At a given instant, a particle with a mass of 4.40×10 3 kg and a charge of 3.60×

Question

At a given instant, a particle with a mass of 4.40×103 kg and a charge of 3.60×108 C has a velocity with a magnitude of 1.80×105 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 ?

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

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

At a given instant, a particle with a mass of 4.40×103 kg and a charge of 3.60×108 C has a velocity with a magnitude of 1.80×105 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 ?

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

Part B

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

F =   N  

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

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

Explanation / Answer

A) the direction of force is

-z direction

B) F= qvB sin(90) = 3.6*10-8*1.8*105*0.8

F = 5.184*10-3 N

C) acceleration, a = F/m = 5.184*10-3 / (4.4*10-3)

a=1.1781 m/s2

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