How long does it take an electron to complete a circular orbit perpendicular to a 1.4 g magnetic field?

Answer

T = 2π / (1.4x10^-4 x 1.76x10^11) = 2.55x10^-7 s

Explanation

1. The centripetal force must be equal.

The centripetal force of circular motion must be equal to the magnetic force hence:

qvB = mv^2/r

v = (qBr) / m

Since angular speed is: ω = v/r = 2π/T

2. v = 2πr/T

Where T is the period of a single orbit. Thus substituting equation 2 in equation 1 and making T the subject:

3. T = 2π / (B (q/m))

Substituting B = 1.4 G = > 1.4 x 10^-4 T

and q/m = 1.76x10^11 C/kg in equation 3

T = 2π / (1.4x10^-4 x 1.76x10^11) = 2.55x10^-7 s

Answer;

=255 ns

Explanation and solution;

The centripetal force of circular motion must be equal to the magnetic force:

qvB = mv^2/r

=> v/r = Bq/m

Remember angular speed is:

ω = v/r = 2π/T

Where T is the period of a single orbit.

Thus:

2π/T = B (q/m)

=> T = 2π / (B (q/m))

q/m = 1.76x10^11 C/kg for an electron, and

B = 1.4 G = 1.4 (1x10^-4) T = 1.4x10^-4 T (T in this case is obviously the SI unit Tesla, unrelated to the period T, sorry 'bout that).

So we have the period:

T = 2π / (1.4x10^-4 x 1.76x10^11)

= 2.55 x10^-7 s

= 255 ns.

Therefore period is 255 ns