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Calculate speed v in crossfield hall effect

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data


    metal strip 6.66 cm long, 1.11 cm wide, and 0.837 mm thick moves with constant velocity through a uniform magnetic field B = 1.55 mT directed perpendicular to the strip, as shown in Fig. 28-37. A potential difference of 2.99 µV is measured between points x and y across the strip. Calculate the speed v.


    [​IMG]


    Ok i know Fnet= qE+q(V x B) and then set equal to zero cause equilibrium and get E=-q x V

    (the x means cross product)

    So now explain to me Why E=Vb in this case. Then i use V=E/B to get my speed velocity. But i dont just understand why the negative sign is dropped of ? Is it because its absolute value or cause Electric Field is point from a + to a - potential, in the x-axis direction ?

    Thanks for your help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 17, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi th3plan,

    The sign is different because the equations are two different things. The equation with the minus sign is a vector equation (it should be [itex]\vec E = - \vec v\times\vec B[/itex]); the other equation is only dealing with the magnitudes.

    For example, suppose I am holding a weight W by applying a force F upwards. The vector equation for equilibrium would be

    [tex]
    \vec F = -\vec W
    [/tex]
    which means that my applied force is equal in magnitude and opposite in direction to the weight. If I wanted to calculate the magnitude of the force, I might write:

    [tex]
    F = W
    [/tex]
    which just means [itex] | \vec F | = | \vec W|[/itex].
     
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