At a given instant, a particle with a mass of 4.00×10 3 kg and a charge of 3.70×

Question

At a given instant, a particle with a mass of 4.00×103 kg and a charge of 3.70×108 C has a velocity with a magnitude of 2.50×105 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.

Part A

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

Part B

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

Part C

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

Part D

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

Explanation / Answer

A) F = Q(V*B)

by using right hand screw rule force is acting in the positive z rirection

b) F = 3.7*10^-8*2.5*10^5*0.8 = 7.4*10^-3 N

c) acceleration is acting in the same direction z diection

c) a = F/m = 7.4*10^-3/4*10^-3 = 1.85 m/sec^2

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