Question

Two protons move parallel to each other with an equal velocity $v=300\text{km/s}$. The ratio of forces of magnetic and electrical interaction of the protons is given as $x\times {10}^{-6}.$ Find $x$

Answer 1

Force of magnetic interaction, ${\overrightarrow{F}}_{mag}=e(\overrightarrow{v}\times \overrightarrow{B})$
Where $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{e(\overrightarrow{v}\times \overrightarrow{r})}{r^3}$
So, ${\overrightarrow{F}}_{mag}=\frac{{\mu }_0}{4\pi }\frac{e^2}{r^3}[\overrightarrow{v}\times (\overrightarrow{v}\times \overrightarrow{r}\left)\right]$
$=\frac{{\mu }_0}{4\pi }\frac{e^2}{r^3}\left[\right(\overrightarrow{v}\cdot \overrightarrow{r})\times \overrightarrow{v}-(\overrightarrow{v}\cdot \overrightarrow{v})\times \overrightarrow{r}]=\frac{{\mu }_0}{4\pi }\frac{e^2}{r^3}(-v^2\overrightarrow{r})$
And ${\overrightarrow{F}}_{ele}=e\overrightarrow{E}=e\frac{1}{4\pi {\epsilon }_0}\frac{e\overrightarrow{r}}{{|\overrightarrow{r}|}^3}$
Hence, $\frac{|{\overrightarrow{F}}_{mag}|}{|{\overrightarrow{F}}_{electric}|}=-v^2{\mu }_0{\epsilon }_0={\left(\frac{v}{c}\right)}^2=1.00\times {10}^{-6}$