An electron orbits in a 3.8 mT field with angular momentum 8.6×10−26kgm2/s.
What is the diameter of the orbit? (Express your answer to two significant figures and include the appropriate units.-> meters)
the electron makes an orbit in a magnetic Field of radius r =
mv/eB
The
angular momentum of a particle in an orbit is L = mvr. So,
multiplying the expression for the radius of the orbit by r
gives
r^2 = mvr/eB = L/eB
r = (L/eB)^1/2
r = (8.6*10^-26/(1.602 * 10^-19*3.8*10^-3))^1/2 = 0.011885
m
or
r = 1.2 cm
DIameter = 2.4 cm or 0.02377 m
An electron orbits in a 3.8 mT field with angular momentum 8.6×10−26kgm2/s. What is the diameter...
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