Question

In a certain region, the earth's magnetic field has a magnitude of 5.1 × 10-5 T and is directed north at an angle of 54° below the horizontal. An electrically charged bullet is fired north and 16° above the horizontal, with a speed of 660 m/s. The magnetic force on the bullet is 3.6 × 10-10 N, directed due east. Determine the bullet's electric charge, including its algebraic sign (+ or -).

Answer 1

Given

magnitude of the magnetic field is B = 5.1*10^-5 T

the direction is 54 degrees below the horizontal

bullet fired at 16 degrrees above the horizontal with a speed v = 660 m/s

the magnetic force is F = 3.6*10^-10 N

we know that the magnetic force is F = q*V*B sin theta

theta is the angle between V and B

here theta = 54+16 = 70 degrees

F = qvB sin theta

q = F /vB sin theta

q = (3.6*10^-10)/(660*5.1*10^-5 sin70) C

q = 1.1381580454288*10^-8 C

q = 1.1381580*10^-9 C

By right hand rule the magnetic force direction to the west for positive charge , but given the force is due east so the charge is -ve  

the charge of the bullet is q = -1.1381580*10^-9 C