Question

a proton moving at right angle in a magnetic field of 0.1 T, experience a force of 2*10^-12N, thus the speed of proton is

Answer 1

Solution. The magnitude of the magnetic force on a charge particle

$$F=qvBsin\theta$$

where F is force; q=1.6 * 10^-19C is charge of the proton; v is speed of proton; Ɵ=90⁰ is angle between magnetic field and velocity.

Therefore, the speed of proton is equal to

$$v=\frac{F}{qBsin\theta}=\frac{2\times10^{-12}N}{1.6\times10^{-19}C\times0.1T}=1.25\times10^8\frac{m}{s}$$

Answer.

$$v=1.25\times10^8\frac{m}{s}$$