91Ó°ÊÓ

We need to find the current which needs to circulate around an $8.0cm$-diameter loop, about the size of a human heart, to produce this field.

As the segment on the loop applies a magnetic field, and the total magnetic field exerted by one loop at any point is given the equation ($29.8$)

$B_{loop}=\frac{{\mathrm{μ}}_{o}I}{2R}$

Need to find the current, therefore, that we can arrange equation ($1$) for $I$

localid="1650966946888" $I=\frac{2R{B}_{loop}}{{\mathrm{μ}}_o}\left(1\right)$

The values for ${\mathrm{μ}}_o,Rand{B}_{loop}$into equation ($2$) to get a current loop

localid="1649202387880" $$\begin{gathered} I=\frac{2RB}{\mathrm{μ}} \\ =\frac{2\left(0.04m\right)\left(90×{10}^{-12}T\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =\boxed{5.7×{10}^{-6}A} \end{gathered}$$

We need to find the magnitude of the heart’s magnetic dipole moment.

The current act has a magnet which has a magnetic dipole moment $\mathrm{μ}$and represents the current loop enclosing an area. $\mathrm{μ}$is perpendicular to the loop and its direction of the right-hand rule from the south pole to the north pole.

localid="1650967159683" $\mathrm{μ}=AI\left(2\right)$

If A is the area. Ring has a diameter $d=8cm$, and the area of the heart is calculated.

$$\begin{gathered} A=\mathrm{π}{\left(\frac{d}{2}\right)}^2 \\ =\mathrm{π}{\left(\frac{8×{10}^{-2}m}{2}\right)}^2 \\ =5×{10}^{-3}{m}^2 \end{gathered}$$

The values for A and I:

$$\begin{gathered} \mathrm{μ}=AI \\ =\left(5×{10}^{-3}{m}^2\right)\left(5.7×{10}^{-6}A\right) \\ =\boxed{28.5×{10}^{-9}A·m^2} \end{gathered}$$