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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

hèllo, sorry i really need some more points so if you need this qustion answered than retype it