An electron that has velocity v = (3 x 105 m/s) and moves along positive x direction through the uniform magnetic field B = (0.8T) which is along the positive z direction.
a. Find the force on the electron (magnitude and direction).
(e = - 1.6 x 10-19 C, m=9.1 x10-31 kg)
b. Calculate the radius of electron’s path in the magnetic field.
(a) F = -e V B
Use the right hand rule for direction
(b) Review the formula for the "Larmor radius" of a charge in a magnetic field.
To find the force on the electron in a magnetic field, we can use the equation:
F = q * v * B * sin(theta)
Where:
F is the force on the electron
q is the charge of the electron (-1.6 x 10^-19 C)
v is the velocity of the electron (3 x 10^5 m/s)
B is the magnetic field (0.8 T)
theta is the angle between the velocity vector and the magnetic field vector
In this case, the electron is moving only in the positive x direction, so the angle between the velocity vector and the magnetic field vector is 90 degrees. Therefore, sin(theta) = sin(90) = 1.
Substituting the values into the equation:
F = (-1.6 x 10^-19 C) * (3 x 10^5 m/s) * (0.8 T) * 1
F = -3.84 x 10^-11 N
The force on the electron is -3.84 x 10^-11 N. The negative sign indicates that the force is in the opposite direction to the velocity, which is the negative x direction.
To calculate the radius of the electron's path in the magnetic field, we can use the equation:
F = (m * v^2) / r
Where:
F is the force on the electron (from the previous calculation)
m is the mass of the electron (9.1 x 10^-31 kg)
v is the velocity of the electron (3 x 10^5 m/s)
r is the radius of the electron's path
Rearranging the equation to solve for r:
r = (m * v) / (F)
Substituting the values into the equation:
r = (9.1 x 10^-31 kg) * (3 x 10^5 m/s) / (-3.84 x 10^-11 N)
r = -2.342 x 10^-6 m
The radius of the electron's path in the magnetic field is approximately 2.342 x 10^-6 m. The negative sign indicates that the path is concave towards the negative x direction.