How Does a Charged Particle Move in Combined Electric and Magnetic Fields?

[zorro]

Homework Statement

A particle of charge q and mass m starts moving from the origin under the action of an electric field \vec{E}=Eo\hat{i} and magnetic field \vec{B}=Bo\hat{k}. Its velocity at (x,3,0) is 4\hat{i} + 3\hat{j}. What is the value of x?

The Attempt at a Solution

The force exerted by the magnetic field should be along negative z-axis so that the y co-ordinate is negative. Why is it positive in the question?

The displacement along the x-axis is due to the electric field. Using equation of motion,
V_x² = 2a_xs_x
Here V_x is 4 and a=E_oq/m
so that x = 8m/qE_o

The answer is incorrect. Any help appreciated.

 

[collinsmark]

Oh, my. This is a great problem!

At first I wrote out a differential equation which ended up looking pretty gnarly. But don't go that route. There is a much easier way! When I figured it out I even felt all giddy. You need to think about this too (I wouldn't want to spoil the fun for you), so I won't give you too much help, but I'll guide you a little.

Homework Statement

A particle of charge q and mass m starts moving from the origin under the action of an electric field \vec{E}=Eo\hat{i} and magnetic field \vec{B}=Bo\hat{k}. Its velocity at (x,3,0) is 4\hat{i} + 3\hat{j}. What is the value of x?

The Attempt at a Solution

The force exerted by the magnetic field should be along negative z-axis so that the y co-ordinate is negative. Why is it positive in the question?

I'm not quite sure what you are saying, but it's not quite right. The force on a particle from a magnetic field is,

\vec F = q \vec v \times \vec B

That cross product is all important here. The cross product means perpendicular to both. The magnetic force is always perpendicular to the magnetic field.

So for this problem, the magnetic field points along the z-axis. That means the magnetic force never points along the z-axis (i.e. the magnetic force always points along some combination of the x- and y-axes).

And as equally as important for this problem, the magnetic force is always perpendicular to the particle's velocity. Is that relevant? Yes, very. It means the magnetic force, by itself, never causes the particle to speed up or slow down. It only causes the particle to change direction. But the magnetic force doesn't affect the particle's speed, thus it doesn't affect its kinetic energy.

The displacement along the x-axis is due to the electric field. Using equation of motion,
V_x² = 2a_xs_x

Sorry, but that equation doesn't apply here. That equation only applies to uniform acceleration. Here the net force is not constant, because the particle's velocity is changing, and the magnetic force is proportional to the particle's velocity.

Sure, the particle starts off accelerating along the x-axis, but at the moment its velocity is non-zero, it veers off the x-axis due to the magnetic force. After that, there are components of the magnetic force which lie on both the x- and y-axes.

The answer is incorrect. Any help appreciated.

Here are my only hints:

• What does the work-energy theorem have to say about this problem?
• W = \int \vec F \cdot \vec {ds}
• You won't need to evaluate the above integral. But you will need to think about what it really means.
• The cross product implies perpendicular. What does the dot product imply?
• What direction is \vec {ds} in relation to the particle's instantaneous velocity?
• The final answer isn't going to be just a number. It's going to be a function of some other parameters given in the problem statement.

 

[zorro]

Thanks for your reply

I'm not quite sure what you are saying, but it's not quite right. The force on a particle from a magnetic field is,

\vec F = q \vec v \times \vec B

The magnetic field is along +ve z-axis. The initial velocity of the particle is along positive x-axis. Using right hand thumb rule, the direction of the magnetic force is along negative y-axis i.e., the particle will move in X-Y plane with changing - positive abcissa and negative ordinate.
But the question says that after some time, the particle is at (x,3,0). Here the y-coordinate is positive, which is against my conclusion.

Using your hints, I figured out the solution.
The change in kinetic energy equals the work done by the electric field.
so 0.5m(25-0) = Eqx
From here x=25m/2Eq which is the correct answer.

 

[collinsmark]

But the question says that after some time, the particle is at (x,3,0). Here the y-coordinate is positive, which is against my conclusion.

The numerical values for q and E₀ were never given in the problem statement. Either one of them might be negative.

