Chapter 7, Problem 43P

(a)

To determine

Find the magnetic flux density $B$ for the given magnetic vector potential.

Answer to Problem 43P

The magnetic flux density $B$ for the given magnetic vector potential is $\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\text{Wb}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Given the magnetic vector potential,

$A=\left(2x^{2}y+yz\right)a_x+\left(x{y}^2-x{z}^3\right)a_y-\left(6xyz-2x^2{y}^2\right)a_z\text{Wb}/\text{m}$

Write the expression to calculate the magnetic flux density.

$B=\nabla \times A$        (1)

Here,

$A$ is the magnetic vector potential.

Substitute $\left(2x^{2}y+yz\right)a_x+\left(x{y}^2-x{z}^3\right)a_y-\left(6xyz-2x^2{y}^2\right)a_z\text{Wb}/\text{m}$ for $A$ in Equation (1).

$\begin{array}{c}B=\nabla \times A\\ =\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \left(2x^{2}y+yz\right)& \left(x{y}^2-x{z}^3\right)& -\left(6xyz-2x^2{y}^2\right)\end{array}\right|\\ =\left[\begin{array}{l}\left(\frac{\partial }{\partial y}\left(-\left(6xyz-2x^2{y}^2\right)\right)-\frac{\partial }{\partial z}\left(x{y}^2-x{z}^3\right)\right)a_x\\ -\left(\frac{\partial }{\partial x}\left(-\left(6xyz-2x^2{y}^2\right)\right)-\frac{\partial }{\partial z}\left(2x^{2}y+yz\right)\right)a_y\\ +\left(\frac{\partial }{\partial x}\left(x{y}^2-x{z}^3\right)-\frac{\partial }{\partial y}\left(2x^{2}y+yz\right)\right)a_z\end{array}\right]\\ =\left[\begin{array}{l}\left(-\left(6xz\left(1\right)-2x^2\left(2y\right)\right)-\left(0-x\left(3z^2\right)\right)\right)a_x\\ -\left(-\left(6yz\left(1\right)-2\left(2x\right)y^2\right)-\left(0+y\left(1\right)\right)\right)a_y\\ +\left(\left(y^2\left(1\right)-z^3\left(1\right)\right)-\left(2x^2\left(1\right)+\left(1\right)z\right)\right)a_z\end{array}\right]\end{array}$

Simplify the above equation.

$\begin{array}{c}B=\left[\begin{array}{l}\left(-6xz+4x^{2}y+3x{z}^2\right)a_x-\left(-6yz+4x{y}^2-y\right)a_y\\ +\left(y^2-z^3-2x^2-z\right)a_z\end{array}\right]\\ =\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\text{Wb}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the magnetic flux density $B$ for the given magnetic vector potential is $\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\text{Wb}/{\text{m}}^{\text{2}}$.

(b)

To determine

Find the magnetic flux for the given magnetic vector potential.

Answer to Problem 43P

The magnetic flux $\left(\psi \right)$ for the given magnetic vector potential is $8\text{Wb}$.

Explanation of Solution

Calculation:

Write the expression to calculate the magnetic flux through a surface.

$\psi ={\displaystyle \underset{S}{\int }B\cdot dS}$

Substitute $\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\text{Wb}/{\text{m}}^{\text{2}}$ for $B$ in above equation.

$\begin{array}{c}\psi ={\displaystyle \underset{S}{\int }\left(\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\right)\cdot dS}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\displaystyle \underset{y=0}{\overset{2}{\int }}\left[\begin{array}{l}\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y\\ +\left(y^2-z^3-2x^2-z\right)a_z\end{array}\right]\cdot dydz{a}_x}}\left\{\because dS=dydz{a}_x\right\}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\displaystyle \underset{y=0}{\overset{2}{\int }}\left[\begin{array}{l}\left(-6xz+4x^{2}y+3x{z}^2\right)a_x\cdot dydz{a}_x+\left(y+6yz-4x{y}^2\right)a_y\cdot dydz{a}_x\\ +\left(y^2-z^3-2x^2-z\right)a_z\cdot dydz{a}_x\end{array}\right]}}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\displaystyle \underset{y=0}{\overset{2}{\int }}\left[\left(-6xz+4x^{2}y+3x{z}^2\right)dydz+0+0\right]}}\left\{\begin{array}{l}\because a_x\cdot a_x=1,\\ a_y\cdot a_x=0,\\ a_z\cdot a_x=0\end{array}\right\}\end{array}$

Simplify the above equation.

