# 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...

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`

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.**

**Sources:**