# Proton in a changing magnetic field

1. Apr 22, 2008

### oh.rry21

sorry i didnt follow the template but its a conceptual question i'm not sure i understand.

if theres a proton in a decreasing magnetic field...

will it move or anything?

according to my deductions, it won't have any forces on it and therefore it won't move. although there will be an induced EMF because of a decreasing magnetic field...it won't have any effect on the proton because theres no "positive" and "negative" pole for it to attract to or be repulsed from.

am i right? o_o?

2. Apr 22, 2008

### dynamicsolo

I think this is where you want to use Maxwell's form of Faraday's Law:

curl E = -dB/dt .

The collapsing magnetic field does generate an electric field, so the proton should be accelerated by it. (Consider that it is the mutual variations in electric and magnetic fields that allow an electromagnetic wave to propagate, in the classical description.)

If the magnetic field has been pointing "upward" and is diminishing, it will produce a local electric field that will push charges perpendicularly to the magnetic field direction. (This is the "cause" of the induced EMF; if there were a closed conductive loop, you'd get a circulating counter-clockwise current, as seen from the view with the magnetic field pointing at you.)

3. Apr 22, 2008

### oh.rry21

so the proton will be pushed to the left because of the local electric field created by a collapsing magnetic field?

but when the proton starts accelerating (assuming its to the left), doesnt the decreasing magnetic field and the velocity vector of the now moving proton exert a force?

if i did my cross product right, the proton moves to the left, and the induced magnetic field exerts a force up? (along the y axis). so i guess if you vector sum the E and B forces it would be diagnolly up and left?

agh im so confused :(

4. Apr 22, 2008

### jandl

You're correct to think that the proton will move due to both the electric field and magnetic field. However, consider Lorentz's equation. If you look at this carefully and remember what a cross product means, you'll see that a magnetic field will not cause a charge to move along the magnetic field lines.

Also remember that a moving charge will create its own magnetic field, but is assumed to not interact with or be affected by that field unless moving at relativistic speeds.

5. Apr 22, 2008

### oh.rry21

right but when i did the cross product of f=q vxb...

the velocity vector is to the left. and i'm assuming that my "b" is the induced magnetic field. the induced magnetic field points out in the same direction as the magnetic field (since its decreasing).

and so i get the magnetic field as having a force down.

summed with the E field that points to the left.

left and down? :(

6. Apr 22, 2008

### Nick89

You might need to consider the lorentz equation in a electric and magnetic field:

$$\vec{F_l} = q( \vec{E} + \vec{v} \times \vec{B})$$

7. Apr 22, 2008

### oh.rry21

well yeah. thats why i added the E and v x b vectors. am i wrong?

8. Apr 22, 2008

### jandl

You can ignore the direction of the induced magnetic field (the field generated by the moving proton). Again, this field will only affect the proton's motion when working at relativistic velocities.

If you are saying that the applied magnetic field is oriented towards the top of the page. Again consider carefully the cross product. Really, this questions asks you to think about the equations and wants you to relate them to the application of the right hand rule.

Also, could you please define the form of your magnetic field, as this might significantly change the problem.

9. Apr 22, 2008

### oh.rry21

sorry :X hahaha okay so its a

uniform magnetic field pointing into the page, decreasing at some rate. its actually a circular magnetic field with some radius R and the proton is in the magnetic field.

the question asks "when it is released from rest, what happens to the proton?"

okay so if i dont take induced magnetic field into consideration...

i still use the cross product v x b. instead i use the original magnetic field. the current created by the magnetic field is in the cw direction using the RHR.

----confused :(

10. Apr 22, 2008

### Nick89

You are contradicting yourself...? First you say it's a uniform magnetic field pointing into the page, then you say it's circular?

Do you mean to say that the radius of the circle is so large you can ignore the 'bend' and approximate it with a uniform field?

11. Apr 22, 2008

### jandl

Alright, your situation changes the problem a lot.

First, if the proton is in the exact center of the magnetic field and the field is truly uniform, the proton won't move. Why?

If the proton is away from center the proton will move. This is when you need to start using the RHR carefully. Place your thumb in the direction that the field is decreasing, your fingers will point in the direction that the E field goes. You should find that it is circular and in the CWC or CW direction? What direction will the proton move? (what if the particle was an electron instead?) When the proton does start moving, use the cross product again and think about direction the of the Lorentz force (v X b), also think about how the direction of the force changes as the location and direction of motion of the proton changes. How could you describe the path of the proton, straight line, spiral, circular, elliptical, etc.?

12. Apr 22, 2008

### oh.rry21

touche. hahaha no its a circular magnetic field :)

13. Apr 22, 2008

### jandl

I forgot to consider another case.

Think about what happens if the proton starts outside of the circle of magnetic field lines.

Nick: I think he means that it is a circle of radius R with a uniform and changing magnetic field pointing into the page.

14. Apr 22, 2008

### oh.rry21

T_T im so confused jandl and nick89. i dont know if i did anything right or wrong

what does the proton actually do? haha :(

i'm the kind of person who can learn from the answer rather than the means of getting there. it'll make a lot more sense to me if i knew what happened and i filled in the gap.

15. Apr 22, 2008

### jandl

Nick and I are having trouble understanding exactly what is going on. The form and orientation of the magnetic field is crucial to this problem. Could you please carefully define the magnetic field? If you can do that, we can easily help you towards an answer.

16. Apr 22, 2008

### Nick89

Consider the following picture:

(I hope this corresponds to what you have in mind)

Let's say the magnetic field B (blue region, going into the picture, -z direction) is a magnetic field induced by the current I in a solenoid (the gray circle around the blue region).
Let's say this magnetic field is increasing.

Due to the increase in magnetic field, an electric field E (shown with red arrows and green circle) will be created.

Is this what you mean?

Now consider a proton on the negative x-axis for example (in the blue region, not in the origin).

Try to find the direction of the lorentz force using the formula:
$$\vec{F} = q( \vec{E} + \vec{v} \times \vec{B} )$$

Last edited: Apr 22, 2008
17. Apr 22, 2008

### oh.rry21

it absolutely corresponds to what i have in mind. except the magnetic field is decreasing its exactly what i have in mind

okay so F=qE i have as going in the negative y direction
and as for the v x b, with its velocty vector in the negative y direction and the b field going into the page...i have the cross product going to the right along the positive x axis.

therefore if we add the vectors it would...go diagnolly down! the positive x axis and the negative x axis!

? how was that? :D

18. Apr 22, 2008

### Nick89

So far so good I think. However I think if the B-field was decreasing the E-field would turn the other way.

What you say however is only valid if you describe the proton starting from rest on the negative x-axis.

Can you describe the path the proton makes after starting from rest? Does it keep going in a circle, maybe an ellipse, parabola??

Actually, I think this depends on the values of the B and E field. Since you are (vector)adding the E and vxB vectors their size matters. If E was VERY large compared to vxB than the motion was effectively downward only.