Question

At the equator, near the surface of the Earth, the magnetic field is approximately 50 ?? northward, and the electric field is about 100 N/C downward in fair weather. Find the magnitude and direction of gravitational, electric, and magnetic forces on an proton in this environment, assuming the proton has an instantaneous velocity of 5?106m/s directed to the west.

Answer 1

Mass of the proton, m = 1.67 x 10^-27 kg

(a) Magnitude of the gravitational force -

F = mg

F = 1.67 x 10^-27 x 9.81

= 1.64 x 10^-26 N

Direction of the gravitational force is in the downward direction towards the earth.

(b) Charge of a proton, q = 1.6 x 10^-19 C

Magnitude of the electric field, E = 100 N/C

So, magnitude of the Electric force -

F = qE

= 1.6 x 10^-19 x 100

= 1.6 x 10^-17 N

The direction of the electric field (which you have said is downward) is the direction a positive charge would move.

Since, the proton is positively charged so the direction of electric force is in the downward direction.

(c) Magnetic force -

F =qv XB

F =qvBsin(theta)

Since the velocity of the proton is to the west and the magnetic field is directed northward, then the two are at right angles (90 deg) Therefore, sin(90) =1 and F reduces to F=qvB

F =1.6 x 10^-19 x 5 x 10^6 x 50 x 10^-6

= 1.6 x 10^-19 x 5 x 50

= 4.0 x 10^-17 N

And the direction of this force is Directed away from you into the paper.