A 3.25 g bullet picks up an electric charge of 1.25 C as it travels down the bar

Question

A 3.25 g bullet picks up an electric charge of 1.25 C as it travels down the barrel of a rifle. It leaves the barrel at a speed of 417 m/s traveling perpendicular to the earth’s magnetic field, which has a magnitude of 5.50×104 T .

A) Calculate the magnitude of the magnetic force on the bullet.

Express your answer with the appropriate units.

B) Calculate the magnitude of the bullet’s acceleration due to the magnetic force at the instant it leaves the rifle barrel.

Express your answer with the appropriate units.

Explanation / Answer

Part A:
Use the formula, F = q*v*B, to calculate the force on the bullet:
q = 1.25 µC = 1.25 x 10^-6 C
v = 417 m/s
B = 5.50 x 10^-4 T
F = q*v*B
F = (1.25 x 10^-6 C)(417)(5.50 x 10^-4)  =2.87*10^-7 N

Part B:
Use the formula, F = ma, to solve for the acceleration:

F = 2.87*10^-7 N
m = 3.25 g = 0.00325 kg
F = ma
a = F/m = 2.87*10^-7/ 0.00325 = 8.83 x 10^-5 m/s^2


Hope this helps! :-)

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