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 thecenter 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

Expert Answers

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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`

From above

`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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