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The area shown is the boundary of a magnetic field directed in the positive z direction. An electron with a velocity along the

x-axis enters the field at coordinates (x, y) =
(−2 m, 0) and exits 0.35 µs later at point A
whose coordinates are (x, y) = (0, 2 m).
What is the magnitude of B~
? The mass
of the electron is 9.109 × 10
−31
kg and the
elemental charge is −1.602 × 10
−19
C.
Answer in units of µT

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jayren
2 answers

  1. Electron enters the magnetic field perpendicular to the magnetic field, therefore, it moves with centripetal acceleration along the curvilinear path due to Lorentz force action
    m•a=q•v•B •sin α.
    Since α=90º, sin α=1; q=e; v=at.=>
    m•a =e•v•B=e•a•t•B =>
    m= e•t•B.
    B=m/e•t=
    =9.109•10^-31/1.602•10^-19•0.35•10^-6 =1.62•10^-5 T =16.2 μT

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    Elena

  2. 16.2

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    Anonymous

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