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?

Arturo O.

From the given information, I get the impression that the charge must be moving toward the left to get the magnetic and electric forces to cancel, even though the problem states it is moving to the right.
 
E points 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 makes v×B point in the -y direction.  That makes E and v×B both point in the -y direction, so the magnetic and electric forces point in the same direction, rather than opposite directions, and do not cancel.  
 
However, if the charge is moving to the left, v×B points in the +y direction, opposite to the direction of E, and the magnetic force can cancel the electric force with an appropriate magnitude for v.  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 
 
v = E/B  (which in this problem is Ey/Bz = (-80000 V/m) / (0.4 T)
 
is a ratio of a negative E to a positive B, which makes v come out negative, and hence moving to the left instead of the right.  If you only want the magnitude of the velocity, you can write
 
|v| = |E| / |B|
 
But the direction of the magnetic force is reversed if the direction of the velocity vector is reversed.
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02/21/17

Arturo O.

After thinking about this a little more, I can see that if +x points left and +y points up, then +z points into the page.  Then B points into the page (in the +z direction, given).  Then v×B points up if the charge is moving to the right (i.e. the charge is moving in the -x direction).  In this situation, the magnetic force opposes the electric force.  So with a right-handed coordinate system of +x pointing left, +y pointing up, and +z pointing into the page, the charge must move in the -x direction (i.e. to the right) to produce a magnetic force opposite the given electric force.  I am predisposed to make +x point toward the right, but that does not work in this problem.  Anyway, Steven's solution is correct with the coordinate system defined as +x to the right, +y up, and +z into the page, with the charge moving in the -x direction, which is to the right.
 
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02/21/17

Arturo O.

I meant +x to the left.
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02/21/17

1 Expert Answer

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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:
 
FE = qE
 
The magnetic force is given by:
 
FB = qvBsinθ
 
though since the particle is traveling along the y axis, and the magnetic field is along the z axis, the angle θ between them is 90o, so that sin(θ) = 1 and FB = qvB.
 
We need:
 
FE = FB
 
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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