Magnetism (Charge, E-field, potential difference, mass) - A bit confused

  • #1

Homework Statement



Image: http://img.photobucket.com/albums/v696/talimtails/PP23.jpg

A particle with unknown mass and charge moves with a constant speed v = 1.9 x 10^6 m/s as it passes undeflected through a pair of parallel plates, as shown above. The plates are separated by a distance d = 6.0 x 10^-3 m, and a constant potential difference V is maintained between them. A uniform magnetic field of magnitude B = .2 T directed into the page exists both between the plates and in a region to the right of them as shown. After the particle passes into the region to the right of the plates where only the magnetic field exists, its trajectory is circular with radius .1 m.

a. What is the sign of the charge of the particle? Justify your answer.
b. On the diagram, clearly indicate the direction of the electric field between the plates. Justify your answer.
c. Determine the magnitude of the potential difference V between th plates.
d. Determine the ratio of the charge to the mass (q/m) of the particle.

Homework Equations


B = kI/r
F = qvBsin*
F = BIL sin*
V = IR
V = Er
F = qE
E = Blv
F = mv^2/r (centripetal force should act on the particle after its trajectory becomes circular)


The Attempt at a Solution



a. I am not sure how I would be able to determine this, honestly. Could anyone give me something to ponder about that might help me solve this?

b. Using the right hand rule, if the B is directed into the page, the force is down and current is to the left. If the force is down, then the electric field will be down as will if the charge is positive, or up if the charge is negative.

c. If I used E = V/r and E = vBL, I could say that V = vBLr, which are the 4 unknown's that I have (L being replaced by d)

d. The mass and charge could be determined by F = qvB = mv^2/r, but that would give me two unknown's. However, I was wondering, if the charge was positive or negative, could I just use the mass of a proton or electron (or even neutron) for the mass and solve for the charge from the above equation (and then finding the ratio)?

Thanks for any help at all :)
 

Answers and Replies

  • #2
312
0
If the current is to the left, then what is the sign of the moving charge? If the magnetic force is down, then in what direction does the electric force act? The electric field?

You should be able to solve for the field and then the voltage for part c

Part d is only asking for the value of q/m, which you can put in terms of everything else

Keep in mind that nowhere does the problem state that the particle is either a proton or electron, it could be anything with charge and mass
 
  • #3
If the current is to the left, and the charge is able to resist the current and go to the right, would it be negative because it is going in the opposite direction? There is no resistance, otherwise I would say the charge is positive since it would go against the current through a resistor.

The electric field should act, if the charge is indeed negative, up above the page. I'm not sure how to find electric force though...

Using E = vBL, I get (1.9 x 10^6 m/s)(.2 T)(6 x 10^-3 m) = 2280 V

And for the last part, I need a ratio. But, even so, I could get q/m on one side of the equation

q/m = v^2/rvB -> v/rB -> (1.9x10^6)/(.1 m x .2 T) = 1.9 x 10^6 / .02, which gives me a 95 mil:1 ratio.. which can't be right. Am I supposed to use the distance between the magnetic plates?

And.. thanks for your help ^^;;
 
  • #4
312
0
The electric field should act, if the charge is indeed negative, up above the page. I'm not sure how to find electric force though...
The sum of the forces is zero, which means that the electric force acts upward. If the charge is negative, which direction is the field?

Using E = vBL, I get (1.9 x 10^6 m/s)(.2 T)(6 x 10^-3 m) = 2280 V
You get the same answer by summing up the forces
Fm = Fe = qvB = qE
E = vB
V = El = vBl

And what makes you think your ratio is wrong? Compare the charge to mass ratio of an electron with what you found
 
  • #5
If the charge is negative, and the electric force is upward, then the electric field is downward. Okay.. I can see why.

I just thought that the ratio was so big I made some huge error.. =/
But looking at the mass of an electron, it's charge to mass ratio is even greater than in this equation.. so I guess I shouldn't complain

Honestly.. thanks again ^^;;
 

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