[Edit: Oh, and B₀ might be negative too. 'Forgot to include that one.]

Using your hints, I figured out the solution.
The change in kinetic energy equals the work done by the electric field.
so 0.5m(25-0) = Eqx
From here x=25m/2Eq which is the correct answer.

Bravo!

 

[zorro]

Thanks alot!

 

[tiny-tim]

The magnetic field is along +ve z-axis. The initial velocity of the particle is along positive x-axis. Using right hand thumb rule, the direction of the magnetic force is along negative y-axis i.e., the particle will move in X-Y plane with changing - positive abcissa and negative ordinate.
But the question says that after some time, the particle is at (x,3,0). Here the y-coordinate is positive, which is against my conclusion.

Yes, but despite its initial "downward" velocity, it can still get to (4,3) …

its motion is a sideways motion combined with a clockwise circular motion of increasing radius …

it will eventually circle round "above" the x-axis, and a suitable choice of parameters should get it to go through (4,3) or any other chosen point in the first or second quadrant.

 

[zorro]

Yes, but despite its initial "downward" velocity, it can still get to (4,3) …

its motion is a sideways motion combined with a clockwise circular motion of increasing radius …

it will eventually circle round "above" the x-axis, and a suitable choice of parameters should get it to go through (4,3) or any other chosen point in the first or second quadrant.

errr...I don't get you...We don't know the x co-ordinate in the question...what is (4,3) ?
Did you mean that irrespective of the sign of charge the particle can have negative y-coordinate ?

 

[tiny-tim]

Sorry, I should have written (x,3) … it can get to (x,3) for any positive value of x, given the right parameters.

I mean that the particle goes in a circle, so although it starts "below the line", eventually it "spirals" round "above the line".

 

[collinsmark]

Yes, tiny-tim brings up a good point. The particle does spiral around with an ever increasing radius, although it has a general inclination to migrate toward one direction due to the electric field. (either the positive or negative x-axis, depending on the signs of q and E₀.)

The particle's kinetic energy starts out at zero. From that point on, over the long run (time average), it gains kinetic energy. Suffice it to say its kinetic energy never goes less than zero (we'd have some serious conservation of energy problems if that happened. )

However, due to the spiraling pattern, sometimes (more than not) energy is transferred from the electric field to the particle*. And also sometimes (less than not) energy is transferred from the particle to the electric field**.

*At places when the velocity and the field are in the same general direction (dot product is positive).
**At places when the velocity and the field are in the opposite general direction (dot product is negative).

But we know for sure that the particle's kinetic energy would never be negative, because that hardly even makes sense. It can be shown, using previously discussed ideas on this thread that the particle's kinetic energy is

K.E. = qE₀x.

So if x happens to be positive, both q and E₀ must be the same sign. If x is negative, q and E₀ must be opposite signs. That's pretty much the only restriction.

[Edit: by that I mean, given the problem statement of constant q and E₀, x will be either always positive or always negative.]

[Edit: deleted my last comment.]

 

[zorro]

Sorry, I should have written (x,3) … it can get to (x,3) for any positive value of x, given the right parameters.

I mean that the particle goes in a circle, so although it starts "below the line", eventually it "spirals" round "above the line".

What are those right parameters?

My mind does not seem to agree with you.
For simplicity, just consider the 2 dimensional plane. The magnetic force acts along neg. Y-axis. So the particle starts to execute a circular motion (remove e.f. for now). The centre of the circle is in xy(negative y) plane. But there is an electric field too. So the motion is spiral with the particle just touching the x-axis and not crossing it, otherwise it will violate the laws. What I am trying to say is that the particle does not cross x-axis once it is acted upon by a perpendicular force provided the particle moves along x-axis initially.

 

[tiny-tim]

… What I am trying to say is that the particle does not cross x-axis once it is acted upon by a perpendicular force provided the particle moves along x-axis initially.

No, although the particle starts along the x-axis and curves away "downward", the magnetic force on it will always be perpendicular to its path, and to the right, so it will eventually be moving directly "downward", then directly parallel to the negative x-axis, then directly "upward", and so on.

