In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50 mC and initially moving northward at 4.75 km>s is deflected toward the east.

(a) What is the sign of the charge of this particle.
(b). Find the magnetic force on the particle.

Explanation:

(a). Since the charge enters the magnetic field and was deflected it carries a +ve charge

(b). The magnetic force on the particle is given as

F = Bqv

Where B = magnetic field in Tesla T

1.25-T

q= charge 8.50 mC

v = average velocity of charge

4.75 km/s converting to

m/s we have 4.75*1000= 4750m/s

Substituting our given corresponding data we have

F= 1.25*8.50*10^-3*4750

F= 50.47N

The magnetic force on the particle is 50.47N

answer; i believe the correct answer is (40,000 j) good luck

the answer would be   dignified - apex