an electron is moving at 2x10^5 m/s through a uniform magnetic field of 1.4x10^-3 T.what is the magnitude of magnetic force if the velocity of the electron and the field make an angle of 45 degree.

To find the magnitude of the magnetic force acting on an electron moving through a magnetic field, we can use the formula:

F = q * v * B * sin(theta)

Where:
F is the magnitude of the magnetic force
q is the charge of the electron (1.6 x 10^-19 C)
v is the velocity of the electron (2 x 10^5 m/s)
B is the magnitude of the magnetic field (1.4 x 10^-3 T)
theta is the angle between the velocity of the electron and the magnetic field (45°)

Substituting the given values into the formula, we get:

F = (1.6 x 10^-19 C) * (2 x 10^5 m/s) * (1.4 x 10^-3 T) * sin(45°)

Now let's calculate the magnitude of the magnetic force:

F = (1.6 x 10^-19) * (2 x 10^5) * (1.4 x 10^-3) * 0.707

F = 4.480 x 10^-19 N

Therefore, the magnitude of the magnetic force acting on the electron is approximately 4.480 x 10^-19 N.