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Circular Mtion and Magnetic field Question This is really giving me a hard time!

  1. Dec 5, 2004 #1
    Circular Mtion and Magnetic field Question ....This is really giving me a hard time!

    1) In the 'Wall of Death' motorcycle stunt,the rider gradually rides higher around a curving wall of a circular arena untill he is travelling around a vertical wall.

    a) show the forces acting on a rider and the bike and explain how their weight is supported.
    b)Before reaching the vertical wall ,the rider travels more slowly round the sloping section of the arena.What additional force provides support at this stage?

    2)A magnetic field of 0.5 T and an electric field of 5000V/m are arranged to produce opposing forces on an elcetron.calculate the speed at which electron is travel in a straight line through these fields.

    I havent got a clue how to do the above questions.Any hints will highly be appriciated.
  2. jcsd
  3. Dec 5, 2004 #2


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    RE: 2 - set [itex]\vec E + \vec v \times \vec B = 0[/itex] and obtain

    [tex]v = \frac {E}{B}[/tex]

    You can figure out the directions
  4. Dec 5, 2004 #3
    1. a) Draw a free-body diagram for the rider/bike: include vectors of all forces acting on the two. What forces are acting on the rider/bike while they approach the vertical wall? Once you do this you will understand how the weight is supported.
    b) This question is easy once you get a.

    2. This type of system is used as a velocity selector in most mass spectrometers. If the electron travels in a straight line, the electric force must be equal in magnitude to the magnetic force. Therefore:
    where [itex]q[/itex] is the charge on the electron, [itex]\vec{E}[/itex] is the electric field strength, [itex]\vec{B}[/itex] is the magnetic field strength, and [itex]\theta[/itex] is the angle between the velocity vector of the electron and the magnetic field vector. Since the electron is traveling in a direction perpendicular to the magnetic field vector, [itex]\theta[/itex] is 90 degrees. Can you find an expression for the speed of the electron? Plug in the values given in the question to get the numerical answer.

    EDIT: I see Tide has developed the expression above. Now you know where it comes from.
    Last edited: Dec 5, 2004
  5. Dec 8, 2004 #4
    | ----@

    I have tried to draw the picture of the track ,the outer boundry(kinda like L shaped curve) is the track while @=the stunt man and ---- = the bike...so,where is the force acting??
    I think the weight of both bike and the man is acting downward so,the horizontal component(I dunno which force is that is trying to act to the centre....i think i am wrong and i dunno how the weight is supported..can u plz gimme some more hint..Any help would b highly appriciated
  6. Dec 8, 2004 #5
    The man is also being accelerated by a normal force to the wall of the ring (otherwise he would be going in a straight line at constant velocity by Newton's laws). This is a centripetal force, which you should already know how to find the magnitude of. :smile:
    Given that the man is experiencing a force normal to the surface, there is one more non-zero force that is present. Do you see what it is ?
    Last edited: Dec 8, 2004
  7. Dec 8, 2004 #6
    is it the frictional force between the tyre and the wall...but which direction is it acting on... refer to the picure above i have drawn..on that,centripetal force is to the right while weight is to the left...so should the frictional force be acting downward...is it how thw weight is supported???

    can u plz help me answer the second part of my question (question is on very top)

    thaks for the help in advance
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