Question

A small, square loop carries a $25\mathrm{A}$current. The on-axis magnetic field strength $50\mathrm{cm}$ from the loop is $7.5\mathrm{nT}$. What is the edge length of the square?

Short answer

The edge length is$13.5\mathrm{mm}.$

Step-by-step solution

Given information

We have given,

Current = $25\mathrm{A}$

Distance = $50\mathrm{cm}$

Magnetic field = $7.5\mathrm{nT}$

We have to find the length of the edge of square.

Simplify

We know the magnetic field strength due to magnetic dipole moment

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\overrightarrow{\mathrm{\mu }}}{2{\mathrm{\pi r}}^3} \\ \overrightarrow{\mathrm{\mu }}=\frac{2{\mathrm{\pi r}}^3\mathrm{B}}{{\mathrm{\mu }}_0} \\ \overrightarrow{\mathrm{\mu }}=\frac{2\mathrm{\pi }\times {(0.5)}^3\times 7.5\times {10}^{-9}}{4\mathrm{\pi }\times {10}^{-7}} \\ \overrightarrow{\mathrm{\mu }}=4.6\mathrm{mA}.{\mathrm{m}}^2 \end{gathered}$$

Then magnetic moment for closed loop will be

$$\begin{gathered} \overrightarrow{\mathrm{\mu }}=AI=L^{2}I \\ L=\sqrt{\frac{\overrightarrow{\mu }}{I}} \\ L=\sqrt{\frac{4.6\times {10}^{-3}}{25}} \\ L=13.5mm \end{gathered}$$