A student makes a short electromagnet by winding of wire $\mathbf{300}\mathbf{}\mathbf{\text{turns}}$around a wooden cylinder of diameter$\mathbf{}\mathit{d}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$. The coil is connected to a battery producing a current of$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{A}}$in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance d will the magnetic field have the magnitude$\mathbf{5}\mathbf{.0}\mathbf{}\mathbf{\text{μT}}$(approximately one-tenth that of Earth’s magnetic field)?

A diameter,$d=5.0\text{cm}=0.05\text{m}$

Number of turns,$N=300$

The current,$i=4\text{A}$

The magnetic field,role="math" localid="1663234250596" $B=5\text{μT=}\begin{array}{rcl}& & 5\end{array}\begin{array}{rcl}& & \times \end{array}\begin{array}{rcl}& & 10\end{array}\begin{array}{rcl}& & -6\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \text{T}\end{array}$

The magnitude of the magnetic dipole moment can be defined as the product of number of turns, current in that loop, and the area of that loop. You can calculate the magnetic field on the axis of a magnetic dipole, a distance$\mathit{Z}$by using the following equation.

The magnetic dipole moment is,

$\mu =NiA$

Here, $N$is the number of turns, $i$is the current, $A$is the area.

The magnetic field is,

role="math" localid="1663233700545" $B=\frac{{\mu }_{0}Ni{R}^2}{2{\left(R^2+Z^2\right)}^{\frac{3}{2}}}$

Here, ${\mu }_0$is the permeability of free space having a value $\begin{array}{rcl}& & 4\pi \times {10}^{-7}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$A^2$}\right.\end{array}$, $R$is the radius, $Z$and is the distance.

By using the formula of magnetic dipole moment,

$\mu =NiA$ ..... (1)

Calculate the area of the circular loop:

$\begin{array}{rcl}A& =& \pi R^2\\ & =& \pi {\left(\frac{d}{2}\right)}^2\\ & =& 3.14\times {\left(\frac{0.05}{2}\right)}^2\\ & =& 1.96\times {10}^{-3}{\text{m}}^2\end{array}$

By substituting this value into equation (1).

$\begin{array}{rcl}\mu & =& 300\times 4\text{A}\times 1.96\times {10}^{-3}{\text{m}}^2\\ & =& 2.356\text{A}\cdot {\text{m}}^{\text{2}}\\ & \approx & 2.4\text{A}\cdot {\text{m}}^{\text{2}}\end{array}$

Hence, the magnitude of the magnetic dipole moment is$2.4\text{A}\cdot {\text{m}}^{\text{2}}$.

The magnetic field of the coil is given bytheequation.

$B=\frac{{\mu }_{0}Ni{R}^2}{2{\left(R^2+Z^2\right)}^{\frac{3}{2}}}$ ….. (2)

But here, $Z\gg d$that means $z\gg R$.

Thus, equation (2) can be written as

role="math" localid="1663233718305" $B=\frac{{\mu }_{0}Ni{R}^2}{2Z^3}$

By multiplying and dividing by $\pi$,you can get

role="math" localid="1663233733788" $B=\frac{{\mu }_{0}Ni\pi R^2}{2\pi Z^3}$

Substitute $NiA$for $\mu$in the above equation.

$B=\frac{{\mu }_0\mu }{2\pi Z^3}$

By solving for $Z$, you can get

$Z^3=\frac{{\mu }_0\mu }{2\pi B}$

role="math" localid="1663234263421" $\begin{array}{rcl}Z& =& {\left(\frac{{\mu }_0\mu }{2\pi B}\right)}^{\frac{1}{3}}\\ & =& {\left(\frac{4\pi \times {10}^{-7}\times 2.36}{2\pi \times 5\times {10}^{-6}}\right)}^{\frac{1}{3}}\\ & =& {(0.0944)}^{\frac{1}{3}}\\ & =& 0.46\text{m}\end{array}$

Hence, the magnetic field have the magnitude $5.0\text{μT}$at axial distance is $0.46\text{m}$.