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 equation:

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

Where:
F is the magnitude of the force
q is the charge of the particle (proton or electron)
v is the magnitude of the velocity of the particle
B is the magnitude of the magnetic field strength
theta is the angle between the velocity vector and the magnetic field vector

For the first question:

Given:
q (charge of a proton) = +1.6 x 10^-19 C
v (velocity of the proton) = 3 km/s = 3 x 10^3 m/s
B (magnetic field strength) = 0.05 T

Since the proton has a positive charge, the force experienced will be in a direction perpendicular to both the velocity and the magnetic field.

Using the equation, we can calculate the magnitude of the force:

F = (1.6 x 10^-19 C) * (3 x 10^3 m/s) * (0.05 T) * sin(90°)

sin(90°) = 1, as the velocity and the magnetic field are perpendicular.

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

F = 2.4 x 10^-16 N

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

For the second question:

Given:
q (charge of an electron) = -1.6 x 10^-19 C
v (velocity of the electron) = 2 x 10^5 m/s
B (magnetic field strength) = 1.4 x 10^-3 T
theta (angle between velocity and magnetic field) = 45°

Since the electron has a negative charge, the force experienced will be in a direction opposite to the velocity vector direction.

Using the equation, we can calculate the magnitude of the force:

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

sin(45°) = √2 / 2

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

F ≈ -2.518 x 10^-14 N

Since the force is negative, it indicates that the force is in the opposite direction of the velocity vector.

Therefore, the magnitude of the force experienced by the electron is approximately 2.518 x 10^-14 Newtons.