Question: In Figure, current $\mathit{I}\mathbf{=}\mathbf{56}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{mA}$is set up in a loop having two radial lengths and two semicircles of radii$\mathit{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{72}\mathbf{}\mathbf{cm}$ and$\mathit{b}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{36}\mathbf{}\mathbf{cm}$ with a common center$\mathit{P}$(a) What are the magnitude and (b) What are the direction (into or out of the page) of the magnetic field at P and the (c) What is the magnitude of the loop’s magnetic dipole moment? and (d) What is the direction of the loop’s magnetic dipole moment?

1. Current is$56.2\mathrm{mA}$
2. Radius$a=5.72\mathrm{cm}$
3. Radius is$b=9.36\mathrm{cm}$

Semicircle wire generates a magnetic field that is half that of circle wire. We use the formula for magnetic field for semicircle, and the direction is given by right hand rule.

Formula:

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Now magnetic field for semicircle is as follows

$$\begin{gathered} B_1=\frac{{\mu }_{o}I\theta }{4\mathrm{\pi a}} \\ =\frac{{\mu }_{o}I\mathrm{\pi }}{4\mathrm{\pi a}} \\ =\frac{{\mu }_{o}I}{4a} \end{gathered}$$

And for radius b it is as follows

$B_2=\frac{{\mu }_{o}I}{4b}$

So total magnetic field is as follows

role="math" localid="1662799565933" $$\begin{gathered} B=B_1+B_2 \\ =\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{4\mathrm{a}}+\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{4\mathrm{b}} \\ =\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{4}(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}) \\ =(4\mathrm{\pi }\times {10}^{-7}{\mathrm{NA}}^{-2})\times (56.2\times {10}^{-3}\mathrm{A})(\frac{1}{0.0572\mathrm{m}}+\frac{1}{0.0936\mathrm{m}}) \\ =4.97\times {10}^{-7}\mathrm{T} \end{gathered}$$

The direction of the magnetic field is given by the right-hand rule, and it is into the page.

Now dipole moment is given as follows

$\mu =IA$

Here

role="math" localid="1662799996915" $$\begin{gathered} A=\frac{\mathrm{\pi }\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}{2} \\ =\frac{\mathrm{\pi }\left({\left(0.0572\mathrm{m}\right)}^2+{\left(0.0936\mathrm{m}\right)}^2\right)}{2} \\ =0.0189{\mathrm{m}}^2 \end{gathered}$$

Now

$$\begin{gathered} \mu =\left(56.2\times {10}^{-3}\mathrm{A}\right)\left(0.0189{\mathrm{m}}^2\right) \\ =1.06\times {10}^{-3}{\mathrm{Am}}^2 \end{gathered}$$

Since the direction is the same as the magnetic field, so it is into the page.