Aproton moving with a velocity of 4.0 × 104 m/s enters a magnetic field of 0.20 t. if the angle between the velocity of the proton and the direction of the magnetic field is 60°, what is the magnitude of the magnetic force on the proton? (e = 1.60 × 10-19
c.

The magnitude of the magnetic force on the proton is 1.1 × 10⁻¹⁵ N

Let's recall magnetic force on moving charge as follows:

where:

F = magnetic force ( N )

B = magnetic field strength ( T )

q = charge of object ( C )

v = speed of object ( m/s )

θ = angle between velocity and direction of the magnetic field

Let's tackle the problem now !

Given:

speed of proton = v = 4.0 × 10⁴ m/s

magnetic field strength = B = 0.20 T

charge of proton = q = 1.60 × 10⁻¹⁹ C

direction of speed = θ = 60°

Asked:

the magnitude of the magnetic force on the proton = F = ?

Solution:

Grade: High School

Subject: Physics

Chapter: Magnetic Field

truck y has a greater velocity than truck x.