Aproton moving eastward with a velocity of 5.0 x 103 mis enters a magnetic field of 0.20 t pointing northward. what is the magnitude and direction of the force that acts on the proton?

1) Magnitude

The magnetic force acting on a moving charge is given by:

where

q is the charge of the particle

v is its speed

B is the magnitude of the magnetic field

is the angle between the directions of v and B.

In this problem, we have a proton, so the charge is ; its speed is , while the magnetic field is . The angle is , because the proton is moving perpendicular to the magnetic field. Therefore, if we substitute these data into the previous equation, we find:

So, the magnitude of the force is .

2) Direction

The direction of the force can be determined using the right-hand rule:

- index finger: direction of the velocity (eastward)

- middle finger: direction of the magnetic field (northward)

- thumb: direction of the force (upward).

So, the direction of the force is upward.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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