An electron is accelerated from rest through a potential difference of 545 V ins

Question

An electron is accelerated from rest through a potential difference of 545 V inside what will call an electron gun. This electric gun happens to be located inside an Helmholtz coil (R= 50cm, N= 109). At the moment the electron is launched, the Helmholtz coil has a current of 1.2 A passing through it and is oriented so that the magnetic field it generates points in the positive z-direction (out of the page). If the electron is launched in the positive x-direction (to the right), then what is the radius of the electrons path?

Explanation / Answer

(4/5)3/2 **N*I/R ) = B (B at the midpoint between the coils for Helmholtz coil)

= 4**10^-7 I = 1.2 A N = 109 R = 50cm. = 0.5 m.

I found B = 0.000235 T


The kinetic energy gained by a charge,

e, after moving through a potential difference V is: Ke=e*V = 1/2*me*v^2    v - velocity


e = 1.6*10^-19 V = 545 Ke  =  8.72*10^-17 J. = 1/2*me*v^2

I found velocity (v) = 13843711.63 m/s

I assume electron is launched 45 degrees in x-z plane and has no initial angle in y - direction. Then using right hand rule we can determine the direction in which electron will travel i.e. in +y direction.
F (on electron by magnetic field) = e(v x B) v-velocity x - cross product
F=evBsin(45)
Substituting values I found F = 3.68*10^-16 N. This force acts on electron and is directed in +y direction

Since B is at 45 angle from velocity and Force The motion of a particle is expected to be a helix.

The axis of a helix  will be directed along z axis

The pitch (p) of a trajectory is a distance moved by electron along z axis during each period T

T = 2**r/v = 2*/w = 2**me /q*B = 152 nanoseconds

p = vz * T = v*cos45 * T = 1.48 m.

Radius of helix is: R=m*v*sin45/q*B = 0.237m.

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