Question

Two protons move parallel to each other, keeping distance $r$ between them, both moving with same velocity $\overrightarrow{v}$. Then the ratio of the electric and magnetic force of interaction between them is:
(c - Velocity of light)

• A. $c^2/v^2$
• B. $2c^2/v^2$
• C. $c^2/2v^2$
• D. none

Answer 1

The correct option is A $c^2/v^2$

There are two forces electric force and magnetic force experienced by proton.

The electric force between the two protons is given as

$F_E=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

Now, the magnetic force is

$F_B=qvB$

$B$ is the magnetic field.

From Biot- savarts law

$B=\frac{{\mu }_0}{4\pi }\frac{IdL}{r^2}$

dL is the length element of the conductor but, the expression will be for charge where is no current.

So,

we know that,

$I=\frac{q}{t}$

$IdL=q\frac{dL}{dt}$

$IdL=qv$

Now, the magnetic field is

$B=\frac{{\mu }_0}{4\pi }\frac{qv}{r^2}$

The magnetic force is

$F_B=qvB$

$F_B=qv\frac{{\mu }_0}{4\pi }\frac{qv}{r^2}$

$F_B=\frac{{\mu }_0}{4\pi }\frac{q^2{v}^2}{r^2}$

Now, the ratio of the electric force and the magnetic force is

$\frac{F_E}{F_B}=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{r^2}}{\frac{{\mu }_0}{4\pi }\times \frac{q^2{v}^2}{r^2}}$

$\frac{F_E}{F_B}=\frac{1}{{\mu }_0{\epsilon }_0{v}^2}$......(I)

We know that, the speed of light

$c=\sqrt{\frac{1}{{\mu }_0{\epsilon }_0}}$

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

Now, put the value of $\frac{1}{{\mu }_0{\epsilon }_0}$ in equation (I)

$\frac{F_E}{F_B}=\frac{c^2}{v^2}$

Hence, the ratio of electric force and magnetic force will be $\frac{F_E}{F_B}=\frac{c^2}{v^2}$.