A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x-direction experiences a force of 2.25x10- 16 N in the +y-direction, and an electron moving at 4.75 km/s in the -z-direction experiences a force of 8.50 X 1O- 16 N.

(a) What are the magnitude and direction of the magnetic field?
(b)What are the magnitude and direction of the magnetic force on an electron moving in the –y direction at 3.2 km/s?

To solve this problem, we can use the equation for the magnetic force on a charged particle moving in a magnetic field:

F = qvBsinθ

where F is the force experienced by the particle, q is its charge, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.

(a) To find the magnitude and direction of the magnetic field in the given scenario, we can use the information about the forces experienced by the proton and the electron.

For the proton:
F = 2.25 x 10^-16 N
q = +1.6 x 10^-19 C (charge of a proton)
v = 1.5 x 10^3 m/s (convert km/s to m/s)

Using the equation F = qvBsinθ, we know that the force experienced by the proton is in the +y-direction. Since the velocity of the proton is in the +x-direction, the angle between the velocity and magnetic field vectors is 90 degrees (sin90 = 1).

2.25 x 10^-16 N = (1.6 x 10^-19 C)(1.5 x 10^3 m/s)(B)(1)

Solving for B gives:
B = 2.25 x 10^-16 N / (1.6 x 10^-19 C x 1.5 x 10^3 m/s)

Calculate B to find the magnitude of the magnetic field.

For the electron:
F = 8.50 x 10^-16 N
q = -1.6 x 10^-19 C (charge of an electron)
v = -4.75 x 10^3 m/s (convert km/s to m/s)

Using the equation again, we know that the force experienced by the electron is opposite to the velocity, in the -z-direction. The angle between the velocity and magnetic field vectors is 90 degrees (sin90 = 1).

8.50 x 10^-16 N = (-1.6 x 10^-19 C)(-4.75 x 10^3 m/s)(B)(1)

Solving for B gives:
B = 8.50 x 10^-16 N / (1.6 x 10^-19 C x 4.75 x 10^3 m/s)

Calculate B to find the magnitude of the magnetic field.

(b) To find the magnitude and direction of the magnetic force on an electron moving in the -y direction at 3.2 km/s, we can again use the equation for magnetic force:

F = qvBsinθ

For the electron:
q = -1.6 x 10^-19 C (charge of an electron)
v = -3.2 x 10^3 m/s (convert km/s to m/s)
B (magnitude of the magnetic field) is already known from part (a).

Since the electron is moving in the -y direction, the angle between its velocity and the magnetic field vectors is 180 degrees (sin180 = 0).

F = (-1.6 x 10^-19 C)(-3.2 x 10^3 m/s)(B)(0)

The force experienced by the electron in this case will be zero, meaning there is no magnetic force acting on it.

So, the magnitude of the magnetic force on an electron moving in the -y direction at 3.2 km/s is zero, and the direction is undefined.