Motion in a magnetic field

[Leo Consoli]

Homework Statement

A particle of mass M, with charge Q is left at the point A. After reaching point P the particle is at the effect of a constant and uniform magnetic field of intensity B reaching point Q.
Find H.

Homework Equations

P=mg
Fm=qvB
Ke=(mv^2)/2
U=mgH

The Attempt at a Solution

The big problem is not solving this question, because I was able to do that, but I didnt understand why during the trajectory from P to Q the particle has no gravitational potential energy.
The solution:- During the trajectory from P to Q the direction of the movement doesn't change so the magnetic force must cancel out with the weight of the particle.

So, using conservation of energy we get that:

This is the right answer but I am still confused.

 

[kuruman]

If the magnetic force exactly opposes gravity, what is the trajectory of the particle across the gap? What is the change in gravitational potential energy?

 

[Leo Consoli]

It would be a circular motion since the magnetic force would be perpendicular to the trajectory of the particle.

 

[kuruman]

It would be a circular motion since the magnetic force would be perpendicular to the trajectory of the particle.

Is the magnetic force the only force acting on the particle when it's in the gap?

 

[Leo Consoli]

There is also the weight of the particle as far as I can tell.

 

[kuruman]

There is also the weight of the particle as far as I can tell.

That is correct. There are two forces acting on the particle. Draw a free body diagram. The acceleration of the particle is in the direction of the net force. What is the net force and in what direction is it?

 

[Leo Consoli]

Weight would be pointing to the ground, but if I don't know if the particle is positive or negative how can I know the direction of the magnetic force? I assumed they would cancel each out in my solution because the trajectory seemed linear from P to Q.

 

[Leo Consoli]

But the magnetic force and the weight are both vertical, so the net force would also be vertical, I don't know to where it would be directed to though.

 

[kuruman]

But the magnetic force and the weight are both vertical, so the net force would also be vertical, I don't know to where it would be directed to though.

You know that the velocity is to the right and that the magnetic field is out of the screen. The magnetic force is given my $\vec F_M=q \vec v \times \vec B$. Use the right hand rule to find its direction. If the particle is positive, you get one direction (what is it?) and if it's negative, you get the opposite direction. Which of the two directions makes sense in view of the given fact that the particle makes it across the gap?

 

[Leo Consoli]

If the particle is positive the direction would be the same as the weight, so since it makes it ouf of the gap it can onby be negative, the magnetic force would be pointing upwards, and the weight would be pointing downwards, the centrifugal acceleration would be in the same direction as the centrifugal resultant force.

 

[kuruman]

If the particle is positive the direction would be the same as the weight, ...

Wrong direction for a positve particle. You need to review your favorite scheme for finding the direction of cross products.

... the centrifugal acceleration would be in the same direction as the centrifugal resultant force.

What on Earth are you talking about? There are two forces, gravity and the magnetic force and they are both vertical. The net force, which provides the acceleration, is also vertical and can point up, down or be zero. What do you think is the case here?

 

[Leo Consoli]

Wrong direction for a positve particle. You need to review your favorite scheme for finding the direction of cross products.

I just used the right hand rule, velocity is to the right, the magnetic field comes out of the screen so the magnetic force would come out of the palm of my hand if the particle is positive and from the back of my hand if the particle is negative.

What on Earth are you talking about? There are two forces, gravity and the magnetic force and they are both vertical. The net force, which provides the acceleration, is also vertical and can point up, down or be zero. What do you think is the case here?

? I just preferred to call the net force as a centrifugal force since its a circular movement.

 

[kuruman]

I just used the right hand rule, velocity is to the right, the magnetic field comes out of the screen so the magnetic force would come out of the palm of my hand if the particle is positive and from the back of my hand if the particle is negative.

OK.

? I just preferred to call the net force as a centrifugal force since its a circular movement.

How about drawing a picture of the path that the particle describes in the gap? It has to start at point P with horizontal velocity $\vec v$ to the right and end up at Q still moving to the right with horizontal velocity as suggested by the drawing you posted. What kind of line would connect points P and Q?

 

[Leo Consoli]

I think it would probably would be a semicircunference with an arc upwards.

 

[kuruman]

I think it would probably would be a semicircunference with an arc upwards.

Can you post a picture? Note that the path cannot be a piece of a circle, To have a piece of a circle, the force must be perpendicular to the velocity and have constant magnitude everywhere along the path. This happens only in zero gravity, i.e. when the magnetic force is the only force acting on the particle.

 

[Leo Consoli]

I think it would be something like this, I didnt get what you said, using the right hand rule the magnetic force is perpendicular to the velocity and the magnetic force is constant in the trajectory since the magnetic field is also constant.

