)
In the region ρ > b , Ampere’s circuital law gives
\oint_{C}{(\pmb{B}/ \mu _{o})\pmb{\cdot} d\pmb{l}} = I + 2\pi a J_{ms1} – 2\pi b J_{ms2}Inserting J_{ms1} and J_{ms2} given in Eqs. (5-86) and (5-88) into the above equation we get
\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)
\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)
\left\lgroup\frac{B_{\phi }}{ \mu _{o}} \right\rgroup 2 \pi \rho = I + 2\pi a \left\lgroup \frac{ \mu _{1}}{ \mu _{o}} – 1 \right\rgroup \frac{I}{2 \pi a} – 2\pi b \left\lgroup \frac{ \mu _{1}}{ \mu _{o}} – 1 \right\rgroup \frac{I}{2 \pi b}The magnetic flux density is therefore
\pmb{B} = B_{\phi} \pmb{a}_{\phi } = \frac{ I \mu _{o}}{2\pi \rho} \pmb{a}_{\phi } ( ρ > b ) (5-89c)
The three results in Eq. (5-89) are the same as those in Eq. (5-83). In each region, H is obtained from B by the relation H = B /μ . An important point to remember is that H cannot be obtained directly from the magnetization currents.
\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)
\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)
\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)
If the magnetization M is induced in a magnetic material of a high permeability, by an externally applied magnetic field, it will in turn distort the magnetic field lines both inside and outside the material. Consider Fig. 5.20(a), which depicts the case when a small cylinder of a high permeability μ is brought into the region of a uniform B. For the sake of argument, we replace the magnetization surface current induced on the surface of the cylinder with a simple circular loop carrying a steady current. Then the magnetic flux lines are obtained, by using numerical methods, by computing and summing the external magnetic flux and the flux due to the current loop, and plotted to scale in Fig. 5.20(b). We see that the magnetic flux lines are concentrated in the interior of the material of a high μ, while the total magnetic flux is conserved inside and outside the material. This example shows that a structure made of a magnetic material of a high μ may be used to confine and guide magnetic flux lines.
