Based on your determination of the radius of the electron path under these conditions, what is your measurement of the ratio e/m in SI units (C/kg)?

PREVIOUS WORK:

Using the given values of accelerating voltage and coil current, we can use the formula for the radius of the circular path of an electron in a magnetic field:

r = (2mV) / (Bq^2)

where m is the mass of the electron, V is the accelerating voltage, B is the magnetic field, and q is the charge of the electron. Plugging in the given values and the constants for m and q, we get:

r = (2 x 9.109 x 10^-31 kg x 100 V) / (1 x 10^-6 T x (1.602 x 10^-19 C)^2)

r 4.9 cm

Therefore, the answer closest to what we found is 4.9 cm.

2..The formula for the magnetic field between two Helmholtz coils with radius r and separation d when a current I is passed through them is:

B = (80/5) * (I/R) * [(R^2/d^2) + 1]^(3/2)

where 0 is the permeability of free space, R is the radius of the coils, and d is the separation between the coils. Plugging in the given values and the constants for 0 and R, we get:

B = (8 x x 10^-7 T m/A) / 5 * (1 A / 0.15 m) * [(0.15^2 / 0.15^2) + 1]^(3/2)

B 26.1 T

Therefore, the magnetic field where the electrons are traveling is approximately 26.1 microtesla.

3.The velocity of the electrons can be calculated using the formula for the kinetic energy of a charged particle accelerated through a potential difference:

KE = qV = 0.5mv^2

Solving for v, we get:

v = sqrt(2qV/m)

Plugging in the given values and the constants for q and m, we get:

v = sqrt(2 x 1.602 x 10^-19 C x 100 V / 9.109 x 10^-31 kg)

v 6.04 x 10^6 m/s

Therefore, the velocity of the electrons at 100 V is approximately 6.04 x 10^6 m/s.

4.To find the percent of the speed of light that the electrons are traveling at, we can divide their velocity by the speed of light and multiply by 100:

percent of speed of light = (v / c) x 100%

Plugging in the given value and the constant for c, we get:

percent of speed of light = (6.04 x 10^6 m/s / 2.998 x 10^8 m/s) x 100%

percent of speed of light 2.01%

Therefore, the electrons are traveling at approximately 2.01% of the speed of light.

Simulation Used: https://virtuelle-experimente.de/en/b-feld/b-feld/experiment.php

https://www.youtube.com/watch?v=jcgWlD9PteM&ab_channel=JohnKielkopf