91Ó°ÊÓ

Write the expression for the magnetic vector potential.

$\stackrel{¯}{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{I}{r}\text{ }dz$ …… (1)

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

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

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{4}\mathbf{π}\mathbf{s}}\mathbf{(}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{θ}}_{\mathbf{2}}\mathbf{−}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{θ}}_{\mathbf{1}}\mathbf{)}$ …… (2)

Here,$B$ is the magnetic field,${\mathrm{μ}}_0$ is the permeability,$I$ is the current and$s$ is the distance,${\mathrm{θ}}_1$ and${\mathrm{θ}}_2$ 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.

$\stackrel{¯}{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{I\widehat{z}}{r}\text{ }dz$

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

$\begin{array}{c}\stackrel{¯}{A}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{I\widehat{z}}{r}\text{ }dz\\ =\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\underset{z_1}{\overset{z_2}{∫}}\frac{dz\widehat{z}}{\sqrt{z^2+s^2}}\\ =\frac{{\mathrm{μ}}_{0}I\widehat{z}}{4\mathrm{π}}\underset{z_1}{\overset{z_2}{∫}}\frac{dz}{\sqrt{z^2+s^2}}\\ =\frac{{\mathrm{μ}}_{0}I\widehat{z}}{4\mathrm{π}}{\left[In(z+\sqrt{z^2+s^2})\right]}_{z_1}^{z_2}\end{array}$

Solve as further,

$\begin{array}{c}\stackrel{¯}{A}=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}[In(z_2+\sqrt{{z_2}^2+s^2})−In(z_1+\sqrt{{z_1}^2+s^2})]\widehat{z}\\ =\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}\end{array}$

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

From equation (2),

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

$B=−\frac{∂A}{∂s}\widehat{\mathrm{ϕ}}$

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

$\begin{array}{c}B=−\frac{∂A}{∂s}\widehat{\mathrm{ϕ}}\\ =−\frac{∂}{∂s}\left[\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\left[\frac{In(z_2+\sqrt{{z_2}^2+s^2})}{In(z_1+\sqrt{{z_1}^2+s^2})}\right]\widehat{z}\right]\widehat{\mathrm{ϕ}}\\ =−\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}}\left[\left(\frac{1}{z_2+\sqrt{{z_2}^2+s^2}}\right)\left(\frac{s}{\sqrt{{z_2}^2+s^2}}\right)−\left(\frac{1}{\sqrt{(z_1+\sqrt{{z_1}^2+s^2})}}\right)\left(\frac{s}{(z_1+\sqrt{{z_1}^2+s^2})}\right)\right]\widehat{\mathrm{ϕ}}\\ =−\frac{{\mathrm{μ}}_{0}Is}{4\mathrm{π}}\left[\begin{array}{l}\left(\frac{1}{z_2+\sqrt{{z_2}^2+s^2}}\right)\left(\frac{z_2−\sqrt{{z_2}^2+s^2}}{z_2−\sqrt{{z_2}^2+s^2}}\right)\left(\frac{1}{\sqrt{{z_2}^2+s^2}}\right)\\ −\left(\frac{1}{\sqrt{(z_1+\sqrt{{z_1}^2+s^2})}}\right)\left(\frac{z_1−\sqrt{{z_1}^2+s^2}}{z_1−\sqrt{{z_1}^2+s^2}}\right)\left(\frac{1}{\left(\sqrt{{z_1}^2+s^2}\right)}\right)\end{array}\right]\widehat{\mathrm{ϕ}}\end{array}$

Solve as further,

$\begin{array}{c}B=−\frac{{\mathrm{μ}}_{0}Is}{4\mathrm{π}}\left[\frac{z_2−\sqrt{{z_2}^2+s^2}}{{\left(z_2\right)}^2−[{z_2}^2+s^2]}\frac{1}{\sqrt{{z_2}^2+s^2}}−\frac{z_1−\sqrt{{z_1}^2+s^2}}{{\left(z_1\right)}^2−[{z_1}^2+s^2]}\frac{1}{\sqrt{{z_1}^2+s^2}}\right]\widehat{\mathrm{ϕ}}\\ =−\frac{{\mathrm{μ}}_{0}Is}{4\mathrm{π}}\left(−\frac{1}{s^2}\right)\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}−\frac{\sqrt{{z_2}^2+s^2}}{\sqrt{{z_2}^2+s^2}}−\frac{z_1}{\sqrt{{z_1}^2+s^2}}+\frac{\sqrt{{z_1}^2+s^2}}{\sqrt{{z_1}^2+s^2}}\right]\widehat{\mathrm{ϕ}}\\ =−\frac{{\mathrm{μ}}_{0}Is}{4\mathrm{π}}\left(−\frac{1}{s^2}\right)\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}−1−\frac{z_1}{\sqrt{{z_1}^2+s^2}}+1\right]\widehat{\mathrm{ϕ}}\end{array}$

Hence, the magnetic field is $B=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}s}\left[\frac{z_2}{\sqrt{{z_2}^2+s^2}}−\frac{z_1}{\sqrt{{z_1}^2+s^2}}\right]\widehat{\mathrm{ϕ}}$

Now, define the two angles,

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

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

$B=\frac{{\mathrm{μ}}_{0}I}{4\mathrm{π}s}\left(\frac{z_2}{\sqrt{z_2^2+s^2}}−\frac{z_1}{\sqrt{z_1^2+s^2}}\right)$

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