Question

An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnetic field of 0.4 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field?

Answer 1

F = 8.6 10⁻¹² N

Step-by-step explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v²

Em₀ = Emf

e ΔV = ½ m v²

v =√ 2 e ΔV / m

v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

v = √(1.8075 10¹⁶)

v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Emo = U = q DV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v2

Emo = Emf

.e DV = ½ m v2

.v = RA 2 e DV / m

.v = RA (2 1.6 10-19 51400 / 9.1 10-31)

.v = RA (1.8075 10 16)

.v = 1,344 108 m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10-19 1,344 108 0.4

F = 8.6 10-12 N