Find the exact magnetic field a distance$\mathit{z}$ above the center of a square loop of side $\mathit{w}$, carrying a current. Verify that it reduces to the field of a dipole, with the appropriate dipole moment, when$\mathit{z}\mathbf{\gg }\mathit{w}$.

The$\omega$ is the side of square loop.

The$z$ is the distance above the center of square loop.

The$I$ is current through loop.

The${\theta }_1$ is the angle made by initial point of the side with the normal.

The${\theta }_2$ is the angle made by end point of the side.

Write the formula ofmagnetic field reduces to the field due to a dipole.

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi }}\frac{{\mathbf{\omega }}^{\mathbf{2}}}{{\mathbf{z}}^{\mathbf{3}}}\stackrel{\mathbf{⏜}}{\mathbf{z}}$ …… (1)

Here,${\mathit{\mu }}_{\mathbf{0}}$ is permeability, $\mathit{\omega }$is the side of square loop,$\mathit{z}$ is the distance above the center of square loop and role="math" localid="1657686397925" $\mathbf{I}$is current through loop.

Draw the circuit diagram for given provided condition.

Consider a point$\mathrm{P}$at a distance$z$above the center of the square loop.

${\theta }_1$and${\theta }_2$signs will be the opposite since they are rotating in the opposite directions with respect to the usual vertical line. Therefore,

$-\sin {\theta }_1=\sin {\theta }_2$

Magnetic field at$\mathrm{P}$due to one side of square loop is,

role="math" localid="1657686862591" $B_1=\frac{{\mu }_{0}I}{4\mathrm{\pi S}}\left(\sin {\theta }_2-\sin {\theta }_1\right)$

Here, $I$is the current flowing through the loop and $S$is the distance of point from the side.

Substitute $\sin {\theta }_{2}for-\sin {\theta }_1$into above equation.

$B_1=\frac{2{\mu }_{0}I\sin {\theta }_2}{4\mathrm{\pi S}}$

Refer to figure, $S$is the hypotenuse, $\frac{\omega }{2}$is the perpendicular and $z$is the base. So from Pythagoras theorem,

$S=\sqrt{z^2+{\left(\frac{\omega }{2}\right)}^2}$

It is understood that the ratio of the perpendicular side to the hypotenuse is the sine of an angle. So,

role="math" localid="1657687065877" ${\sin }_2=\frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}}$

Substitute $\sqrt{z^2+{\left(\frac{\omega }{2}\right)}^2}$for $S$and $\frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}}$for $\sin {\theta }_2$into equation (1).

$$\begin{gathered} B_1=\frac{2{\mu }_{0}I\left({\displaystyle \frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}}}\right)}{4\mathrm{\pi }\left(\sqrt{{\mathrm{z}}^2+{\left({\displaystyle \frac{\mathrm{\omega }}{2}}\right)}^2}\right)} \\ =\frac{{\mu }_{0}I}{4\mathrm{\pi }}\frac{\omega }{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2} \end{gathered}$$

Similarly, magnetic field at$\mathrm{P}$due to all the sides of square loop is,

$B_1=4\left(\frac{{\mu }_{0}I}{4\mathrm{\pi }}\frac{\omega }{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}\right)$

From the figure, horizontal components of magnitude field will cancel each other.

Determine the vertical component of magnetic field at point P is,

$B_{1ver}=B_1\sin \varphi$ …… (2)

Here,$\varphi$is the angle formed by the line from a point P to the centre of a side and the normal on the plane of a square is such that$\sin \varphi =\frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\displaystyle \frac{{\omega }^2}{4}}}}$

Substitute $\frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\displaystyle \frac{{\omega }^2}{4}}}}$for $\sin \varphi$and $4\left(\frac{{\mu }_{0}I}{4\mathrm{\pi }}\frac{\omega }{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}\right)$for$B_1$into equation (2).

$B_{1ver}=4\left(\left(\frac{{\mu }_{0}I}{4\mathrm{\pi }}\frac{\omega }{z^2+{\left({\displaystyle \frac{\omega }{2}}\right)}^2}\right)\right)\left(\frac{{\displaystyle \frac{\omega }{2}}}{\sqrt{z^2+{\displaystyle \frac{{\omega }^2}{4}}}}\right)$

Hence, magnetic field at point P is$B=\frac{{\mu }_{0}I}{2\mathrm{\pi }}\frac{{\omega }^2}{{\left(z^2+{\displaystyle \frac{{\omega }^2}{4}}\right)}^{{\displaystyle \frac{3}{2}}}}\stackrel{⏜}{z}$

For $z\gg \omega$, write the above equation as,

$B=\frac{{\mu }_{0}I}{2\mathrm{\pi }}\frac{\omega }{z^3}\stackrel{⏜}{z}$

Determine the value of magnetic field reduces to the field due to a dipole, when $\omega$is

Magnetic dipole moment is,

$m=IA$

Here, $A$is the area of square loop.

Substitute ${\omega }^2$for $A$(as$\omega$is the side of the square loop) into above equation.

$m=I{\omega }^2$

Substitute $m$for $I{\omega }^2$into equation (1).

$B=\frac{{\mu }_{0}m}{2{\mathrm{\pi z}}^3}\stackrel{⏜}{z}$

Hence, the value of magnetic field reduces to the field due to a dipole, when $z\gg \omega$is $B=\frac{{\mu }_{0}m}{2{\mathrm{\pi z}}^3}\stackrel{⏜}{z}$.