91Ó°ÊÓ

The given data is listed below as:

• The radius of the loop of wire is,R
• The current in the wire is, I

Themagnetic field is described as the region inside a magnetic material that is beneficial for an object to exert force on another object. The exerted force is the magnetism force exerted.

The equation of the magnetic dipole moment is expressed as:

$m=AI$ …(¾±)

Here, A is the enclosed area and I is the current in the wire.

The equation of the enclosed area is expressed as:

$A=\mathrm{Ï€}{R}^2\hat{z}$

Here, R is the radius of the loop of wire and $\hat{z}$is the position vector in the z direction.

Substituterole="math" localid="1657524876988" $\mathrm{Ï€}{R}^2\hat{z}$ for A in the equation (i).

$m=I\mathrm{Ï€}{R}^2\hat{z}$

Thus, the magnetic dipole moment is role="math" localid="1657524833782" $I\mathrm{Ï€}{R}^2\hat{z}$.

The magnetic field far from the origin is described as the magnetic field of a point dipole. Hence, the equation of the magnetic field is expressed as:

$B≈\frac{{\mathrm{μ}}_0\mathrm{m}}{4{\mathrm{Ï€°ù}}^3}[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}]$

Here, ${\mathrm{μ}}_0$is the permeability, m is the magnetic dipole moment of the wire, r is the radius of the wire, and$\mathrm{θ}$ is the angle subtended by the wire.

Substitute${\mathrm{\pm õÏ€¸é}}^2\hat{z}$ for in the above equation.

role="math" localid="1657525736604" $B≈\frac{{\mathrm{μ}}_0{\mathrm{\pm õÏ€¸é}}^2}{4{\mathrm{Ï€°ù}}^3}[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}]$ …(¾±¾±)

Thus, the magnetic field at points far from the origin is$\frac{{\mathrm{μ}}_0{\mathrm{\pm õÏ€¸é}}^2}{4{\mathrm{Ï€°ù}}^3}[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}]$ .

The equation of the exact magnetic field along the z axis is expressed as:

$B\left(z\right)≈\frac{{\mathrm{μ}}_{0}I}{2}\frac{R^2}{{(R^2+z^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Here,R is the radius of the loop of the wire and z is the point along the z axis.

When ,$z>>R$ the above equation reduces to:

$B\left(z\right)≈\frac{{\mathrm{μ}}_{0}I}{2}\frac{R^2}{z^3}\hat{z}$

As the point lies along the z axis, then substitute 0 for$\mathrm{θ}$ ,z for r and $\hat{z}$for $\hat{r}$in the equation (ii).

$$\begin{gathered} B≈\frac{{\mathrm{μ}}_0{\mathrm{\pm õÏ€¸é}}^2}{4\mathrm{Ï€}{z}^3}[2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}] \\ ≈\frac{{\mathrm{μ}}_0{\mathrm{\pm õÏ€¸é}}^2}{4\mathrm{Ï€}{z}^3}\left(2\hat{z}\right) \\ ≈\frac{{\mathrm{μ}}_0{\mathrm{IR}}^2}{2z^3}\hat{z} \end{gathered}$$

Thus, the answer is consistent with the exact field.