Magnetism and Electric Potential Problem

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SUMMARY

The discussion centers on calculating the uniform magnetic field required for an electron to travel in a straight line between two parallel plates with a potential difference of 120 V and a separation of 20.0 mm. The electron is initially accelerated through a potential difference of 1.6 kV, resulting in a velocity of 2.37e7 m/s. The magnetic field was calculated using the formula B = V/(distance)(v), yielding a value of 2.53e-4 T. However, the user, Beverly, encountered issues with the direction of the magnetic field and the correctness of the calculated value.

PREREQUISITES
  • Understanding of electric potential and kinetic energy relationships
  • Familiarity with the Lorentz force equation (F = qvB)
  • Knowledge of the right-hand rule for magnetic fields
  • Basic principles of electromagnetism, specifically regarding charged particles in electric and magnetic fields
NEXT STEPS
  • Review the right-hand rule and its application in determining magnetic field direction
  • Study the relationship between electric fields and magnetic fields in charged particle motion
  • Explore the effects of varying potential differences on electron trajectories
  • Investigate the implications of fringing fields in practical applications
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as engineers working with charged particle dynamics in electric and magnetic fields.

bbbbbev
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Hi. I'm having some trouble with this problem:

An electron accelerated from rest through potential difference 1.6 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 120 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?

I was thinking that all I really needed to do was figure out the velocity of the electron and plug all the values into the equation

F = qvBsin(phi)

phi = 90 degrees in this problem, so the equation can just be written as

F = qvB

To find v, I used the fact that change in kinetic energy = change in potential energy.

KE = PE

(1/2)mv^2 = q(deltaV)

v = sqrt((2q(deltaV))/m)

After plugging in all the numbers I got:

v = sqrt(2(1.6e-19C)(1600V)/(9.11e-31kg))

v = 2.37e7 m/s

When I plug that into the magnetic field equation I get:

B = F/qv, so B = Eq/qv. Since E = V/d and the q's cancel, I then used

B = V/(distance)(v)

B = (120 V)/(0.02 m)(2.37e7m/s) = 2.53e-4 T

But that was not the correct answer. I know this is a lot of work to look through, but I'd appreciate any help anyone could give me because I'm really not sure I'm looking at this problem correctly. What am I missing.

Thanks,

Beverly
 
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you got the value of B correct, but you must be careful about hte direction of the magnetic field. Since the magenetic field obeys the right hand rule which way should hte magnetic field point?
 
I think the magnetic field is in the k direction, which is to the left, but I don't think that value for B is right. I put that in as my answer (on webassign), and it counted it wrong. I don't know what else to try! Thanks for you help! Beverly
 

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