Solutions to Assignment 5
1.
a) From the relations F =mv 2 /r and F =qvB (sin(θ) = 1 because the velocity and field are
perpendicular to each other), we get the relation r =mv /qB , which tells us that the radius is directly
proportional to the speed. From our above equations we also see that the Force is directly proportional
to the speed, so the radius must be proportional to the force. Since the particles have equal charge and
mass, we can rank the particles by radius size, which gives us B>A>C. This is also the ranking for the
force.
b) From our above relations we can see that the radius of a particle's path is inversely proportional to its
charge, but that only tells us the magnitude. To get the sign we must use the right hand rule (RHR).
For a velocity pointed right and a field into the screen the RHR tells us that the force on a positive
particle will be pointed up. Since A and B curve up, we can tell that they are positive while C is
negative. A has a smaller radius than B, telling us that it has a larger magnitude of charge (m and v are
equal), meaning it is the most positive. So we can rank the particles A>B>C.
2.
a) We know that particle 1 has a positive charge, so we can apply the right hand rule to all of the
regions it visits (A, B, & D). In region A, the particle starts off moving right and curves down. RHR:
If we point our index finger right and our thumb down, our middle finger points out of the screen,
which must be the direction of the field because our particle is positive. In region D the particle starts
off moving down and curves right, so applying the RHR tells us the field points into the screen.
(Alternative method: The direction that the particle curves changes between A and D, making an S
pattern, which tells us the field must be in the opposite direction. If the field were in the same
direction, but say had a different magnitude, the particle would continue curving the same way but the
radius of its circle would change). Going back into region A we already know the field is pointed out,
so we move to region B, where applying the RHR or noticing that the particle's path makes an S tells us
that the field in B must also be pointed into the screen. We then look at particle 2, which has an
unknown charge, but we see that it curves in region D but not in region C. Since in curves in a region
where we know there is a magnetic field, we know it must be charged. The only way for it not to curve
in region C must be for there to be no field.
b) We know that particle 1's speed charge and mass are constant throughout its path, so we can rank the
field strength based on the curvature of each section of the path. We only care about the magnitude, so
we ignore the direction of each field. Remember, r =mv /qB , so the larger the field strength, the
tighter the particle curves. In region A it makes a quarter turn in 1 square, so its radius is 1 square. In
region D it makes a half turn in 1 square, giving it a radius of ½. In region B it makes a quarter turn in
2 squares, giving it a radius of 2. We already know the field is zero in region C. Thus D>A>B>C.
c) KE=1 / 2mv 2 The mass doesn't change, and we know that magnetic forces only change the
direction of velocity, not its magnitude. Thus KE is constant throughout the path.
d) They give us the size of the side of a region, so each box is 1/3 of that. Let's call the side of a
box a and the side of an entire region 3a. In region a we see that particle 1 has a radius of a.
Going back to our equation r =mv /qB , we can easily rearrange this to solve for B,
B=mv /qr . We must convert microcoulombs into coulombs, milligrams into kilograms, and
centimeters into meters. Then we plug them into our equation to get the solution.
e) We are looking for the time for the entirety of particle 1's path. Each segment of the path can be
broken up into a section of a circle. So we can solve for the time for each of these fractions of a circle
and add them up. If we didn't know the speed of the particle, we would have to use the period
equation, T =2 m/qB , but since we know the speed and we know it doesn't change, we can just
use our basic formula t=d /v . The total distance it travels is the sum of all the arcs it goes through.
It travels 2 quarter circles (or 1 semicircle) in region A, a semicircle in region D, and a quarter circle in
region B. So we can see that d =1/2 2 a1/ 22 a / 21 /4 2 2a  , where we plugged in
the radii of regions A, D, and B as a, a/2, and 2a, respectively. Then we get the time by t=d /v
f) We can use the relation r =mv /qB and solve for q since we know m,v, and B, or we can observe
that m, v and B are the same for both particles and the radius of particle 2 is 4 times that of particle 1,
meaning it must have ¼ the charge.
3.
a) The magnetic force is given by F =qvBsin The particles all have identical mass and charge. B
is pointed to the right, so we can see that there will be an angle between B and v. vsin  gives us
the magnitude of the component of the velocity perpendicular to the field, and since in this case the
field points right, that is the y-component of the velocity. Therefore F =qBv y The magnitude of vy
for particles 1 and 3 is 2, while the magnitude of vy for particles 2 and 4 is 1. So the answer is
1=3>2=4.
b) At the instant shown in the picture, the direction of the magnetic force exerted on particle 1 is
exactly the same as the direction of the magnetic force exerted on particle 2.
True. Applying the RHR shows us that the force points into the page for both particles.
At the instant shown in the picture, the direction of the magnetic force exerted on particle 1 is in
the opposite direction of the magnetic force exerted on particle 4.
False. vy for particles 1 and 4 both point up, so the force must be in the same direction.
Particle 2 will travel in a circular path because of the influence of the magnetic field.
False. Particle 2 has a component parallel to the field as well as perpendicular to it. The perpendicular
component will trace out a circle while the parallel component is unaffected. The particle will trace out
a spiral in space.
As time goes by, the magnitude of the force exerted by the field on particle 3 will change, but the
direction of the force will remain the same.
False. In a uniform magnetic field, the perpendicular component of the velocity traces out a circle.
Uniform circular motion demands a force of constant magnitude that is perpendicular to the velocity at
all times. Since the direction of velocity vector changes around the circle, so must the force.
As time goes by, the direction of the force exerted by the field on particle 3 will change, but the
magnitude of the force will remain the same.
True. Same reason as the previous statement.