Question

Two protons move parallel to each other, keeping distance r between them, both moving with same velocity v. Then the ratio of the electric and magnetic force of interaction between them is:

• A. $\frac{c^2}{v^2}$
• B. $\frac{2c^2}{v^2}$
• C. $\frac{c^2}{2v^2}$
• D. None

Answer 1

The correct option is A

$\frac{c^2}{v^2}$

Electric and magnetic force

Electric force and magnetic force both exists between moving charges.

1. These forces are interrelated, a moving magnet can make electrical charges move, cause an electric current, and in turn cause magnetism.
2. When a charged particle do not deflect in a field, the magnitude of the magnetic force and electric force is equal, hence the ratio is 1.

Explanation

Correct option is option A.

1. Given two similar charges that is protons are moving parallel to each other with same velocity v.
2. Electric force between them (F_E) is the function of product of charges (q²) and distance between them (r). That is $F_E=\frac{1}{4{\epsilon }_0\pi }\frac{q^2}{r^2}$
3. Magnetic force between the protons (F_M) is given as a function of charges, distance between them and velocity (v) with which they are moving. $F_M=\frac{{\mu }_0}{4\pi }\frac{q^2{v}^2}{r^2}$
4. Therefore the ratio is $\frac{F_E}{F_M}\frac{{\displaystyle \frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}}}{\frac{{\mu }_0}{4\pi }\frac{q^2{v}^2}{r^2}}=\frac{1}{{\mu }_0{\epsilon }_0{v}^2}$
5. If C is the velocity of light then, $\frac{F_E}{F_M}=\frac{C^2}{v^2}----\left(\frac{1}{{\mu }_0{\epsilon }_0}=\frac{1}{C^2}\right)$

Therefore ration of electric and magnetic force is $\frac{C^2}{v^2}$. Option A is correct.