Question: In Fig 29-76 a conductor carries$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{A}$along the closed path abcdefgharunning along $\mathbf{8}$of the $\mathbf{12}$edges of a cube of edge length $\mathbf{10}\mathbf{}\mathbf{cm}$. (a)Taking the path to be a combination of three square current loops (bcfgb, abgha, and cdefc), find the net magnetic moment of the path in unit-vector notation.(b) What is the magnitude of the net magnetic field at the xyzcoordinates of$(0,5.0m,0)$?

• Current is $i=0.6\mathrm{A}$
• Length is $l=10\mathrm{cm}$

We use the formula for the magnetic moment in terms of current and area in all three directions, and from that, we find the net magnetic moment. Then, we have to find the magnetic field from the magnetic moment.

Formula:

$\overrightarrow{\mathbf{B}}\mathbf{=}\frac{\mathbf{\mu }\overrightarrow{\mathbf{\mu }}}{\mathbf{2}{\mathbf{\pi y}}^{\mathbf{3}}}$

$\mathit{\mu }\mathbf{=}\mathit{i}\mathit{A}$

There is no enclosed area in x and z directions. There are only three edges with the current flowing through them for both x and z directions. Therefore, net magnetic field moment for x and z components of the patch are zero.

That means

${\overrightarrow{\mu }}_{abgha}=0$

${\overrightarrow{\mu }}_{cdefc}=0$

There is an enclosed area for the current flowing through y components. Therefore, the net magnetic field for y component of patch $bcfgb$is

role="math" localid="1662806597353" $$\begin{gathered} {\overrightarrow{\mu }}_{bcfgb}=iA\hat{\mathrm{j}} \\ =\left(6.00\mathrm{A}\right){\left(10\times {10}^{-2}\mathrm{m}\right)}^2\hat{\mathrm{j}} \\ =\left(0.06{\mathrm{Am}}^2\right)\hat{\mathrm{j}} \end{gathered}$$

Hence, the net magnetic moment of path in unit vector notation is $(0.06{\mathrm{Am}}^2)\hat{\mathrm{j}}$.

The magnetic field is only in y direction, so

$$\begin{gathered} {\overrightarrow{B}}_y=\frac{\mu \overrightarrow{\mu }}{2{\mathrm{\pi y}}^3} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(0.06{\mathrm{Am}}^2\right)\hat{\mathrm{j}}}{2\mathrm{\pi }\times {\left(5.0\mathrm{m}\right)}^3} \\ =\left(9.6\times {10}^{-11}\mathrm{T}\right)\hat{\mathrm{j}} \end{gathered}$$

So the magnitude of magnetic field is is $(9.6\times {10}^{-11}\mathrm{T})\hat{\mathrm{j}}$.