Charged particles in magnetic field

In summary: Actually, I see. You're saying that the force between the particles in the Y-direction is the sum of the forces in the X and Z-directions?The force between the particles in the Y-direction is the sum of the forces in the X and Z-directions.
  • #1
utkarshakash
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Homework Statement


Two particles have equal masses m and electric charges of equal magnitude and opposite sign (+q and –q). The particles are held at rest in uniform magnetic field B. The direction of the field is perpendicular to the line connecting the charges. The particles are released simultaneously. What is the minimum initial separation L that allows the particles not to collide after they are released? Neglect the effects of gravity.

The Attempt at a Solution



Let the distance of closest approach be d.
Applying conservation of energy,
[itex]\dfrac{-kq^2}{L} = mv^2 - \dfrac{kq^2}{d} [/itex]

At distance of closest approach,
Fe=Fb
[itex]vd^2=kq/B [/itex]

But it looks like I will need one more equation. If the trajectory of the particles were circular, I could have applied the formula r=mv/qB but this isn't the case here.
 

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  • #2
Something weird here: if a charged particle is at rest in a magnetic field, it does not experience any force. So releasing it should not bring about any motion. Are we missing an electric field in the exercise formulation ?

(I do see an Fe in the picture...)

[edit] total miss. The E field is from the particles themselves. Sorry :redface:. Have to rethink this one.
 
  • #3
BvU said:
Something weird here: if a charged particle is at rest in a magnetic field, it does not experience any force. So releasing it should not bring about any motion. Are we missing an electric field in the exercise formulation ?

(I do see an Fe in the picture...)

Pay close attention to this line
"The particles are held at rest in uniform magnetic field B".
They are being held at rest initially and when they are released, the particles start moving towards each other due to interaction force and thus, they also experience magnetic force. Here Fe denotes the force due to mutual interaction of the particles.
 
  • #4
Ah, I've had some coffee and now I wanted to ask: what is the limiting case ?
But you already described it, adding you need another eqn. But: [itex]\dfrac{-kq^2}{L} = mv^2 - \dfrac{kq^2}{d} [/itex] and [itex]vd^2=kq/B [/itex] have how many unknowns ?

[edit] More coffee: 3 unknowns. Have to rethink some more.
 
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  • #5
BvU said:
Ah, I've had some coffee and now I wanted to ask: what is the limiting case ?
But you already described it, adding you need another eqn. But: [itex]\dfrac{-kq^2}{L} = mv^2 - \dfrac{kq^2}{d} [/itex] and [itex]vd^2=kq/B [/itex] have how many unknowns ?

3 unknowns- L,v and d.
 
  • #6
##vd^2=kq/B## should probably be ##vd^2 \ge kq/B## ? But in the limiting case Fe=Fb, as you (also!) already describe niicely.

Anyway, I never see d=0 appearing: v goes to infinity if d goes to 0. We end up in relativistic quantum electrodynamics!

We must be doing something wrong: for every L there comes an equation in d that looks solvable (albeit not easily, but we don't need d): ##km/B^2 = d^3(1-d/L)##.

If I could have my undemocratic way, I would say: there is no lower limit to L !

(maybe someone more qualified could bring in the Maxwell equations here and
 
  • #7
See if you can use Newton's 2nd law in the y-direction to find a general relation between the y-component of velocity of one of the particles and its x coordinate. Then see what you can deduce if you use this relation in the energy equation at the point of closest approach.

Edit: There is no reason that the electric force magnitude Fe should equal the magnetic force magnitude FB at the point of closest approach, in general. But I think it does turn out that for the minimum value of L for which the particles avoid collision, then the two forces do have equal magnitudes at the closest approach. The motion is then interesting.
 
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  • #8
TSny said:
See if you can use Newton's 2nd law in the y-direction to find a general relation between the y-component of velocity of one of the particles and its x coordinate. Then see what you can deduce if you use this relation in the energy equation at the point of closest approach.

Edit: There is no reason that the electric force magnitude Fe should equal the magnetic force magnitude FB at the point of closest approach, in general. But I think it does turn out that for the minimum value of L for which the particles avoid collision, then the two forces do have equal magnitudes at the closest approach. The motion is then interesting.

Let at any instant velocity of the charged particle on the left side be v and it makes an angle θ with the vertical.
In Y- direction

[itex] F_b \sin \theta = mv \cos \theta \dfrac{d(v \cos \theta)}{dy} \\
F_e - F_b \cos \theta = mv \sin \theta \dfrac{d(v \sin \theta)}{dx}[/itex]

The expression does not seem nice, though :uhh:. I don't see how does this help.
 
  • #9
Express Cartesian components of the positively charged particle as ##x##, ##\dot{x}##, ##\ddot{x}##, and similarly for ##y##.

Can you use the 2nd law to relate ##\ddot{y}## to ##\dot{x}##? Can you integrate it once to relate ##\dot{y}## to ##x##?
 
  • #10
TSny said:
Express Cartesian components of the positively charged particle as ##x##, ##\dot{x}##, ##\ddot{x}##, and similarly for ##y##.

Can you use the 2nd law to relate ##\ddot{y}## to ##\dot{x}##? Can you integrate it once to relate ##\dot{y}## to ##x##?

The direction of Fb changes continuously. How do I express its components along X and Y axis? Without this, it isn't possible to apply Newton's second law.
 
  • #11
##\vec{F} = q\vec{V}\times \vec{B}##

Note ##\vec{V} = \dot{x} \, \hat{i} + \dot{y} \, \hat{j}## and ##\vec{B} = B\, \hat{k}##.

