Posted by jayren on Monday, July 30, 2012 at 10:22pm.
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
physics - Elena, Tuesday, July 31, 2012 at 2:40am
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
physics - Anonymous, Thursday, January 21, 2016 at 5:10pm
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