Magnetostatics treats the "source current" (the one that sets up the field) and the "recipient current" (the one that experiences the force) so asymmetrically that it is by no means obvious that the magnetic force between two current loops is consistent with Newton's third law. Show, starting with the Biot-Savart law (Eq. 5.34) and the Lorentz force law (Eq. 5.16), that the force on loop 2 due to loop 1 (Fig. 5.61) can be written as

${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}\mathbf{\oint }\mathbf{\oint }\frac{\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}{\mathbf{dl}}_{\mathbf{1}}\mathbf{‐}{\mathbf{dl}}_{\mathbf{2}}$

Figure 5.60

Figure 5.61

In this form, it is clear that ${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{F}}_{\mathbf{1}}$, since role="math" localid="1657622030111" $\hat{\mathbf{r}}$changes direction when the roles of 1 and 2 are interchanged. (If you seem to be getting an "extra" term, it will help to note that${\mathbf{dl}}_{\mathbf{2}}\mathbf{‐}\hat{\mathbf{r}}\mathbf{=}\mathbf{dr}$.)

Consider the given force on loop 2 due to loop 1 (Fig. 5.61) can be written as

$F_2=-\frac{{\mu }_0}{4\mathrm{\pi }}{l}_1{l}_2\oint \oint \frac{r}{r^2}(d{l}_1‐d{l}_2)$

Consider the given condition it is clear that ${\mathrm{F}}_2=-{\mathrm{F}}_1$.

Write the formula of Lorentz force on loop 2.

${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{o}}}{\mathbf{4}\mathbf{\pi }}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}\mathbf{\oint }{\mathbf{dl}}_{\mathbf{1}}\mathbf{\oint }\left({\mathrm{dl}}_2‐\frac{r}{r^2}\right)\mathbf{-}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}\mathbf{\oint }\mathbf{\oint }\frac{\mathbf{r}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{(}{\mathbf{dl}}_{\mathbf{1}}\mathbf{‐}{\mathbf{dl}}_{\mathbf{2}}\mathbf{)}$ …… (1)

Here, ${\mathit{\mu }}_{\mathbf{o}}$is permeability, is radius of spherical shell, ${\mathbf{l}}_{\mathbf{1}}$is current on loop 1 , ${\mathbf{l}}_{\mathbf{2}}$is current on loop.

We know that magnetic field of a steady line current given by the Biot-Savart law.

$$\begin{gathered} B\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\int \frac{l\times r}{r^2}dl\text{'} \\ =\frac{{\mu }_o}{4\mathrm{\pi }}l\int \frac{dl\text{'}\times r}{r^2} \end{gathered}$$

Now for the Biot Savart’s law, the field of loop 1 is

$B_1=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1\oint \frac{d{l}_1\times r}{r^2}$

From the Lorentz force law, the force on loop 2 is

$$\begin{gathered} F_2=l_2\oint (d{l}_2\times B_1) \\ =\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2\oint \oint \frac{d{l}_2\times (dl\times r)}{r^2} \end{gathered}$$ …… (2)

Now assume that

$d{l}_2\times (d{l}_1\times r)=d{l}_1(d{l}_2‐r)-r(d{l}_1‐d{l}_2)$

Now substitute $d{l}_2\times (d{l}_1\times r)=d{l}_1(d{l}_2‐r)-r(d{l}_1‐d{l}_2)$into equation (2).

$$\begin{gathered} F_2=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\left(\frac{1}{r^2}\right)\left[d{l}_1(d{l}_2‐r)-r(d{l}_1‐d{l}_2)\right] \\ \left[=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\left(\frac{1}{r^2}\right)d{l}_1(d{l}_2‐r)-\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\left(\frac{1}{r^2}\right)r(d{l}_1‐d{l}_2)\right] \\ =\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2\oint d{l}_1\oint \left(d{l}_2‐\frac{r}{r^2}\right)-\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\frac{r}{r^2}(d{l}_1‐d{l}_2) \end{gathered}$$

Now $r=r_2-r_1$

Here, is position vector of source point and is position vector of field point

Then $r=(x_2-x_1)\hat{x}+(y_2-y_1)\hat{y}+(z_2-z_1)\hat{z}$

Then$\left|r\right|=l=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$

Now determine the

${\nabla }_2\left(\frac{1}{r}\right)=\left\{\begin{array}{c}\frac{\partial }{\partial x_2}{\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{1}{2}}\\ +\frac{\partial }{\partial y_2}{\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{1}{2}}\\ +\frac{\partial }{\partial z_2}{\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{1}{2}}\end{array}\right\}$

$=\left\{\begin{array}{c}\left(-\frac{1}{2}\right){\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{1}{2}}\left[2(x_2-x_1)\right]\hat{x}\\ +\left(-\frac{1}{2}\right){\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{3}{2}}\left[2(y_2-y_1)\right]\hat{y}\\ +\left(-\frac{1}{2}\right){\left[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2\right]}^{-\frac{3}{2}}\left[2(z_2-z_1)\right]\hat{z}\\ \end{array}\right\}$

$$\begin{gathered} =-\frac{(x_2-x_1)}{r^3}\hat{x}-\frac{(y_2-y_1)}{r^3}\hat{y}-\frac{(z_2-z_1)}{r^3}\hat{z} \\ =-\frac{r}{r^3}=-\frac{r}{r^2} \end{gathered}$$

Thus, $\frac{r}{r^3}=-{\nabla }_2\left(\frac{1}{r}\right)$

Now substituting the value of ${\nabla }_2\left(\frac{1}{r}\right)$into equation (1).

Then, role="math" localid="1657687197487" $F_2=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2\oint d{l}_1\oint d{l}_2-\left(1-{\nabla }_2\left(\frac{1}{r}\right)\right)-\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\frac{r}{r^2}(d{l}_1‐d{l}_2)$

Now, $\oint d{l}_2-\left(1-{\nabla }_2\left(\frac{1}{r}\right)\right)=0$

Because line integral over a closed loop is zero.

Then $F_2=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\frac{r}{r^2}(d{l}_1‐d{l}_2)$

Therefore, the value of force on loop by loop is $F_2=\frac{{\mu }_o}{4\mathrm{\pi }}{l}_1{l}_2∯\frac{r}{r^2}(d{l}_1‐d{l}_2)$.