Answer: Therefore, the magnitude of the magnetic field is 5.24 × 10^-9 T and its direction is into the page.
Explanation: The force on a charged particle moving through a magnetic field is given by the formula F = qvB, where F is the force, q is the charge, v is the velocity of the particle, and B is the magnetic field strength.
In this problem, the particle has a charge of -1.9 × 10^-6 C and is traveling with a velocity of 8.1 × 10^3 m/s. The force acting on the particle is perpendicular to both the velocity vector and the magnetic field vector. Therefore, the force acting on the particle is responsible for the circular motion of the particle, and the radius of the circle is related to the velocity, magnetic field, and the mass of the particle.
The radius of the circular path can be calculated using the formula R = mv/qB, where m is the mass of the particle, v is the velocity of the particle, q is the charge on the particle, and B is the magnetic field strength.
Plugging in the given values, we get:
R = (3.1 × 10^-12 kg) × (8.1 × 10^3 m/s) / (-1.9 × 10^-6 C × B)
Simplifying, we get:
R = - 13.11 m^2 / (C kg s B)
Rearranging the terms, we get:
B = - 13.11 m^2 / (C kg s R)
Plugging in the given values, we get:
B = - 13.11 m^2 / (C kg s × 0.05 m) = - 5.24 × 10^-9 T
The magnitude of the magnetic field is 5.24 × 10^-9 T.
The direction of the magnetic field can be found using the right-hand rule. If we point our right thumb in the direction of the velocity vector and our fingers in the direction of the magnetic field vector, then the direction of the force vector is perpendicular to both and can be found using our right hand. In this case, the force vector points upward, so the magnetic field must point into the page (i.e., in the negative z-direction).