posted by jayren on .
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
kg and the
elemental charge is −1.602 × 10
Answer in units of µT
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 =>
=9.109•10^-31/1.602•10^-19•0.35•10^-6 =1.62•10^-5 T =16.2 μT