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 find the magnitude of the force experienced by a charged particle moving in a magnetic field, you can use the formula:

F = q * v * B * sinθ

where:
F is the force experienced by the charged particle,
q is the charge of the particle (in coulombs),
v is the velocity of the particle (in meters per second),
B is the magnetic field strength (in teslas), and
θ is the angle between the velocity vector and the magnetic field vector.

Let's solve the first problem:

1.) The proton has a charge of +1.6 x 10^-19 coulombs and is moving at a velocity of 3 km/s (which is equal to 3,000 m/s) inside a magnetic field of strength 0.05 T.

Using the formula, we have:

F = (1.6 x 10^-19 C) * (3,000 m/s) * (0.05 T) * sin(0°)
= (1.6 x 10^-19 C) * (3,000 m/s) * (0.05 T) * 0
= 0

Therefore, the magnitude of the force experienced by the proton is 0 N.

Let's solve the second problem:

2.) The electron has a charge of -1.6 x 10^-19 coulombs and is moving at a velocity of 2 x 10^5 m/s through a magnetic field of strength 1.4 x 10^-3 T. The angle between the velocity and the magnetic field is 45 degrees.

Using the formula, we have:

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

Simplifying the expression:

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

Now, we can calculate the magnitude of the force using the given values:

F = 2.24 x 10^-17 N

Therefore, the magnitude of the force experienced by the electron is 2.24 x 10^-17 N.