Question

A 3.50 g bullet moves with a speed of 115 m/s perpendicular to the earth's magnetic field of 5.00 x 10^-5 T. If the bullet possesses a net charge of 16.5 x 10^-9 C, by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.50 km? Express your answer in three significant figures.

Answer 1

F = q*(V x B)

This will cause the bullet to accelerate at right angles to its velocity at the rate a = F/m:

a = q*V*B/m

After a time t, the deflection will be

s = 0.5*a*t² = 0.5*q*V*B*t²/m

The velocity in the direction of motion remains V, so the time it take the bullet to travel a distance D is t = D/V

Put it all together to get

s = 0.5*q*V*B*(D/V)²/m

s = 3.305900621 *10 ^-6  m

plz chek calculation before submission