 

[kuruman]

You know that the velocity is always tangent to the path. This is not the case in your drawing at points P and Q. From your drawing it is clear that you are assuming that the direction of the force is up. Furthermore, the magnitude is constant. This means that the horizontal component of the velocity is constant while the vertical component of the velocity increases at a constant rate from an initial value of zero. I'll give you a generous hint: Another situation where the force is constant in the vertical direction but zero in the horizontal direction is gravity. What path does an object follow when it rolls off the top of a table? Turn it upside down and you have the particle's path if the net force is up. Don't turn it upside down, and you have the particle's path when the net force is down. So what do you think? What must be true for the particle to go from point P to point Q?

 

[Leo Consoli]

I think the net force is up.

 

[haruspex]

The problem statement contains a blunder. H ought to be specified as the starting height above points P and Q.
Also, the slope should be given as frictionless, or some KE will go into rotation.

I assumed they would cancel each out in my solution because the trajectory seemed linear from P to Q.

Clearly that is a solution to the problem. if released at exactly the right height the velocity will be such that the two forces cancel.
If we assume the problem is well posed (apart from the blunder) then there are no other solutions.
If you do not wish to make that assumption, can you show there are no funny loop-the-loop solutions?

 

[Leo Consoli]

Clearly that is a solution to the problem. if released at exactly the right height the velocity will be such that the two forces cancel.
If we assume the problem is well posed (apart from the blunder) then there are no other solutions.
If you do not wish to make that assumption, can you show there are no funny loop-the-loop solutions?

My problem isn't exactly with solving it, but my question comes from the fact that, in the second step of my solution, I didnt need to consider the gravitational potential energy of the body traveling from P to Q while equating the mechanical energies, and I don't understand why I can do that.

 

[kuruman]

The problem statement contains a blunder. H ought to be specified as the starting height above points P and Q.

Nice catch.

My problem isn't exactly with solving it, but my question comes from the fact that, in the second step of my solution, I didnt need to consider the gravitational potential energy of the body traveling from P to Q while equating the mechanical energies, and I don't understand why I can do that.

Look at the equation you posted in #1, $q v B = m g$. It says that the magnetic force (up) is equal to the gravitational force (down). If that's the case, the net force is zero, the particle travels in a straight line and keeps traveling in a straight line. Because the straight line is horizontal, there is no change in gravitational potential energy. Even if the particle went from P to Q following some other path, the change in gravitational potential energy would still be zero because there is no vertical height change.

 

[haruspex]

can you show there are no funny loop-the-loop solutions?

Update...
I believe I can show that other solutions only exist if the width PQ is a multiple of $2\pi g(\frac{m}{qB})^2$. For those widths, the height of the descent can be anything, provided PQ is sufficiently high above the ground.

 

[Leo Consoli]

Look at the equation you posted in #1, qvB=mgqvB=mgq v B = m g. It says that the magnetic force (up) is equal to the gravitational force (down). If that's the case, the net force is zero, the particle travels in a straight line and keeps traveling in a straight line. Because the straight line is horizontal, there is no change in gravitational potential energy. Even if the particle went from P to Q following some other path, the change in gravitational potential energy would still be zero because there is no vertical height change.

I understood, but the comparison is between the particle at point A and at a random point during the motion from P to Q, and during this the particle still has some height from the black line that is used as H`s referential, as I understand it the gravitational potential energy would be equal to the work needed to bring that particle from the "ground" to the height h where it is at against the gravitational acceleration g, I still don't get why it would be 0.

 

[haruspex]

I understood, but the comparison is between the particle at point A and at a random point during the motion from P to Q, and during this the particle still has some height from the black line that is used as H`s referential, as I understand it the gravitational potential energy would be equal to the work needed to bring that particle from the "ground" to the height h where it is at against the gravitational acceleration g, I still don't get why it would be 0.

Potential energies are always relative. You can set your ground level wherever you like.
As I noted, the question as given makes no sense. You cannot find the value of the H in the diagram. H ought to have been defined as the height the particle starts above the PQ level. So it is logical to set the PQ level as your reference. That makes the GPE zero there.

 

[Leo Consoli]

I see, thank you and kuruman for the patience and the help.

 

[jbriggs444]

Update...
I believe I can show that other solutions only exist if the width PQ is a multiple of $2\pi g(\frac{m}{qB})^2$. For those widths, the height of the descent can be anything, provided PQ is sufficiently high above the ground.

Oh, right. A sag in the trajectory cannot proceed infinitely downward with a radial acceleration that scales directly with velocity. Then symmetry assures us that the eventual high point must be level with P. So it is clear that the trajectory is periodic and the only difficulty is tuning the period.

 

[haruspex]

Oh, right. A sag in the trajectory cannot proceed infinitely downward with a radial acceleration that scales directly with velocity. Then symmetry assures us that the eventual high point must be level with P. So it is clear that the trajectory is periodic and the only difficulty is tuning the period.

Yes, it's very interesting. The motion is the sum of a constant horizontal velocity and a circular motion at constant rate. Varying the entry speed changes the radius but not the period nor the horizontal 'drift' velocity.