Question

1.

A particle carrying a charge of 0.50 µC enters a magnetic field of strength 0.045 T, with a velocity of 350 m/s.
The velocity is perpendicular to the magnetic field. What is the magnetic force acting on the charged particle?

Answer 1

Step-by-step explanation:

The magnetic force (F) acting on a charged particle moving in a magnetic field is given by the equation:

F = qvBsinθ

where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity of the particle and the direction of the magnetic field.

In this case, the charge of the particle is 0.50 µC, the velocity is 350 m/s, and the magnetic field strength is 0.045 T. Since the velocity of the particle is perpendicular to the magnetic field, θ = 90°. Therefore, we can substitute the values into the equation and solve for the magnetic force:

F = qvBsinθ

F = (0.50 x 10^-6 C)(350 m/s)(0.045 T)(sin 90°)

F = 0.0084375 N

Therefore, the magnetic force acting on the charged particle is 0.0084375 N.