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Old Jul 24th 2011, 10:33 PM   #1
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velocity selection. Speed of particle.

A velocity selector has an electric field of magnitude 2950 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.57 x 10^3 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +3.60 x 10^-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.53 x 10^-9 N, pointing directly upward. What is the speed of this particle?


electric field is: E = 2950 N/C
Speed of a particle is v = 5.57x10^3
B = E / v
2950N/C / 5.57x10^3

=529622 T

F^2 =(Eq)^2 + (Bqv)^2

i must solve for velocity and I'm not sure how to manipulate this formula for what I am solving for.
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Old Jul 26th 2011, 12:38 PM   #2
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Smile try this

Originally Posted by rcmango View Post
A velocity selector has an electric field of magnitude 2950 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.57 x 10^3 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +3.60 x 10^-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.53 x 10^-9 N, pointing directly upward. What is the speed of this particle?


electric field is: E = 2950 N/C
Speed of a particle is v = 5.57x10^3
B = E / v
2950N/C / 5.57x10^3

=529622 T

F^2 =(Eq)^2 + (Bqv)^2

i must solve for velocity and I'm not sure how to manipulate this formula for what I am solving for.
First, I think you made an error in finding the B field. Remember the force on a moving charged particle from a magnetic field is given by qVXB. Initially charged particles move through without being deflected, so the net force must be zero, so:

F = Eq - qVB, since the forces directly oppose each other, using a sign difference is good enough. Setting the force to zero yields:


0 = Eq - qVB, now divide through by the non zero charge,

0 = E - VB, and solve for B, you know they are in the opposite directions.

0 = E - VB
E/V = B = 2950/5.57*10^3 = 0.529623 T, put that back in.

F = 2950q - 0.529623qV

Plug in the known values and solve for the velocity:

1.53*10^-9 = 2950*3.6*10^-12 - 0.529623*3.6*10^-12*V

Now since they only ask for the speed, they do not care about the direction it may head in, so this makes it much easier.

[(1.53*10^-9)-(2950*3.6*10^-12)]/[-(0.529623)*(3.6*10^-12)] = V

V = 4767.54 m/s,
or simply 4767 m/s

I ignored a couple of things. Because the E field acts to move charge upward and the B field acts to move a charge that is moving at a right angle to it downward, they oppose each other. Really the force from a B field is qVXB, that qV "cross" B; which at a right angle is simply qVB. They only give you the resultant force that is acting "directly upward", so you can assume that is the force acting on the charge when it is still moving at a right angle to the B field.

So the charge must be moving about 4767 m/s at a right angle into the B field such that the net force is directly upward on the charge. Basically it is the speed it has when it first enters both fields. At that speed, it would feel a net upward force of what is given.

Also, I chose to call upward the positive direction. Since the E field points upward and the charge is positive, that is the direction a positive charge would be forced from the E field. They gave the force as positive also, so I am assumming the E field is what is causing the charge to move upwards. It would work out to be the same if I had called downward to be positive because knowing the force is acting upward, I would have to change the sign to a negative value, so:

F = -qE + qvXB

-1.53*10^-9 = -qE + qvB, since the cross product is one.
(qE-1.53*10^-9)/qB = (1.53*10^-9-qE)/-qB

Craig
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Old Jul 31st 2011, 11:18 PM   #3
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Thanks for your response, it was very informative, that problem was hard for me to understand, much better now.
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