Chapter 1, Problem 1.32P

To determine

To show that $F_{12}^{\text{mag}}\le \frac{v_1{v}_2}{c^2}{F}_{12}^{\text{el}}$.

Answer to Problem 1.32P

It is showed that $F_{12}^{\text{mag}}\le \frac{v_1{v}_2}{c^2}{F}_{12}^{\text{el}}$.

Explanation of Solution

Write the expression for the Coulomb’s law.

  ${\overrightarrow{F}}_{12}^{\text{el}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{s^2}\widehat{s}$        (I)

Here, ${\overrightarrow{F}}_{12}^{\text{el}}$ is the electrostatic force between the charges, ${\epsilon }_0$ is the absolute permittivity, $q_1,q_2$ are the charges, $s$ is the distance between the charges and $\widehat{s}$ is the unit vector.

Write the equation for the electric force on a charged particle.

  $\overrightarrow{F}=\overrightarrow{E}q$        (II)

Here, $\overrightarrow{F}$ is the electric force, $\overrightarrow{E}$ is the electric field and $q$ is the charge.

Write the equation for the electric force on charge $q_1$ due to charge $q_2$ using equation (II).

  ${\overrightarrow{F}}_{12}^{\text{el}}=\overrightarrow{E}\left({\overrightarrow{r}}_1\right)q_1$        (III)

Here, $\overrightarrow{E}\left({\overrightarrow{r}}_1\right)$ is the electric field at ${\overrightarrow{r}}_1$ due to charge $q_2$ .

Write the equation for the electric field at ${\overrightarrow{r}}_1$ due to charge $q_2$ .

  $\overrightarrow{E}\left({\overrightarrow{r}}_1\right)=\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{s^2}\widehat{s}$

Put the above equation in equation (III).

  ${\overrightarrow{F}}_{12}^{\text{el}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{s^2}\widehat{s}$

Write the expression for the magnitude of the above force.

  $F_{12}^{\text{el}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{s^2}$        (IV)

Write the expression for the magnetic force on a charged particle.

  $\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

Here, $\overrightarrow{F}$ is the magnetic force, $\overrightarrow{v}$ is the velocity of the particle and $\overrightarrow{B}$ is the magnetic field.

Write the equation for the magnetic force on charge $q_1$ due to charge $q_2$ using the above equation.

  ${\overrightarrow{F}}_{12}^{\text{mag}}=q_1\left({\overrightarrow{v}}_1\times \overrightarrow{B}\left({\overrightarrow{r}}_1\right)\right)$        (V)

Here, ${\overrightarrow{F}}_{12}^{\text{mag}}$ is the magnetic force on charge $q_1$ due to charge $q_2$, ${\overrightarrow{v}}_1$ is the velocity of charge $q_1$ and $\overrightarrow{B}\left({\overrightarrow{r}}_1\right)$ is the magnetic field at ${\overrightarrow{r}}_1$ due to charge $q_2$.

Write the equation for the magnetic field at ${\overrightarrow{r}}_1$ due to charge $q_2$.

  $\overrightarrow{B}\left({\overrightarrow{r}}_1\right)=\frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}{\overrightarrow{v}}_2\times \widehat{s}$

Here, ${\mu }_0$ is the absolute permeability and ${\overrightarrow{v}}_2$ is the velocity of charge $q_2$ .

Put the above equation in equation (V).

  $\begin{array}{c}{\overrightarrow{F}}_{12}^{\text{mag}}=q_1\left({\overrightarrow{v}}_1\times \frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}{\overrightarrow{v}}_2\times \widehat{s}\right)\\ =q_1\left[\left(\frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}\right){\overrightarrow{v}}_1\times \left({\overrightarrow{v}}_2\times \widehat{s}\right)\right]\end{array}$        (VI)

Write the identity of vector triple product.

  $\overrightarrow{A}\times \left(\overrightarrow{B}\times \overrightarrow{C}\right)=\left(\overrightarrow{A}\cdot \overrightarrow{C}\right)\overrightarrow{B}-\left(\overrightarrow{A}\cdot \overrightarrow{B}\right)\overrightarrow{C}$

Use the above identity in equation (VI).

  ${\overrightarrow{F}}_{12}^{\text{mag}}=q_1\left[\left(\frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}\right)\left[\left({\overrightarrow{v}}_1\cdot \widehat{s}\right){\overrightarrow{v}}_2-\left({\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2\right)\widehat{s}\right]\right]$        (VII)

The vectors ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$ are perpendicular so that their scalar product must be zero.

Write the expression for the scalar product of ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$.

  ${\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2=0$

Put the above equation in equation (VII).

  $\begin{array}{c}{\overrightarrow{F}}_{12}^{\text{mag}}=q_1\left[\left(\frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}\right)\left[\left({\overrightarrow{v}}_1\cdot \widehat{s}\right){\overrightarrow{v}}_2-0\right]\right]\\ =q_1\left[\left(\frac{{\mu }_0}{4\pi } \frac{q_2}{s^2}\right)\left({\overrightarrow{v}}_1\cdot \widehat{s}\right){\overrightarrow{v}}_2\right]\\ =\frac{{\mu }_0}{4\pi } \frac{q_1{q}_2}{s^2}\left({\overrightarrow{v}}_1\cdot \widehat{s}\right){\overrightarrow{v}}_2\end{array}$

Write the expression for the magnitude of the above magnetic force.

  $\begin{array}{c}{F}_{12}^{\text{mag}}=\frac{{\mu }_0}{4\pi } \frac{q_1{q}_2}{s^2}{v}_1{v}_2\\ =\left(v_1{v}_2\right)\left({\mu }_0{\epsilon }_0\right)\frac{1}{4\pi {\epsilon }_0} \frac{q_1{q}_2}{s^2}\end{array}$        (VIII)

Here, $F_{12}^{\text{mag}}$ is the magnitude of the magnetic force on charge $q_1$ due to charge $q_2$, $v_1$ is the magnitude of the velocity of charge $q_1$ and $v_2$ is the magnitude of the velocity of charge $q_2$ .

Write the expression for the speed of light.

  $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$

Here, $c$ is the speed of light.

Rewrite the above equation for ${\mu }_0{\epsilon }_0$ .

  ${\mu }_0{\epsilon }_0=\frac{1}{c^2}$

Put the above equation in equation (VIII).

  $F_{12}^{\text{mag}}=\left(v_1{v}_2\right)\left(\frac{1}{c^2}\right)\frac{1}{4\pi {\epsilon }_0} \frac{q_1{q}_2}{s^2}$        (IX)

Put equation (IV) in equation (IX).

  $F_{12}^{\text{mag}}=\left(v_1{v}_2\right)\left(\frac{1}{c^2}\right)F_{12}^{\text{el}}$        (X)

Conclusion:

From equation (VII) and (X), in which case the velocities are not perpendicular, the term $\left(v_1{v}_2\right)F_{12}^{\text{el}}$ becomes larger than $F_{12}^{\text{mag}}$ .

Rewrite equation (X) including the case velocities are not perpendicular.

  $F_{12}^{\text{mag}}\le \left(v_1{v}_2\right)\left(\frac{1}{c^2}\right)F_{12}^{\text{el}}$

Thus, it is proved that $F_{12}^{\text{mag}}\le \frac{v_1{v}_2}{c^2}{F}_{12}^{\text{el}}$ in nonrelativistic domain.