Question

At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of 4.97x10-5T and points upward and toward the north at an angle 55º above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of 1.9x10-28 kg is moving directly down toward Earth's surface wiht a speed of 5.68x107 m/s. What is the Force on the muon?

Answer 1

there would be two forces acting on the muon

first,
gravitational force towards the center of earth = 1.9*10^-28*10 = 1.9*10^-27 N

second magnetic force

Force on proton, F = q(v × B)

F = qvB*cos $\small \theta$

use cos $\theta$ instead of sin, this is because the perpendicular component of magnetic field is used and to calculate perpendicular component

here given:

the magnitude of magmetic field B =4.97*10^-5T

charge of the cosmic ray muon q =1.60*10^-19C

speed of cosmic ray muon v = 5.68*10⁷m/s

= 1.60*10^-19 *5.68*10⁷*4.97*10^-5*cos(55) = 2.59*10^-16 N

as gravitation force is much lesser than magnetic force we can ignore that..
so magnitude of force is 2.59*10^-16 N