When a particle of charge q moves with a velocity v⃗ in a magnetic field b⃗ , the particle is acted upon by a force f⃗ exerted by the magnetic field. to find the direction and magnitude of this force, follow the steps in the following tactics box. keep in mind that the right-hand rule for forces shown in step 2 gives the direction of the force on a positive charge. for a negative charge, the force will be in the opposite direction. if the magnetic field of the wire is 4.0×10−4 t and the electron moves at 6.0×106 m/s , what is the magnitude f of the force exerted on the electron? express your answer in newtons to two significant figures.

Explanation:

For a charged particle moving perpendicularly to a magnetic field, the magnitude of the force exerted on the particle is:

where

q is the magnitude of the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

In this problem, we have

is the charge of the electron

is the velocity

is the magnetic field

Substituting into the formula, we find the force:

15.3 cos 65 degrees trig

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: