An electron moves in a circular path perpen-
dicular to a magnetic field that has a magni-
tude of 0.883 × 10−3 T.
The angular momen-
tum of the electron as it moves around the
center of the circle is 1.40 × 10−25J·s.
Find the radius of the circular path.
electron charge is 1.6022e-19
electron mass is 9.109e-31
1 Answer | Add Yours
The magnetic force on a charge `q` moving with speed `v` in a magnetic field `B` is (Lorentz force):
`F_m =q*(v xx B)` (written as vectors).
The condition for stable circular path in a plane is the equality of magnetic force (centripetal force) with the centrifugal force. Since `v_|_ B` one has:
`e*v*B =m*a_(cf)` or equivalent `evB =m*(v^2/R)`
`eBR =mv` (1)
Now the angular momentum in a circular path is the same for all the trajectory:
`L =|L| =|R xx p| = Rmv` since `R_|_v`
`mv = L/R` which combined with (1) gives
`eBR =L/R` or equivalent
`R = sqrt(L/(eB)) =sqrt((1.4*10^-25)/(1.6*10^-19*0.883*10^-3)) =0.0315 m= 3.15 cm`
The radius of the circular path is 3.15 cm.
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