It is the fact that the direction of the magnetic force is changing that causes all this.

(and the parameters are E and B … even if B stays positive, they can be adjusted so as to make the path go through any point (+x,3,0))

 

[collinsmark]

(Pardon my poor graphics by the way.) I haven't worked out the math, but this idea seems reasonable to me, even though I haven't backed it up. I was thinking something like this for q, E₀ and B₀ all being positive.

We know that the y component of the position is positive however, given the problem statement (but the y-component in my above figure are always negative or zero. So the assumptions in the above figure are not correct together. Something needs to be negative). So if q if E₀ are the same sign, B₀ must be negative. If q and E₀ are opposite signs, B₀ must be positive.

But as I've said I haven't worked out the differential equation to know for sure, but the above idea seems reasonable to me at the moment (Particularly for a Friday night after getting no sleep -- And it is a Friday night after all, so if I'm wrong, cut me a little slack for that if for nothing else).

But you know, I really need to say "thank goodness" for the work-energy theorem. Using that we were able to solve the problem statement's requirement without having to solve the gnarly differential equation; and we found the found the answer in a few simple steps. Thank goodness for that.

 

[zorro]

(Pardon my poor graphics by the way.) I haven't worked out the math, but this idea seems reasonable to me, even though I haven't backed it up. I was thinking something like this for q, E₀ and B₀ all being positive.

I really don't understand what kind of path the particle will follow. Is the path along the loops shown by your figure in X-Y plane or is it this way-

Where the path is spiral around the x-axis with the locus of centre of 'circles' on it.

Please prove which one is right.

 

[tiny-tim]

I'm pretty sure it's the second one, since I see no reason why the average position should move away from the x-axis.

But if you want it proved … try it yourself!

 

[zorro]

Let me try...
The velocity of the particle will always have 2 components- along x and y axes.
No component along z-axis because no force is acting along it.
the component along y-axis is perpendicular to the magnetic field always and that is responsible for the circular motion of the particle. The component along x-axis makes the circular path spiral. So the particle executes spiral motion with centre on the x-axis.

Is it correct?

 

[tiny-tim]

Let me try...
The velocity of the particle will always have 2 components- along x and y axes.
No component along z-axis because no force is acting along it.
the component along y-axis is perpendicular to the magnetic field always and that is responsible for the circular motion of the particle. The component along x-axis makes the circular path spiral. So the particle executes spiral motion …

Correct so far.

But you haven't proved the last bit …

… with centre on the x-axis.

 

[zorro]

Correct so far.

But you haven't proved the last bit …

Thats simple.
The motion starts from x-axis. So the initial centre is along x-axis.

 

[tiny-tim]

No, the initial centre is clearly below the x-axis.

 

[zorro]

Oh yes!...what causes the centre to shift back then??

 

[tiny-tim]

Because the radius of curvature is getting larger …

essentially, the centre of curvature is moving away from the path.

 

[zorro]

Because the radius of curvature is getting larger …

essentially, the centre of curvature is moving away from the path.

Does it mean that the centre of curvature keeps moving away from the path always? In that case the centre will lie on x-axis for a small time interval.

 

[collinsmark]

Actually, now I think we were all wrong. Doh!

Upon thinking about it further, the patter might actually look like this for q, E₀ and B₀ all being positive. (To make it match the data given in the problem statement, flip it up to the positive y-axis by making B₀ negative, or by making both q and E₀ negative.)

https://www.physicsforums.com/attachment.php?attachmentid=30367&stc=1&d=1291498687

This shape is a consequence of the velocity starting out at zero at the origin. The shape would be quite different if the initial velocity was non-zero.

My logic is this. The particle will accelerate along the x-axis, skewed along the y-axis until it reaches some maximum velocity. At this point the magnetic force is greater than the electric force, so it still accelerates in the negative x direction. 'Same thing as saying it "decelerates." As it goes more negative in the x-direction, toward the y-axis, the velocity slows. But there is some symmetry here. The particle's slowing down in the second half of the first hump is symmetric to it speeding up in the first half of the first hump.

