A 3.50g bullet moves with a speed of 110m/s perpendicular to the Earth\'s magnet

Question

A 3.50g bullet moves with a speed of 110m/s perpendicular to the Earth's magnetic field of 5x10^-5 T. If the bullet possesses a net charge of 11.5x10^-9 C, by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.50km? A 3.50g bullet moves with a speed of 110m/s perpendicular to the Earth's magnetic field of 5x10^-5 T. If the bullet possesses a net charge of 11.5x10^-9 C, by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.50km?

Explanation / Answer

As we know that the lorentz force is given by

F = qvB

F = 11.5 x 10-9 x 110 x 5 x 10-5

F = 1.26 x 10-11 N

Now force is also be given by

F = ma

1.26 x 10-11 = 3.5 x 10-3 x a

a = 3.6 x 10-9 m/s2

now given the horizontal distance covered by the bullet is 1.5 KM

hence

T = distance / speed

T = 1.5 x 103 / 110

T = 13.63 s

Now the distance it deflected at the same T

Y = 0.5 aT2

Y = 0.5 x 3.6 x 10-9 x (13.63)2

Y = 3.34 x 10-7 m

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