Question

Two protons a and b move parallel to the x-axis in opposite directions with equal speeds v. At the instant shown, the ratio of magnetic force and electric force acting on the proton a is (c = speed of light in vacuum)

Answer 1

Answer:

The ratio of magnetic force and electric force acting on the proton is given as

$F_b : F_e = \frac{v^2}{c^2}$

Explanation:

Magnetic field due to moving proton at a distance "r" from it is given as

$B = \frac{\mu_o ev}{4\pi r^2}$

now we know that force on a moving charge is given as

$F_b = evB$

so we have

$F_b = ev(\frac{\mu_o ev}{4\pi r^2})$

$F_b = \frac{\mu_o e^2v^2}{4\pi r^2}$

now electrostatic force between two charges is given as

$F_e = \frac{e^2}{4\pi \epsilon_0 r^2}$

now ratio of magnetic force and electric force is given as

$F_b : F_e = \mu_o\epsilon_o v^2$

also we know that

$c^2 = \frac{1}{\mu_o\epsilon_o}$

so we will have

$F_b : F_e = \frac{v^2}{c^2}$

#Learn

Topic : Magnetic force on moving charge

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