1.)what is the magnitude of the force experienced by a proton moving at 3 km/s inside magnetic field of strength 0.05T

2.)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 calculate the magnitude of the force experienced by a charged particle moving through a magnetic field, you can use the formula:

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

where:
F is the magnitude of the force,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength,
theta is the angle between the velocity vector and the magnetic field vector.

Let's apply this formula to the given questions:

1.) For the proton moving at 3 km/s inside a magnetic field of 0.05 T, we know that the charge of a proton is +1.6 x 10^-19 C. Since the question does not provide the angle, we can assume that the angle between the proton's velocity vector and the magnetic field vector is 90 degrees (perpendicular), therefore, sin(theta) = sin(90) = 1.

Using the formula:

F = (1.6 x 10^-19 C) * (3 x 10^3 m/s) * (0.05 T) * 1

F = 2.4 x 10^-16 N

The magnitude of the force experienced by the proton is 2.4 x 10^-16 Newtons.

2.) For the electron moving at 2 x 10^5 m/s through a magnetic field of 1.4 x 10^-3 T with an angle of 45 degrees, we can apply the formula:

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

sin(45) = 0.70710678

F = -4.4827 x 10^-17 N

Note: The negative sign indicates that the force acting on the electron is in the opposite direction of the velocity.

The magnitude of the magnetic force experienced by the electron is 4.4827 x 10^-17 Newtons.