Chapter 8, Problem 8.33P

To determine

(a)

The expression of $H_{\varphi }(\rho )$ for $z>0$.

Answer to Problem 8.33P

The required expression is, $H_{\varphi }(\rho )=\frac{23.9}{\rho }\text{A/m}$.

Explanation of Solution

Given Information:

The toroid is having square cross section, $2.5\text{cm}<\rho <3.5\text{cm}$ , $-0.5\text{cm}<z<0.5\text{cm}$ . The flux in ${\text{a}}_{\varphi }$ direction is established by an mmf of $150\text{A}\cdot \text{t}$.

   ${\mu }_r=\{\begin{array}{cc}20& -0.5\text{cm}<z<0\\ 10& 0<z<0.5\text{cm}\end{array}$

Calculation:

The magnetic field intensity,

   $\begin{array}{l}{H}_{\varphi }(\rho )=\frac{NI}{2\pi \rho }=\frac{150}{2\pi \rho }\\ =\frac{23.9}{\rho }\text{A/m}\end{array}$

Conclusion:

The required expression is, $H_{\varphi }(\rho )=\frac{23.9}{\rho }\text{A/m}$.

To determine

(b)

The expression of $B_{\varphi }(\rho )$ for $z>0$.

Answer to Problem 8.33P

The required expression is, $B_{\varphi }(\rho )=\frac{3\times {10}^{-4}}{\rho }{\text{Wb/m}}^2$.

Explanation of Solution

Given Information:

The toroid is having square cross section, $2.5\text{cm}<\rho <3.5\text{cm}$ , $-0.5\text{cm}<z<0.5\text{cm}$ . The flux in ${\text{a}}_{\varphi }$ direction is established by an mmf of $150\text{A}\cdot \text{t}$.

   $\begin{array}{l}{\mu }_r=\{\begin{array}{cc}20& -0.5\text{cm}<z<0\\ 10& 0<z<0.5\text{cm}\end{array}\\ H_{\varphi }(\rho )=\frac{23.9}{\rho }\text{A/m}\end{array}$

Calculation:

The magnetic flux density,

   $\begin{array}{l}{B}_{\varphi }(\rho )={\mu }_0{\mu }_r{H}_{\varphi }=\left(4\pi \times {10}^{-7}\right)\left(10\right)\frac{23.9}{\rho }\\ =\frac{3\times {10}^{-4}}{\rho }{\text{Wb/m}}^2\end{array}$

Conclusion:

The required expression is, $B_{\varphi }(\rho )=\frac{3\times {10}^{-4}}{\rho }{\text{Wb/m}}^2$.

To determine

(c)

The expression of ${\varphi }_{z>0}$.

Answer to Problem 8.33P

The required value is, ${\varphi }_{z>0}=5\times {10}^{-7}\text{Wb}$.

Explanation of Solution

Given Information:

The toroid is having square cross section, $2.5\text{cm}<\rho <3.5\text{cm}$ , $-0.5\text{cm}<z<0.5\text{cm}$ . The flux in ${\text{a}}_{\varphi }$ direction is established by an mmf of $150\text{A}\cdot \text{t}$.

   $\begin{array}{l}{\mu }_r=\{\begin{array}{cc}20& -0.5\text{cm}<z<0\\ 10& 0<z<0.5\text{cm}\end{array}\\ B_{\varphi }(\rho )=\frac{3\times {10}^{-4}}{\rho }{\text{Wb/m}}^2\end{array}$

Calculation:

The magnetic flux,

   $\begin{array}{l}{\varphi }_{z>0}={\displaystyle \underset{S}{\int }\text{B}\cdot d\text{S}}={\displaystyle \underset{0}{\overset{0.5\times {10}^{-2}}{\int }}{\displaystyle \underset{2.5\times {10}^{-2}}{\overset{3.5\times {10}^{-2}}{\int }}\frac{3\times {10}^{-4}}{\rho }d\rho }dz}\\ =3\times {10}^{-4}\left(0.5\times {10}^{-2}\right)\ln \left(\frac{3.5\times {10}^{-2}}{2.5\times {10}^{-2}}\right)\\ =5\times {10}^{-7}\text{Wb}\end{array}$

Conclusion:

The required value is, ${\varphi }_{z>0}=5\times {10}^{-7}\text{Wb}$.

