Question

We are now in a position to treat the example in Sect. 8.2.1 quantitatively. Suppose${\mathit{q}}_{\mathbf{1}}$ is at${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathit{v}\mathit{t}$ and${\mathit{q}}_{\mathbf{2}}$ is at$\mathit{y}\mathbf{=}\mathbf{-}\mathit{v}\mathit{t}$ (Fig. 8.3, with$\mathit{t}\mathit{<}\mathit{0}$ ). Find the electric and magnetic forces on${\mathit{q}}_{\mathbf{1}}$ and${\mathit{q}}_{\mathbf{2}}$ . Is Newton’s third law obeyed?

Short answer

The electric and magnetic forces on $q_1$and $q_1$are

$F_1=\left[\begin{array}{c}\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\\ \left(\hat{x}-\hat{y}+\frac{v^2}{c^2}\hat{x}\right)\end{array}\right]$and$F_2=\left[\begin{array}{c}\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1-{\displaystyle \frac{v^2}{c^2}}}{{\left(1-{\displaystyle \frac{v^2}{2c^2}}\right)}^{3/2}}\frac{1}{2\sqrt{2}{\left(vt\right)}^2}\\ \left(-\hat{x}+\hat{y}+\frac{v^2}{c^2}\hat{x}\right)\end{array}\right]$ respectively, and Newton’s third law is not obeyed due to an equal magnitude with opposite direction of charges.

Step-by-step solution

Expression for the Lorentz force law for force on charge q2 :

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

${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathit{q}}_{\mathbf{2}}\left(E_1+v_2\times B_1\right)$ …… (1)

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

Determine the electric field and magnetic field of charge q1 at q2 :

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

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

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

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

Write the expression for the magnetic field.

$B_1=\frac{1}{c^2}\left(v_1\times E\right)$

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

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

Substitute the value of E in the above expression.

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

Determine the electric and magnetic forces on q1 and q2 :

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

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

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,

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

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

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

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

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