91影视

Write the expression for the Lorentz force law for force on the charge$q_2$ .

${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathit{q}}_{\mathbf{2}}\left(E_1+v_2脳B_1\right)$ 鈥︹�� (1)

Here, q is the charge, E is the magnetic field, v is the velocity, and B is the magnetic field.

Write the expression for the electric field of charge$q_1$at$q_2$.

$E_1=\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2{\sin }^2胃}{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}$

Substitute$胃=45掳$and$R=\left(-vt\hat{x}+vt\hat{y}\right)$in the above expression.

$E_1=\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\left(-\hat{x}+\hat{y}\right)$

Write the expression for the magnetic field.

$B_1=\frac{1}{c^2}\left(v_1脳E\right)$

Here, $v_1=-v\hat{x}$.

$B_1=-\frac{v}{c^2}\left(\hat{x}脳E\right)$

Substitute the value of E in the above expression.

$$\begin{gathered} B_1=-\frac{v}{c^2}\left(\hat{x}脳E_1=\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\left(-\hat{x}+\hat{y}\right)\right) \\ {B}_1=-\frac{v}{c^2}\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\hat{z} \end{gathered}$$

Substitute the known value of $E_1,v_2=-v\hat{y}$and$B_1$in equation (1).

$$\begin{gathered} F_2=q_2\left(\begin{array}{c}\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\left(-\hat{x}+\hat{y}\right)\\ \left(-v\hat{y}\right)脳-\frac{v}{c^2}\frac{q_1}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\hat{z}\end{array}\right) \\ {F}_2=\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\left(-\hat{x}+\hat{y}+\frac{v^2}{c^2}\hat{x}\right) \end{gathered}$$

The electric field of charge $q_2$at $q_1$is reversed, i.e., $E_2=-E_1$. So, the magnetic field will be $B_2=-B_1$ and also the electric force is reversed. In the case of reversion, as the magnetic force now points in the y-direction instead of the x-direction, the force on the charge $q_1$is,

$F_1=\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\left(\hat{x}-\hat{y}+\frac{v^2}{c^2}\hat{x}\right)$

As the forces are equal in magnitude and opposite in direction, due to the concept of Newton鈥檚 third law is not obeyed.

Therefore, the electric and magnetic forces on $q_1$and $q_2$are

$F_1=\left[\begin{array}{c}\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\\ \left(\hat{x}-\hat{y}+\frac{v^2}{c^2}\hat{x}\right)\end{array}\right]$and$F_2=\left[\begin{array}{c}\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\\ \left(\hat{x}+\stackrel{鈫�}{y}+\frac{v^2}{c^2}\hat{x}\right)\end{array}\right]$

respectively, and Newton鈥檚 third law is not obeyed due to an equal magnitude with opposite direction of charges.