Question

An electron moves at 2.40×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.10×10−2 T .

a. What is the largest possible magnitude of the force on the electron due to the magnetic field? Express your answer in newtons to two significant figures. Fmax = nothing N.
(b) If the actual acceleration of theelectron is one-fourth of the largest magnitude in part (a), whatis the angle between the electron velocity and the magneticfield?

Answer 1

a) F = 2.7 10⁻¹⁴ N , b) a = 2.97 10¹⁶ m / s² c) θ = 14º

Step-by-step explanation:

The magnetic force on the electron is given by the expression

F = q v x B

Which can be written in the form of magnitude and the angle found by the rule of the right hand

F = q v B sin θ

where θ is the angle between the velocity and the magnetic field

a) the maximum magnitude of the force occurs when the velocity and the field are perpendicular, therefore, without 90 = 1

F = e v B

F = 1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²

F = 2.73 10⁻¹⁴ N

F = 2.7 10⁻¹⁴ N

b) Let's use Newton's second law

F = m a

a = F / m

a = 2.7 10⁻¹⁴ / 9.1 10⁻³¹

a = 2.97 10¹⁶ m / s²

The actual acceleration (a1) is a quarter of this maximum

a1 = ¼ a

a1 = 7.4 10¹⁵ m / s²

With this acceleration I calculate the force that is executed on the electron

F = ma

e v b sin θ= ma

sin θ = ma / (e v B)

sin θ = 9.1 10⁻³¹ 7.4 10¹⁵ / (1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²)

sin θ = 6.734 10⁻¹⁵ / 27.26 10⁻¹⁵

sin θ = 0.2470

θ = 14.3º