To determine

(d)

The expressions of $H_{\varphi }(\rho )$ , $B_{\varphi }(\rho )$ , and ${\varphi }_{z<0}$ for $z<0$.

Answer to Problem 8.33P

The required expressions are,

   $\begin{array}{l}{H}_{\varphi }(\rho )=\frac{23.9}{\rho }\text{A/m}\\ B_{\varphi }(\rho )=\frac{6\times {10}^{-4}}{\rho }{\text{Wb/m}}^2\\ {\varphi }_{z<0}=1.0\times {10}^{-6}\text{Wb}\end{array}$.

Explanation of Solution

Given Information:

The toroid is having square cross section, $2.5\text{cm}<\rho <3.5\text{cm}$ , $-0.5\text{cm}<z<0.5\text{cm}$ . The flux in ${\text{a}}_{\varphi }$ direction is established by an mmf of $150\text{A}\cdot \text{t}$.

   ${\mu }_r=\{\begin{array}{cc}20& -0.5\text{cm}<z<0\\ 10& 0<z<0.5\text{cm}\end{array}$

Calculation:

The magnetic field intensity,

   $\begin{array}{l}{H}_{\varphi }(\rho )=\frac{NI}{2\pi \rho }=\frac{150}{2\pi \rho }\\ =\frac{23.9}{\rho }\text{A/m}\end{array}$

The magnetic flux density,

   $\begin{array}{l}{B}_{\varphi }(\rho )={\mu }_0{\mu }_r{H}_{\varphi }=\left(4\pi \times {10}^{-7}\right)\left(20\right)\frac{23.9}{\rho }\\ =\frac{6\times {10}^{-4}}{\rho }{\text{Wb/m}}^2\end{array}$

The magnetic flux,

   $\begin{array}{l}{\varphi }_{z<0}={\displaystyle \underset{S}{\int }\text{B}\cdot d\text{S}}={\displaystyle \underset{-0.5\times {10}^{-2}}{\overset{0}{\int }}{\displaystyle \underset{2.5\times {10}^{-2}}{\overset{3.5\times {10}^{-2}}{\int }}\frac{6\times {10}^{-4}}{\rho }d\rho }dz}\\ =6\times {10}^{-4}\left(0.5\times {10}^{-2}\right)\ln \left(\frac{3.5\times {10}^{-2}}{2.5\times {10}^{-2}}\right)\\ =1.0\times {10}^{-6}\text{Wb}\end{array}$

Conclusion:

The required expressionsare,

   $\begin{array}{l}{H}_{\varphi }(\rho )=\frac{23.9}{\rho }\text{A/m}\\ B_{\varphi }(\rho )=\frac{6\times {10}^{-4}}{\rho }{\text{Wb/m}}^2\\ {\varphi }_{z<0}=1.0\times {10}^{-6}\text{Wb}\end{array}$.

To determine

(e)

The value of ${\varphi }_{\text{total}}$.

Answer to Problem 8.33P

The required value is, ${\varphi }_{\text{total}}=1.5\times {10}^{-6}\text{Wb}$.

Explanation of Solution

Given Information:

The toroid is having square cross section, $2.5\text{cm}<\rho <3.5\text{cm}$ , $-0.5\text{cm}<z<0.5\text{cm}$ . The flux in ${\text{a}}_{\varphi }$ direction is established by an mmf of $150\text{A}\cdot \text{t}$.

   $\begin{array}{l}{\mu }_r=\{\begin{array}{cc}20& -0.5\text{cm}<z<0\\ 10& 0<z<0.5\text{cm}\end{array}\\ {\varphi }_{z<0}=1.0\times {10}^{-6}\text{Wb}\\ {\varphi }_{z>0}=5\times {10}^{-7}\text{Wb}\end{array}$

Calculation:

The totalflux,

   $\begin{array}{l}{\varphi }_{\text{total}}={\varphi }_{z<0}+{\varphi }_{z>0}=1.0\times {10}^{-6}+5.0\times {10}^{-7}\\ =1.5\times {10}^{-6}\text{Wb}\end{array}$

Conclusion:

The required value is, ${\varphi }_{\text{total}}=1.5\times {10}^{-6}\text{Wb}$.