Question

A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of the magnetic field if there is a 45o angle between it and the proton's velocity

Answer 1

the strength of the magnetic field is 3 x 10⁻⁵ T

Step-by-step explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

`F = qvB \ sin(\theta)\\\\B = (F)/(qv* sin(\theta)) \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 * 10^(-19) \ C\\\\B = (1.7* 10^(-16))/((1.602 * 10^(-19))* (5* 10^7) * sin \ (45)) \\\\B = 3 * 10^(-5) \ T`

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T