At a given instant, a particle with a mass of 4.30×10 3 kg and a charge of 3.80×

Question

At a given instant, a particle with a mass of 4.30×103 kg and a charge of 3.80×108 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.

A) What is the direction of the magnetic force on the particle ? (+x.-x.+y.-y.+z.-z)

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

C) What is the direction of the resulting acceleration of the particle? +x.-x.+y.-y.+z.-z)

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

Explanation / Answer

A)

The magnetic force problem uses the right hand rule. First, point your fingers in the direction of the velocity (in this case, +x). Then, curl your fore fingers in the direction of the negative field (negative z). This would give us a positive y direction, however, a proton has a positive charge (the right hand rule works with the electron). In this case, we need to rotate the result and say its direction is actually negative y direction.

The proton, once it enters the magnetic field, will move in a circular motion while it is in the magnetic field. In a field that is uniform, the particle will make a full 180 degree circle and have the same initial velocity when it leaves the magnetic field.

So direction of magnetic force on particle is in -z direction

B)

Magnetic Force = qvB
  

= 7.6*10^-3 N

C)As we know that F = ma so acceleration also have direction in +z direction

D)

As we know that F = m*a

so ,

a = F/m

a =  7.6*10^-3 N/4.30*10^-3

a = 1.767 m/s^2

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