A particle (q = –4.0 µC, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s2. What is the magnitude of the magnetic field?

The magnitude of the magnetic field is 4.73 T.

Explanation:

Given that,

Charge of the particle,

It is moving with a velocity of 2 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s². We need to find the magnitude of magnetic field.

When it enters in magnetic field,

B is magnetic field

So, the magnitude of the magnetic field is 4.73 T.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.