A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/s. What is the force experienced by the proton if it enters parallel to the direction of the magnetic field with a speed of 200 m/s?



8 N

4 N

2 N

0 N

My answer would have to be... 8N?

F=qvBsinα

If sinα=0 F=0

0N man!

To determine the force experienced by a charged particle moving through a magnetic field, we can use the formula:

F = q * v * B * sin(theta)

Where:
F is the force experienced by the charged particle
q is the charge of the particle (in this case, a proton with a charge of +1.6 x 10^-19 C)
v is the velocity of the particle
B is the magnetic field strength
theta is the angle between the velocity vector and the magnetic field vector

In this case, the proton is entering the magnetic field parallel to the direction of the magnetic field, which means the angle between the velocity vector and the magnetic field vector (theta) is 0 degrees. When theta is 0 degrees, sin(theta) is also 0, which means the force experienced by the proton is 0 N.

So, the correct answer is: 0 N.