16. An electron moves with a speed of 8.0 x 106 m/s along the +x-axis. It enters a region where
there is a magnetic field of 2.5 T, directed at an angle of 60° to the +x-axis and lying in the
xy-plane. (1 eV = 1.60 x 10^-19C, mel = 9.11 x 10^-31kg)
a. Calculate the magnitude of the magnetic force on the electron.
b. Calculate the magnitude of the acceleration of the electron.
a. The formula for the magnetic force on a moving charge in a magnetic field is:
F = qvb sin(theta)
Where:
- F is the magnetic force (in N)
- q is the charge of the particle (in C)
- v is the velocity of the particle (in m/s)
- b is the magnitude of the magnetic field (in T)
- theta is the angle between the velocity and the magnetic field (in degrees)
Plugging in the values given:
F = (-1.6 x 10^-19 C)(8.0 x 10^6 m/s)(2.5 T)sin(60°)
F = -6.4 x 10^-14 N (Note: the negative sign indicates that the force is in the opposite direction to the motion of the electron)
Therefore, the magnitude of the magnetic force on the electron is 6.4 x 10^-14 N.
b. The formula for the acceleration of a charged particle moving in a magnetic field is:
a = (q/m)vb sin(theta)
Where:
- a is the acceleration of the particle (in m/s^2)
- q is the charge of the particle (in C)
- m is the mass of the particle (in kg)
- v is the velocity of the particle (in m/s)
- b is the magnitude of the magnetic field (in T)
- theta is the angle between the velocity and the magnetic field (in degrees)
Plugging in the values given:
a = (1.6 x 10^-19 C / 9.11 x 10^-31 kg)(8.0 x 10^6 m/s)(2.5 T)sin(60°)
a = 6.1 x 10^15 m/s^2 (Note: the direction of acceleration is perpendicular to both the velocity and magnetic field, and can be calculated using the right-hand rule)
Therefore, the magnitude of the acceleration of the electron is 6.1 x 10^15 m/s^2.