Chapter 8, Problem 35P

(a)

To determine

Calculate the magnetic flux density $B_1$.

Answer to Problem 35P

The magnetic flux density $B_1$ is $4a_{\rho }+15a_{\varphi }-8a_z\text{m}{\text{Wb/m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Given that,

$B_2=10a_{\rho }+15a_{\varphi }-20a_z{\text{m Wb/m}}^{\text{2}}$

Consider,

$B_2=\left(B_{\rho },B_{\varphi },B_z\right)$ in $\text{m}{\text{Wb/m}}^{\text{2}}$.

Consider the expression for the magnetic boundary condition.

$B_{1n}=B_{2n}$        (1)

Therefore, the equation (1) becomes,

$B_{1n}=B_{2n}=15a_{\varphi }\left\{\because \text{magentic}\text{flux}\text{density}\text{along}\varphi \text{-direction}\right\}$

Thus, the tangential component of the magnetic flux density for region 2 is,

$B_{2t}=10a_{\rho }-20a_z$

The tangential component of magnetic field intensity for region 1 is,

$H_{1t}=K+H_{2t}$

Since K is the free current density which is equal to zero then,

$\begin{array}{l}{H}_{2t}=H_{1t}\\ \frac{B_{2t}}{{\mu }_2}=\frac{B_{1t}}{{\mu }_1}\left\{\because H=\frac{B}{\mu }\right\}\\ B_{1t}=\frac{{\mu }_1}{{\mu }_2}{B}_{2t}\end{array}$

Substitute $5{\mu }_{\text{o}}$ for ${\mu }_2$, $2{\mu }_{\text{o}}$ for ${\mu }_1$, and $10a_{\rho }-20a_z$ for $B_{2t}$ in above equation.

$\begin{array}{c}{B}_{1t}=\frac{2{\mu }_{\text{o}}}{5{\mu }_{\text{o}}}\left(10a_{\rho }-20a_z\right)\\ =0.4\left(10a_{\rho }-20a_z\right)\\ =4a_{\rho }-8a_z\end{array}$

Consider the expression for the magnetic flux density $B_1$ in region 1.

$B_1=B_{1t}+B_{1n}$        (2)

Here,

$B_{1t}$ is the tangential component of magnetic flux density for region 1, and

$B_{1n}$ is the normal component of magnetic flux density for region 1.

Substitute $4a_{\rho }-8a_z$ for $B_{1t}$ and $15a_{\varphi }$ for $B_{1n}$ in equation (2).

$\begin{array}{c}{B}_1=4a_{\rho }-8a_z+15a_{\varphi }\\ =4a_{\rho }+15a_{\varphi }-8a_z\text{m}{\text{Wb/m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the magnetic flux density $B_1$ is $4a_{\rho }+15a_{\varphi }-8a_z\text{m}{\text{Wb/m}}^{\text{2}}$.

(b)

To determine

Calculate the energy densities in the two medium 1 and 2.

Answer to Problem 35P

The energy densities in the two medium 1 and 2 are $60.68\text{J}/{\text{m}}^{\text{3}}$ and $57.7\text{J}/{\text{m}}^{\text{3}}$ respectively.

Explanation of Solution

Calculation:

The energy density in medium 1 is calculated as follows,

$W_{m1}=\frac{1}{2}\frac{{\left|B_1\right|}^2}{{\mu }_1}$        (3)

Here,

${\mu }_1$ is the permeability of region 1, and

$B_1$ is the magnetic flux density of region 1.

Substitute $2{\mu }_{\text{o}}$ for ${\mu }_1$ and $\left(4a_{\rho }+15a_{\varphi }-8a_z\right)\times {10}^{-3}$ for $B_1$ in equation (3).

$\begin{array}{c}{W}_{m1}=\frac{1}{2}\frac{{\left|\left(4a_{\rho }+15a_{\varphi }-8a_z\right)\times {10}^{-3}\right|}^2}{\left(2{\mu }_{\text{o}}\right)}\\ =\frac{\left[4^2+{15}^2+{\left(-8\right)}^2\right]\times {10}^{-6}}{4{\mu }_{\text{o}}}\\ =\frac{305\times {10}^{-6}}{4{\mu }_{\text{o}}}\\ =\frac{76.25\times {10}^{-6}}{{\mu }_{\text{o}}}\end{array}$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$ in above equation.

$\begin{array}{c}{W}_{m1}=\frac{76.25\times {10}^{-6}}{4\pi \times {10}^{-7}}\\ =60.68\frac{\text{J}}{{\text{m}}^{\text{3}}}\end{array}$

The energy density in medium 2 is calculated as follows,

$W_{m2}=\frac{1}{2}\frac{{\left|B_2\right|}^2}{{\mu }_2}$        (4)

Here,

${\mu }_2$ is the permeability of region 2, and

$B_2$ is the magnetic flux density of region 2.

Substitute $5{\mu }_{\text{o}}$ for ${\mu }_2$ and $\left(10a_{\rho }+15a_{\varphi }-20a_z\right)\times {10}^{-3}$ for $B_2$ in equation (4).

$\begin{array}{c}{W}_{m2}=\frac{1}{2}\frac{{\left|\left(10a_{\rho }+15a_{\varphi }-20a_z\right)\times {10}^{-3}\right|}^2}{\left(5{\mu }_{\text{o}}\right)}\\ =\frac{\left[{10}^2+{15}^2+{\left(-20\right)}^2\right]\times {10}^{-6}}{10{\mu }_{\text{o}}}\\ =\frac{725\times {10}^{-6}}{10{\mu }_{\text{o}}}\\ =\frac{72.5\times {10}^{-6}}{{\mu }_{\text{o}}}\end{array}$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$ in above equation.

$\begin{array}{c}{W}_{m2}=\frac{72.5\times {10}^{-6}}{4\pi \times {10}^{-7}}\\ =57.7\frac{\text{J}}{{\text{m}}^{\text{3}}}\end{array}$

Conclusion:

Thus, the energy densities in the two medium 1 and 2 are $60.68\text{J}/{\text{m}}^{\text{3}}$ and $57.7\text{J}/{\text{m}}^{\text{3}}$ respectively.