How do you write the y-component of the cross product in terms of ##\dot{x}## and ##B \;##?
 
  • #12
TSny said:
##\vec{F} = q\vec{V}\times \vec{B}##

Note ##\vec{V} = \dot{x} \, \hat{i} + \dot{y} \, \hat{j}## and ##\vec{B} = B\, \hat{k}##.

How do you write the y-component of the cross product in terms of ##\dot{x}## and ##B \;##?

Fnet = ma

[itex] \left( qB \dot{y} + \dfrac{kq^2}{x^2} \right) \hat{i} -qB\dot{x} \hat{j}=m(\ddot{x}\hat{i}+\ddot{y}\hat{j})[/itex]

Equating respective components

[itex] qB \dot{y} + \dfrac{kq^2}{x^2}=m\ddot{x} \\
-qB\dot{x} =m\ddot{y} [/itex]

Integrating the last equation I get
[itex]m\dot{y}=-qBx[/itex]
 
  • #13
utkarshakash said:
Fnet = ma

[itex] \left( qB \dot{y} + \dfrac{kq^2}{x^2} \right) \hat{i} -qB\dot{x} \hat{j}=m(\ddot{x}\hat{i}+\ddot{y}\hat{j})[/itex]

Equating respective components

[itex] qB \dot{y} + \dfrac{kq^2}{x^2}=m\ddot{x} \\
-qB\dot{x} =m\ddot{y} [/itex]

Note that x is not the distance between the two charges. It's the x coordinate of the positively charged particle. So, your expression for the electrostatic force in terms of x needs to be modified.

Integrating the last equation I get
[itex]m\dot{y}=-qBx[/itex]

What about a constant of integration? You need to satisfy the initial conditions.
 
  • #14
TSny said:
Note that x is not the distance between the two charges. It's the x coordinate of the positively charged particle. So, your expression for the electrostatic force in terms of x needs to be modified.
What about a constant of integration? You need to satisfy the initial conditions.

At x=L/2, vy = 0

[itex]m\dot{y}=-qBx+L/2 [/itex]

For the electrostatic force, the x should be replaced by 2x.
 
  • #15
utkarshakash said:
At x=L/2, vy = 0

[itex]m\dot{y}=-qBx+L/2 [/itex]

OK. [EDIT: But you need some parentheses to make this correct. Note that qBx does not have the same dimensions as L/2.]

Can you use this to get an expression for [itex]\dot{y} [/itex] when the particles are at their closest approach distance d?

What is [itex]\dot{x}[/itex] at the closest approach?

What does your energy equation in your first post look like at the closest approach?

For the electrostatic force, the x should be replaced by 2x.

Good.
 
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  • #16
TSny said:
OK. [EDIT: But you need some parentheses to make this correct. Note that qBx does not have the same dimensions as L/2.]

Can you use this to get an expression for [itex]\dot{y} [/itex] when the particles are at their closest approach distance d?

What is [itex]\dot{x}[/itex] at the closest approach?

What does your energy equation in your first post look like at the closest approach?



Good.

I get this equation
[itex]2m\dot{y} = -qB(d+L) [/itex]
At distance of closest approach [itex]\dot{x}=0[/itex]. So I can replace v with [itex]\dot{y}[/itex] in my energy equation. But the two equations aren't easy to solve.
 
  • #17
utkarshakash said:
I get this equation
[itex]2m\dot{y} = -qB(d+L) [/itex]

That's not quite correct. Check all of your signs on the right hand side.

At distance of closest approach [itex]\dot{x}=0[/itex]. So I can replace v with [itex]\dot{y}[/itex] in my energy equation.

Good.

But the two equations aren't easy to solve.

It's not too bad if you note that 1/d - 1/L = (L-d)/(Ld). Look for a common factor of (L-d) in the equation. See if you can end up with a quadratic equation for d.
 
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  • #18
TSny said:
That's not quite correct. Check all of your signs on the right hand side.



Good.



It's not too bad if you note that 1/d - 1/L = (L-d)/(Ld). Look for a common factor of (L-d) in the equation. See if you can end up with a quadratic equation for d.

You are a genius! :smile: Thank you so much for helping. I got the correct answer

[itex]L^3 = \dfrac{16mk}{B^2} [/itex]
 

What are charged particles in a magnetic field?

Charged particles in a magnetic field refer to particles that have an electric charge and are subjected to the force of a magnetic field. This force causes the particles to move in a circular or helical path.

How do charged particles interact with a magnetic field?

Charged particles interact with a magnetic field through the Lorentz force, which is the force exerted on a charged particle by an electric and magnetic field. This force causes the particle to move in a curved path perpendicular to both the electric and magnetic field.

What is the difference between positive and negative charged particles in a magnetic field?

The difference between positive and negative charged particles in a magnetic field is the direction in which they are deflected. Positive particles are deflected in one direction, while negative particles are deflected in the opposite direction. This is due to the fact that positive particles have a positive charge and negative particles have a negative charge, and the direction of the Lorentz force depends on the direction of the magnetic field.

How does the strength of the magnetic field affect the motion of charged particles?

The strength of the magnetic field affects the motion of charged particles by determining the radius of their circular or helical path. A stronger magnetic field will result in a smaller radius, while a weaker magnetic field will result in a larger radius. This is because the Lorentz force is directly proportional to the strength of the magnetic field.

What is the application of charged particles in magnetic fields?

Charged particles in magnetic fields have numerous applications in various fields such as particle accelerators, mass spectrometers, and magnetic resonance imaging (MRI) machines. They are also used in plasma physics research and in the study of the Earth's magnetic field.

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