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Finding Magnitude of Magnetic Field

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron accelerated from rest through a voltage of 420 V enters a region of constant magnetic field. If the electron follows a circular path with a radius of 29 cm, what is the magnitude of the magnetic field?

    2. Relevant equations

    r=mv/lqlB
    K=VQ
    Conservation of Energy: Total Energy = KE + PE
    KE = 1/2 mv^2

    3. The attempt at a solution
    So I know I need to use r=mv/qB to find the Magnitude of the Magnetic Field.
    r= .29m
    m=9.11x10-31 kg
    q= 1.6x10-19
    v= ???
    B= what we're solving for.

    If I can find the velocity of the electron, then I will be able to solve for the Magnitude of the Magnetic field.

    I tried looking online for other solutions and someone said this, but I can't really understand it that well:

    From conservation of energy we can calculate the velocity

    K = V*q so 1/2*m*v^2 = V*q...so v = sqrt(2*V*q/m) = sqrt(2*410*1.60x10^-19/9.11x10^-31)
    = 1.20x10^7m/s

    Now we have F = m*a ....q*v*B = m*v^2/r

    What is this K=vq equation? I couldn't find it when I was looking for it. It's the Voltage x The Charge?

    Because we know the particle starts at rest, we know that the potential energy = kinetic energy. But I really don't know how to find the velocity. Help please!
     
  2. jcsd
  3. Mar 2, 2016 #2
    What work is done on a charged particle, on passing through a p.d.?
     
  4. Mar 3, 2016 #3
    What is a p.d? We also haven't talked about work on charged particles at all. I'm taking the general algebra based physics, not the calculus based one.
     
  5. Mar 3, 2016 #4
    P.d. stands for potential difference; my bad.

    Ok, here - do you know the magnitude of the force that acts on a charge q in an electric field E?
     
  6. Mar 3, 2016 #5
    Ah right!!! Of course. The Potential Difference.
    The equations I know for Electric Field are...
    E= Force/q0
    E= kq/r
    But where does the potential difference come into play? The P.d is essentially a voltage, right?
     
  7. Mar 3, 2016 #6
    Yes, it is.
    Travelling along an electric field causes a change in p.d. - do you know of any relation between this potential and the existing electric field?
     
  8. Mar 3, 2016 #7
    Also, coming back to Felectric = qE -
    so if knowing that work done by a force is F.s (where s is the displacement along the force), can we write the work done in terms of the electrc field?
     
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