(a) a cosmic ray proton moving toward the earth at 5.00×107m/s experiences a magnetic force of 1.70×10−16n. what is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) is the value obtained in part (a) consistent with the known strength of the earth’s magnetic field on its surface? discuss.

a) B = 3 × 10⁻⁵ T

b) The earth's magnetic field is around 5 × 10⁻⁵ T. thus, the obtained magnetic field is consistent with the earth's magnetic field

Explanation:

Given:

Speed of proton, v = 5.00 × 10⁷ m/s

Magnetic force, F = 1.70×10⁻¹⁶ N

Angle between the force and the velocity, Θ = 45°

a) Now,

the magnetic force is given as:

F = qvBsinΘ

where,

q is the charge = 1.6 × 10⁻¹⁹ C for proton

B is the Magnetic field

on substituting the values in the above equation, we get

1.70×10⁻¹⁶ N = 1.6 × 10⁻¹⁹  5.00 × 10⁷ × B × sin45°

or

B = 3 × 10⁻⁵ T

b) The earth's magnetic field is around 5 × 10⁻⁵ T. thus, the obtained magnetic field is consistent with the earth's magnetic field

gas and liquid

the maxium height is 34m/s