An electrically charged bullet of mass 0.010 kg leaves the barrel of a gun trave

Question

An electrically charged bullet of mass 0.010 kg leaves the barrel of a gun traveling parallel to the ground and perpendicular to the Earth's magnetic field at 2.000e+3 m/s. In theory, what charge could the bullet carry in order to remain at a constant height? Assume that the Earth is flat, and that the magnetic field (0.500 gauss) is horizontal and points left if you are looking in the direction of the bullet's motion.

I was initially going to use the equations r=mv/qB but I dont have anything to plug in for r. I dont know where to go from here. Thanks ahead for any help

Explanation / Answer

m = 0.010 kg, v = 2.000e+3 m/s. B = 0.500 gauss = 5.0*10^-5 T
B is horizontal and points left if you are looking in the direction of the bullet's motion. By the right hand rule, v x B is upward. The magnetic force should be upward to balance the weight so that the charge remains at a constant height. Therefore, the charge is positive
qvB = mg q = mg/(vB) = 0.98 C

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