But I think what we all forgot before was that the first half of the first hump is symmetrical to the second half of the first hump. The end result is a repeating pattern along the y-axis. It will go positive or negative along the y-axis depending on the signs of q, E₀ and B₀.

That mistake corrected (well, sort of. I haven't proved anything yet), I still claim, that the particle will be restricted to one quadrant.

I have to run some errands, so I can't prove this right now, but I might be able to get back to it later. In the mean time, I offer you this simulation that corroborates what I'm saying:

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=36"

and this:
http://cnx.org/content/m31345/latest/"

 

[zorro]

The animation in the link you gave is awesome!
These are my observations-
The path of the particle you drew in the figure is true if the electric and magnetic fields are comparable.
However in the situations of physics problems, I have observed that the magnetic field is very less compared to the electric field (in terms of magnitude).
If you set up smaller values of B like 3,10T and E as 100,200 etc, the path is almost parabolic with negligible zig-zag movements in the x-z plane.

So the path can be very complex or simple depending upon the relative values of the two fields and offcourse the initial velocity of the particle.

P.S. - The modules written by Sunil Kumar Singh on cnx.org are very good and provide indepth knowledge on topics that are not even covered by university level books!

 

[collinsmark]

Just so we're all on the same page my attached drawing (the newer one) figure mysteriously disappeared somehow that goes with my last post. I'll attach it again here for reference.

 

[collinsmark]

The animation in the link you gave is awesome!
These are my observations-
The path of the particle you drew in the figure is true if the electric and magnetic fields are comparable.
However in the situations of physics problems, I have observed that the magnetic field is very less compared to the electric field (in terms of magnitude).
If you set up smaller values of B like 3,10T and E as 100,200 etc, the path is almost parabolic with negligible zig-zag movements in the x-z plane.

So the path can be very complex or simple depending upon the relative values of the two fields and offcourse the initial velocity of the particle.

P.S. - The modules written by Sunil Kumar Singh on cnx.org are very good and provide indepth knowledge on topics that are not even covered by university level books!

Yes, the shapes can vary quite a bit depending on the strengths. My only real points are as follows.

• Given the way the problem statement was phrased, the particle is confined to one quadrant. And which quadrant that is depends on the signs of q, E₀ and B₀. Since the problem statement implies a positive y position, it means that at least one of the numerical values for q, E₀ and/or B₀ is negative.
• Isn't it nice that we have the work-energy theorem. Using the work-energy theorem, we found a rather simple solution to the problem statement that would have otherwise have been very complicated.

 

[tiny-tim]

hi collinsmark!

thanks for the useful link and the correction!

clearly my instinct was completely wrong …

the "drift" is in the y direction not the x direction

here's a simple proof of why:

if E < cB, put r' = r - (E/B)tj

then v' = v - (E/B)j,

so dv'/dt = dv/dt = (q/m)(Ei - Bv x k)

= (q/m)(Bv' x k) …

in other words, in a frame of reference moving at speed E/B in the y-direction, the electromagnetic field becomes a pure magnetic field of strength B

(we could deduce the same … apart from a "relativistic" correction … from the standard transformation equations, see http://en.wikipedia.org/wiki/Mathem...formation_of_fields_in_vector_field_approach")

and since the initial velocity in that frame is E/B in the negative y-direction, the particle moves at that constant speed round a stationary circle of radius mE/qB², exactly as if it were on the rim of a stationary wheel touching the y-axis

so in the original frame of reference, the particle moves exactly as if it were on the rim of a wheel rolling along the y-axis at speed E/B, ie a cycloid!

surprisingly, this is exactly the same motion as that of a bead fixed to a weightless circular wire stuck in a groove on the ceiling but free to roll along it …

the bead feels a constant vertical force, mg (equivalent to E),

and if you do all the maths (there must be a simple way of doing it, but I can't see one ), you find that the angular speed of the wire is a constant, √(g/r), and that, in addition to mg, there is a force, 2sin(θ/2)mg (equivalent to B), directed towards the instantaneous centre of rotation at the top of the circle and proportional to the instantaneous speed, 2sin(θ/2)√(gr) !​