A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x-direction experiences a force of 2.25 \times 10^{-16} N2.25×10 −16 N in the +y-direction, and an electron moving at 4.75 km/s in the -z-direction experiences a force of 8.50 \times 10^{-16} \mathrm{N}8.50×10 −16 N in the +y-direction.
(a) What are the magnitude and direction of the magnetic field?
(b) What are the magnitude and direction of the magnetic force on an electron moving in the -y-direction at 3.20 km/s?

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moya1316 moya1316 · Answer 1 · 2021-05-17T04:00:50+02:00

Answer:

a) B = 0.9375 T -z, b) B = 1.54 T

Explication

a) The magnetic force is

F = q v x B

f = q v B sin θ

bold indicates vectors

For direction let's use the right hand rule.

If the charge is positive

the flea in the direction of velocity, toward + x

the fingers extended in the direction of B

the palm the direction of the force + and

therefore the magnetic field goes in the direction of -z

F = q v B

2.25 10-16 = 1.6 10-19 1.50 103 B

B = 2.25 10-16 / 2.4 10-16

B = 0.9375 T -z

b) now an electron

thumb speed direction, -z

opposite side of palm force + y

therefore the direction of the magnetic field is + x

B = F / qv

B = 8.5 10-16 / 1.16 10-19 4.75 103

B = 1.54 T