MARY B.

asked • 02/20/17# Physics electricity question

A particle of charg, q, enters a region where an electric field is uniform, E = 80000 V/m, and directed downward (-y direction). Perpendicular to the electric field, and directed in the + z direction, is a magnetic field, B = 0.4 T. If the particle is moving to the right with appropriate speed, v, the particle will not be deflected by these crossed electric and magnetic fields. What speed should be selected to achieve this effect?

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## 1 Expert Answer

Steven W. answered • 02/20/17

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Hi Mary!

The speed must be such that the magnitudes of the (oppositely directed) electric and magnetic forces on the charge are equal (and thus they cancel out, since they are pointing in opposite directions).

The electric force on the charge q is given by:

F

_{E}= qEThe magnetic force is given by:

F

_{B}= qvBsinθthough since the particle is traveling along the y axis, and the magnetic field is along the z axis, the angle θ between them is 90

^{o}, so that sin(θ) = 1 and F_{B}= qvB.We need:

F

_{E}= F_{B}qE = qvB

E = vB

v = E/B

This is the velocity for the particle needed to travel through this region undeflected. Note that this is independent of charge.

I hope this helps! Just let me know if you have more questions about this.

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Arturo O.

Epoints in the -y direction (given). Assuming the x and y axes are in the plane of the page, +x points right, -y points down, so in a right-handed system, +z points point up from the page. It is given that the magnetic field points in the +z direction. Then if the particle is moving to the right, as stated, it is moving in the +x direction, which makesv×Bpoint in the -y direction. That makesEandv×Bboth point in the -y direction, so the magnetic and electric forces point in the same direction, rather than opposite directions, and do not cancel.v×Bpoints in the +y direction, opposite to the direction ofE, and the magnetic force can cancel the electric force with an appropriate magnitude forv. I concur with Steven's solution to the problem, but with the additional condition that the charge be moving in the -x direction (i.e. to the left), contrary to the problem statement. Note also that Steven's algebraically correct solution of_{y}/B_{z}= (-80000 V/m) / (0.4 T)v| = |E| / |B|02/21/17