91影视

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

$F_2=-\frac{{渭}_0}{4\mathrm{蟺}}{l}_1{l}_2鈭�鈭�\frac{r}{r^2}(d{l}_1鈥�d{l}_2)$

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

Write the formula of Lorentz force on loop 2.

${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{o}}}{\mathbf{4}\mathbf{蟺}}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}\mathbf{鈭�}{\mathbf{dl}}_{\mathbf{1}}\mathbf{鈭�}\left({\mathrm{dl}}_2鈥�\frac{r}{r^2}\right)\mathbf{-}\frac{{\mathbf{渭}}_{\mathbf{0}}}{\mathbf{4}\mathbf{蟺}}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}\mathbf{鈭�}\mathbf{鈭�}\frac{\mathbf{r}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{(}{\mathbf{dl}}_{\mathbf{1}}\mathbf{鈥�}{\mathbf{dl}}_{\mathbf{2}}\mathbf{)}$ 鈥︹�� (1)

Here, ${\mathit{渭}}_{\mathbf{o}}$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.

$$\begin{gathered} B\left(r\right)=\frac{{渭}_0}{4\mathrm{蟺}}鈭�\frac{l脳r}{r^2}dl\text{'} \\ =\frac{{渭}_o}{4\mathrm{蟺}}l鈭�\frac{dl\text{'}脳r}{r^2} \end{gathered}$$

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

$B_1=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1鈭�\frac{d{l}_1脳r}{r^2}$

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

$$\begin{gathered} F_2=l_2鈭�(d{l}_2脳B_1) \\ =\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�鈭�\frac{d{l}_2脳(dl脳r)}{r^2} \end{gathered}$$ 鈥︹�� (2)

Now assume that

$d{l}_2脳(d{l}_1脳r)=d{l}_1(d{l}_2鈥�r)-r(d{l}_1鈥�d{l}_2)$

Now substitute $d{l}_2脳(d{l}_1脳r)=d{l}_1(d{l}_2鈥�r)-r(d{l}_1鈥�d{l}_2)$into equation (2).

$$\begin{gathered} F_2=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\left(\frac{1}{r^2}\right)\left[d{l}_1(d{l}_2鈥�r)-r(d{l}_1鈥�d{l}_2)\right] \\ \left[=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\left(\frac{1}{r^2}\right)d{l}_1(d{l}_2鈥�r)-\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\left(\frac{1}{r^2}\right)r(d{l}_1鈥�d{l}_2)\right] \\ =\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�d{l}_1鈭�\left(d{l}_2鈥�\frac{r}{r^2}\right)-\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\frac{r}{r^2}(d{l}_1鈥�d{l}_2) \end{gathered}$$

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

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

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

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

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

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

Then, role="math" localid="1657687197487" $F_2=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�d{l}_1鈭�d{l}_2-\left(1-{鈭�}_2\left(\frac{1}{r}\right)\right)-\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\frac{r}{r^2}(d{l}_1鈥�d{l}_2)$

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

Because line integral over a closed loop is zero.

Then $F_2=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\frac{r}{r^2}(d{l}_1鈥�d{l}_2)$

Therefore, the value of force on loop by loop is $F_2=\frac{{渭}_o}{4\mathrm{蟺}}{l}_1{l}_2鈭�\frac{r}{r^2}(d{l}_1鈥�d{l}_2)$.