An electron enters a magnetic field of 0.35 T , moving perpendicular to it with

Question

An electron enters a magnetic field of 0.35 T, moving perpendicular to it with a speed of 3·105 m/s.

a.) Find the radius of the orbit it will make.
4.875  µm (micrometers) CORRECT!

II.) Helical motion:
If an electron were sent at the same speed, but at an angle of 20o to the magnetic field:

c.) What would be the radius of the helical motion?
______?_______ µm (micrometers) **Help!***The answer is not 14.235 or 1.4253!*****

d.) How long would it take for the electron to travel 3 m along ("downstream" to) the magnetic field?
_____________ µs (microseconds)

Explanation / Answer

a)

The force of the magnetic field on the electron is:
F = Q(V) X B
This will equal the centrifugal force of the electron in its orbit
F = m(V^2)R
m(V^2)/R = Q(V)B
V = Q(B)R/m = (1.6 x 10^-19)(0.73)(R)/(9.1 x 10^-31)
V/R = 6.153846154 x 10^11

R = V/(6.153846154 * 10^10)

R = 4.875 * 10^-6 m

c)

As we know that the

radius of charged particle if it enters in a magnetic field = (m*v*sin (theta))/q*B

= (9.1*10^-31*3*10^5*sin20)/(1.6*10^-19*0.35)

= 1.667348199 * 10^-6 m

d)

The period of the motion (the time that the particle takes to complete one revolution) is equal = 2*pi*R/(v*sin(theta))

T = 2*pi*1.667348199 * 10^-6/(3*10^5*sin20) =

1.021017612 * 10^-10 seconds to cover 1.047625771 * 10^-5 m

So time required to cover 3 m = (1.021017612 * 10^-10 seconds/1.047625771 * 10^-5 m) *(3)

   = 2.92 micro seconds

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