A positively charged Lithium ion (2 electrons, 3 protons, and 3 neutrons) is accelerated in a vacuum between two parallel plates with a potential difference of 3.0 x 102 V between them. The ion then moves into a uniform external field with a strength of 0.20 T [into the page]. See the diagram below (not to scale):
(a) Determine the speed of the ion as it leaves the parallel plates. (Hint: consider the number of excess or deficit electrons of the ion)
(b) Determine the magnetic force (magnitude and direction) experienced by the ion as it enters the magnetic field.
(c) What is the radius that of the motion of the ion within the field?
Thank you!
To solve these problems, we need to apply the principles of electromagnetism and motion of charged particles in magnetic fields. Let's go step by step:
(a) To determine the speed of the ion as it leaves the parallel plates, we need to consider the potential difference applied to it. The electric potential energy gained by the ion will be converted into kinetic energy. The formula to calculate the kinetic energy of a moving object is:
Kinetic Energy = (1/2) * m * v^2
where m is the mass of the ion and v is its velocity. We know that the ion has a charge of +2e (two excess protons), so it experiences a force from the electric field between the plates. The electrical force can be calculated using the formula:
Force = q * E
where q is the charge and E is the electric field strength. By equating this force with the equation for Force (Force = mass * acceleration), we can solve for the acceleration:
a = q * E / m
Since the acceleration is constant, we can use the kinematic equation:
v^2 = u^2 + 2 * a * x
Where v is the final velocity, u is the initial velocity (which is 0 as the ion starts from rest), a is the acceleration, and x is the distance traveled between the plates. The distance traveled can be calculated from the potential difference applied:
x = V / E
Substituting the values, we can find the speed of the ion as it leaves the parallel plates.
(b) To determine the magnetic force experienced by the ion as it enters the magnetic field, we need to use the formula for the magnetic force on a charged particle in a magnetic field:
Force = q * v * B * sin(theta)
where q is the charge, v is the velocity of the ion, B is the magnetic field strength, and theta is the angle between the velocity of the ion and the magnetic field. In this case, theta is 90 degrees since the ion moves perpendicular to the magnetic field.
(c) To find the radius of the motion of the ion within the field, we can use the equation for the centripetal force:
F_c = m * (v^2 / r)
where F_c is the centripetal force exerted on the ion, m is its mass, v is its velocity, and r is the radius of the circular path. We know that the magnetic force on the charged particle acts as the centripetal force, so we can equate them:
q * v * B = m * (v^2 / r)
Now, we can solve for r.
I hope this explanation helps you solve the given problems!