A proton moves eastward in the plane of
Earth’s magnetic equator so that its distance from the ground remains constant. What is the speed of the proton if Earth’s magnetic field points north and has a magnitude of 5.2 × 10−5 T? The acceleration of gravity is 9.81 m/s2 and the charge on a proton is 1.60×10−19. Mass of proton is 1.676e^-27
Answer in units of m/s
To solve this problem, we can use the formula for the magnetic force experienced by a moving charged particle:
F = qvB
where
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle, and
B is the magnetic field strength.
In this case, the proton is moving eastward, perpendicular to the magnetic field, so the angle between the velocity vector and the magnetic field vector is 90 degrees.
Since the proton is moving at a constant distance from the ground, we know that the gravitational force and the magnetic force must balance each other. Therefore, we can set up the equation:
F_gravity = F_magnetic
m * g = q * v * B
where
m is the mass of the proton,
g is the acceleration due to gravity,
q is the charge of the proton,
v is the velocity of the proton, and
B is the magnetic field strength.
Plugging in the given values:
(1.676e^-27 kg) * (9.81 m/s^2) = (1.60e^-19 C) * v * (5.2e^-5 T)
Now, we can solve for v:
v = [(1.676e^-27 kg) * (9.81 m/s^2)] / [(1.60e^-19 C) * (5.2e^-5 T)]
v ≈ 3258.3 m/s
So, the speed of the proton is approximately 3258.3 m/s.