Question

A proton 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?

Answer 1

1) Magnitude

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

`F=qvB \sin \theta`

where

q is the charge of the particle

v is its speed

B is the magnitude of the magnetic field

`\theta` is the angle between the directions of v and B.

In this problem, we have a proton, so the charge is
`q=1.6 \cdot 10^(-19) C`; its speed is
`v=5.0 \cdot 10^3 m/s`, while the magnetic field is
`B=0.20 T`. The angle is
`\theta=90^(\circ)`, because the proton is moving perpendicular to the magnetic field. Therefore, if we substitute these data into the previous equation, we find:

`F=(1.6 \cdot 10^(-19)C)(5.0 \cdot 10^3 m/s)(0.20 T)(\sin 90^(\circ))=1.6 \cdot 10^(-16) N`

So, the magnitude of the force is
`1.6 \cdot 10^(-16) N`.

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.