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Write the expression for the magnetic vector potential.

$\stackrel{炉}{A}=\frac{{渭}_0}{4蟺}鈭�\frac{I}{r}\text{鈥�}dz$ 鈥︹�� (1)

Here,$\stackrel{炉}{A}$is the magnetic vector potential, ${渭}_0$is the permeability, $I$ is the current and $r$is the distance.

Write the expression for magnetic field due to straight current carrying conductor.

$\mathit{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{I}}{\mathbf{4}\mathbf{蟺}\mathbf{s}}\mathbf{(}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{胃}}_{\mathbf{2}}\mathbf{鈭�}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{胃}}_{\mathbf{1}}\mathbf{)}$ 鈥︹�� (2)

Here,$B$ is the magnetic field,${渭}_0$ is the permeability,$I$ is the current and$s$ is the distance,${胃}_1$ and${胃}_2$ are the angles.

Consider a straight wire carrying current $I$, placed along the z-axis within the boundaries $z_1$and $z_2$shown in below figure.

Consider a point at a particular distance$s$ from the wire. From this point at a distance$r$ consider a small current element $dl$shown in figure.

Using Pythagoras theorem to find the value of $r$.

Write the expression for $r$.

$r=\sqrt{s^2+z^2}$

Let $\widehat{z}$be the unit vector indicating the direction of current. Then write the expression for the magnetic vector potential.

$\stackrel{炉}{A}=\frac{{渭}_0}{4蟺}鈭�\frac{I\widehat{z}}{r}\text{鈥�}dz$

Substitute$\sqrt{s^2+z^2}$ for $r$in above equation.

$\begin{array}{c}\stackrel{炉}{A}=\frac{{渭}_0}{4蟺}鈭�\frac{I\widehat{z}}{r}\text{鈥�}dz\\ =\frac{{渭}_{0}I}{4蟺}\underset{z_1}{\overset{z_2}{鈭�}}\frac{dz\widehat{z}}{\sqrt{z^2+s^2}}\\ =\frac{{渭}_{0}I\widehat{z}}{4蟺}\underset{z_1}{\overset{z_2}{鈭�}}\frac{dz}{\sqrt{z^2+s^2}}\\ =\frac{{渭}_{0}I\widehat{z}}{4蟺}{\left[In(z+\sqrt{z^2+s^2})\right]}_{z_1}^{z_2}\end{array}$

Solve as further,

$\begin{array}{c}\stackrel{炉}{A}=\frac{{渭}_{0}I}{4蟺}[In(z_2+\sqrt{{z_2}^2+s^2})鈭�In(z_1+\sqrt{{z_1}^2+s^2})]\widehat{z}\\ =\frac{{渭}_{0}I}{4蟺}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}\end{array}$

Thus, the magnetic vector potential is $\frac{{渭}_{0}I}{4蟺}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}$.

From equation (2),

Let$蠒$be the direction of magnetic field and perpendicular to the$\hat{z}$. Then the magnetic field id given by,

$B=鈭�\frac{鈭�A}{鈭�s}\widehat{蠒}$

Substitute $\frac{{渭}_{0}I}{4蟺}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}$for $A$in above equation.

$\begin{array}{c}B=鈭�\frac{鈭�A}{鈭�s}\widehat{蠒}\\ =鈭�\frac{鈭�}{鈭�s}\left[\frac{{渭}_{0}I}{4蟺}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}\right]\widehat{蠒}\\ =鈭�\frac{{渭}_{0}I}{4蟺}\left[\left(\frac{1}{z_2+\sqrt{{z_2}^2+s^2}}\right)\left(\frac{s}{\sqrt{{z_2}^2+s^2}}\right)鈭�\left(\frac{1}{\sqrt{(z_1+\sqrt{{z_1}^2+s^2})}}\right)\left(\frac{s}{(z_1+\sqrt{{z_1}^2+s^2})}\right)\right]\widehat{蠒}\\ =鈭�\frac{{渭}_{0}Is}{4蟺}\left[\begin{array}{l}\left(\frac{1}{z_2+\sqrt{{z_2}^2+s^2}}\right)\left(\frac{z_2鈭�\sqrt{{z_2}^2+s^2}}{z_2鈭�\sqrt{{z_2}^2+s^2}}\right)\left(\frac{1}{\sqrt{{z_2}^2+s^2}}\right)\\ 鈭�\left(\frac{1}{\sqrt{(z_1+\sqrt{{z_1}^2+s^2})}}\right)\left(\frac{z_1鈭�\sqrt{{z_1}^2+s^2}}{z_1鈭�\sqrt{{z_1}^2+s^2}}\right)\left(\frac{1}{\left(\sqrt{{z_1}^2+s^2}\right)}\right)\end{array}\right]\widehat{蠒}\end{array}$

Solve as further,

$\begin{array}{c}B=鈭�\frac{{渭}_{0}Is}{4蟺}\left[\frac{z_2鈭�\sqrt{{z_2}^2+s^2}}{{\left(z_2\right)}^2鈭�[{z_2}^2+s^2]}\frac{1}{\sqrt{{z_2}^2+s^2}}鈭�\frac{z_1鈭�\sqrt{{z_1}^2+s^2}}{{\left(z_1\right)}^2鈭�[{z_1}^2+s^2]}\frac{1}{\sqrt{{z_1}^2+s^2}}\right]\widehat{蠒}\\ =鈭�\frac{{渭}_{0}Is}{4蟺}\left(鈭�\frac{1}{s^2}\right)\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}鈭�\frac{\sqrt{{z_2}^2+s^2}}{\sqrt{{z_2}^2+s^2}}鈭�\frac{z_1}{\sqrt{{z_1}^2+s^2}}+\frac{\sqrt{{z_1}^2+s^2}}{\sqrt{{z_1}^2+s^2}}\right]\widehat{蠒}\\ =鈭�\frac{{渭}_{0}Is}{4蟺}\left(鈭�\frac{1}{s^2}\right)\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}鈭�1鈭�\frac{z_1}{\sqrt{{z_1}^2+s^2}}+1\right]\widehat{蠒}\end{array}$

Hence, the magnetic field is $B=\frac{{渭}_{0}I}{4蟺s}\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}鈭�\frac{z_1}{\sqrt{{z_1}^2+s^2}}\right]\widehat{蠒}$

Now, define the two angles,

$\sin {胃}_1=\frac{z_2}{\sqrt{z_2^2+s^2}}$ and $\sin {胃}_2=\frac{z_1}{\sqrt{z_1^2+s^2}}$

Substitute$\frac{z_2}{\sqrt{z_2^2+s^2}}$ for$\sin {胃}_1$ and$\frac{z_1}{\sqrt{z_1^2+s^2}}$ for$\sin {胃}_2$ in equation (2)

$B=\frac{{渭}_{0}I}{4蟺s}\left(\frac{z_2}{\sqrt{z_2^2+s^2}}鈭�\frac{z_1}{\sqrt{z_1^2+s^2}}\right)$

Thus, from equation (2) and (3) it is clear that, the magnetic vector potential is consistent.