$\begin{array}{c}\psi ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\displaystyle \underset{y=0}{\overset{2}{\int }}\left(-6xz+4x^{2}y+3x{z}^2\right)dydz}}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\left[-6xzy+4x^2\left(\frac{y^2}{2}\right)+3x{z}^{2}y\right]}_0^{2}dz}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}{\left[-6xyz+2x^2{y}^2+3xy{z}^2\right]}_0^{2}dz}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}\left[\left(-6x\left(2\right)z+2x^2{\left(2\right)}^2+3x\left(2\right)z^2\right)-\left(-6x\left(0\right)z+2x^2{\left(0\right)}^2+3x\left(0\right)z^2\right)\right]dz}\end{array}$

Simplify the above equation.

$\begin{array}{c}\psi ={\displaystyle \underset{z=0}{\overset{2}{\int }}\left[\left(-12xz+8x^2+6x{z}^2\right)-\left(0\right)\right]dz}\\ ={\displaystyle \underset{z=0}{\overset{2}{\int }}\left(-12xz+8x^2+6x{z}^2\right)dz}\\ ={\left[-12x\left(\frac{z^2}{2}\right)+8x^{2}z+6x\left(\frac{z^3}{3}\right)\right]}_0^2\\ ={\left[-6x{z}^2+8x^{2}z+2x{z}^3\right]}_0^2\end{array}$

Simplify the above equation.

$\begin{array}{c}\psi =\left[\left(-6x{\left(2\right)}^2+8x^2\left(2\right)+2x{\left(2\right)}^3\right)-\left(-6x{\left(0\right)}^2+8x^2\left(0\right)+2x{\left(0\right)}^3\right)\right]\\ =\left[\left(-24x+16x^2+16x\right)-\left(0\right)\right]\\ =-24\left(1\right)+16{\left(1\right)}^2+16\left(1\right)\left\{\because x=1\right\}\\ =8\text{Wb}\end{array}$

Conclusion:

Thus, the magnetic flux $\left(\psi \right)$ for the given magnetic vector potential is $8\text{Wb}$.

(c)

To determine

Show that the relation $\nabla \cdot A$ and $\nabla \cdot B$ is equal to zero.

Explanation of Solution

Calculation:

Substitute $\left(2x^{2}y+yz\right)a_x+\left(x{y}^2-x{z}^3\right)a_y-\left(6xyz-2x^2{y}^2\right)a_z\text{Wb}/\text{m}$ for $A$ to find $\nabla \cdot A$.

$\begin{array}{c}\nabla \cdot A=\frac{\partial }{\partial x}{A}_x+\frac{\partial }{\partial y}{A}_y+\frac{\partial }{\partial z}{A}_z\\ =\frac{\partial }{\partial x}\left(2x^{2}y+yz\right)+\frac{\partial }{\partial y}\left(x{y}^2-x{z}^3\right)+\frac{\partial }{\partial z}\left(-\left(6xyz-2x^2{y}^2\right)\right)\\ =\left(2\left(2x\right)y+0\right)+\left(x\left(2y\right)-0\right)-\left(6xy\left(1\right)-0\right)\\ =4xy+2xy-6xy\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \cdot A=6xy-6xy\\ =0\end{array}$

Substitute $\left(-6xz+4x^{2}y+3x{z}^2\right)a_x+\left(y+6yz-4x{y}^2\right)a_y+\left(y^2-z^3-2x^2-z\right)a_z\text{Wb}/{\text{m}}^{\text{2}}$ for $B$ to find $\nabla \cdot B$.

$\begin{array}{c}\nabla \cdot B=\frac{\partial }{\partial x}{B}_x+\frac{\partial }{\partial y}{B}_y+\frac{\partial }{\partial z}{B}_z\\ =\frac{\partial }{\partial x}\left(-6xz+4x^{2}y+3x{z}^2\right)+\frac{\partial }{\partial y}\left(y+6yz-4x{y}^2\right)+\frac{\partial }{\partial z}\left(y^2-z^3-2x^2-z\right)\\ =\left(-6\left(1\right)z+4\left(2x\right)y+3\left(1\right)z^2\right)+\left(\left(1\right)+6\left(1\right)z-4x\left(2y\right)\right)+\left(0-3z^2-0-1\right)\\ =\left(-6z+8xy+3z^2\right)+\left(1+6z-8xy\right)+\left(-3z^2-1\right)\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \cdot B=-6z+8xy+3z^2+1+6z-8xy-3z^2-1\\ =0\end{array}$

Conclusion:

Thus, the relation $\nabla \cdot A$ and $\nabla \cdot B$ is equal to zero and it is shown.