A 2.3-keV proton is moving horizontally and passes perpendicular to the Earth’s magnetic field at a location where the field magnitude is 52.0 T.
i) Determine the magnetic force on the proton?
ii) Find the ratio of this force to the gravitation force on the proton?
iii) What energy of proton would be needed to make the magnetic and gravitational forces to be equal?
iv) What direction proton would be, to keep the motion in horizontal?
i) To determine the magnetic force on the proton, we can use the equation:
F = q(v x B)
where F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field.
In this case, the charge of the proton is given as 1.6 x 10^-19 C, the velocity is not given, but since it is moving horizontally, we can assume it is perpendicular to the magnetic field. Therefore, the velocity and the magnetic field vectors are at right angles to each other, resulting in their cross product having a maximum magnitude of vB.
We can plug in the given values:
F = (1.6 x 10^-19 C)(v)(52.0 μT)
ii) To find the ratio of the magnetic force to the gravitational force on the proton, we can use the equation:
ratio = F_magnetic / F_gravitational
The magnitude of the gravitational force on a particle is given by:
F_gravitational = mg
where m is the mass of the proton and g is the acceleration due to gravity.
The ratio is then:
ratio = (1.6 x 10^-19 C)(v)(52.0 μT) / mg
iii) To determine the energy of the proton needed for the magnetic and gravitational forces to be equal, we can equate the two forces:
(1.6 x 10^-19 C)(v)(52.0 μT) = mg
Solving for v:
v = mg / (1.6 x 10^-19 C)(52.0 μT)
Then, to find the energy, we can use the kinetic energy formula:
E = (1/2)mv^2
iv) To keep the motion in a horizontal plane, the magnetic force should be in the vertical direction. This means that the magnetic field and velocity vectors should be perpendicular to each other, resulting in the maximum magnitude of the magnetic force. Therefore, the proton should move perpendicular to the Earth's magnetic field.