An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 * 10-19 c and its mass is 9.10939 * 10-31 kg. how strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.9 cm? answer in units of t.

By definition, the potential energy is:

U = qV

Where,

q: load

V: voltage.

Then, the kinetic energy is:

K = mv ^ 2/2

Where,

m: mass

v: speed.

As the power energy is converted into kinetic energy, we have then:

U = K

Equating equations:

qV = mv ^ 2/2

From here, we clear the speed:

v = root (2qV / m)

Substituting values we have:

v = root ((2 * (1.60218 * 10 ^ - 19) * 3600) / 9.10939*10^-31))

v = 3.56 * 10 ^ 7 m / s

Then, the centripetal force is:

Fc = Fm

mv ^ 2 / r = qvB

By clearing the magnetic field we have:

B = mv / qr

Substituting values:

B = (9.10939 * 10 ^ - 31) * (3.56 * 10 ^ 7) / (1.60218 * 10 ^ - 19) * 0.059

B = 3.43 * 10 ^ - 3 T

Answer:

A magnetic field that must be experienced by the electron is:

B = 3.43 * 10 ^ - 3 T