At a given instant, a particle with a mass of 4.10×10 3 kg and a charge of 3.70×

Q: At a given instant, a particle with a mass of 4.10×10 3 kg and a charge of 3.70×

part1 : j x - i = k so , +z part2 : F = qvB = 3.70×10^-8 * 1.9×10^5 * 0.800 = 0.005624 N part3 : same as F that is +z part 4 : a = F/m = 0.005624/(4.10×103) = 1.3717 m /s^2

ID: 1423038 • Letter: A

Question

At a given instant, a particle with a mass of 4.10×103 kg and a charge of 3.70×108 C has a velocity with a magnitude of 1.90×105 m/s in the +y direction. It is moving in a uniform magnetic field that has magnitude 0.800 T and is in the -xdirection.

A) What is the direction of the magnetic force on the particle ?

B) What is the magnitude of the magnetic force on the particle ?

C) What is the direction of the resulting acceleration of the particle?

D) What is the magnitude of the resulting acceleration of the particle?

+x. -x. +y. -y. +z. -z.

Explanation / Answer

part1 : j x - i = k so , +z

part2 : F = qvB = 3.70×10^-8 * 1.9×10^5 * 0.800 = 0.005624 N

part3 : same as F that is +z

part 4 : a = F/m = 0.005624/(4.10×103) = 1.3717 m /s^2

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At a given instant, a particle with a mass of 4.10×10 3 kg and a charge of 3.70×

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