A proton moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What is the magnitude of the force that the magnetic field exerts on the proton? (e = 1.60 × 10-19 C)
The magnetic force experienced by a moving charged particle is given by the formula:
F = q * v * B * sin(theta)
Where F is the force, q is the charge of the particle, v is the velocity, B is the magnetic field strength, and theta is the angle between the velocity and the magnetic field.
In this case, the proton has a charge of q = +1.60 × 10^-19 C, a velocity of v = 5.0 km/s = 5000 m/s, and the magnetic field strength is B = 0.20 T. The angle between the velocity (eastward) and the magnetic field (northward) is 90 degrees, so sin(theta) = sin(90°) = 1.
Now we can plug these values into the formula:
F = (1.60 × 10^-19 C) * (5000 m/s) * (0.20 T) * 1
F = 1.60E-19 * 5000 * 0.20
F = 1.6e-16 N
The force exerted by the magnetic field on the proton is 1.6 × 10^-16 N.