An electron accelerates from rest to the right, in a horizontally directed electric field. The electron then leaves the electric field at a speed of 4.0 × 106, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and direction of the magnetic force on the electron. (Charge on an electron: q = 1.6 × 10-19)
To calculate the magnitude and direction of the magnetic force on the electron, we can use the formula:
F = q * v * B * sin(theta)
Where:
F is the magnetic force
q is the charge of the electron
v is the velocity of the electron
B is the magnitude of the magnetic field
theta is the angle between the velocity vector and the magnetic field vector
In this case, the electron is moving horizontally to the right, which means the velocity vector is horizontal and perpendicular to the direction of the magnetic field.
Since the electron leaves the electric field and enters the magnetic field, we can assume that the angle between the velocity vector and the magnetic field vector is 90 degrees.
Now, let's plug in the given values into the formula:
q = 1.6 × 10^-19 C
v = 4.0 × 10^6 m/s
B = 0.20 T
theta = 90 degrees
F = (1.6 × 10^-19 C) * (4.0 × 10^6 m/s) * (0.20 T) * sin(90 degrees)
sin(90 degrees) = 1
F = (1.6 × 10^-19 C) * (4.0 × 10^6 m/s) * (0.20 T) * 1
F = 1.28 × 10^-11 N
Therefore, the magnitude of the magnetic force on the electron is 1.28 × 10^-11 N.
Since the angle between the velocity and magnetic field is 90 degrees, the force will be perpendicular to both and can be determined using the right-hand rule. In this case, the force will be directed into the screen (towards the bottom).
So, the direction of the magnetic force on the electron is into the screen.