Question

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.

Answer 1

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