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.
Answer in units of m/s mass of proton is 1.673 e^-27
To find the speed of the proton, we can use the concept of centripetal force.
When a charged particle moves in a magnetic field, it experiences a force known as the magnetic force. The magnetic force on a charged particle moving perpendicular to a magnetic field is given by the formula:
F = q * v * B
Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
In this case, the proton is moving eastward in the magnetic field, so the magnetic force acts as the centripetal force to keep the proton at a constant distance from the ground. The centripetal force is given by the formula:
F = (m * v^2) / r
Where:
m is the mass of the proton
v is the speed of the proton
r is the radius of the circular path
Since the proton is at a constant distance from the ground, we can assume it moves in a circular path. Thus, we can equate the magnetic force and the centripetal force:
q * v * B = (m * v^2) / r
Rearranging the equation, we can solve for the speed v:
v = (q * r * B) / m
Given:
Charge of proton, q = 1.60 × 10^(-19) C
Mass of proton, m = 1.673 × 10^(-27) kg
Magnetic field strength, B = 5.2 × 10^(-5) T
We need to find the radius, r, to calculate the speed. However, the radius is not provided in the information given. If you have the value of the radius or any other additional information, please provide it so that we can calculate the speed of the proton.