Total Force on a Point Charge in Motion

In summary, the conversation discusses the need to use relativistic transformation equations and find the fields that may exert a force on a particle in the rest frame of the observer. The electric and magnetic forces on a point charge are calculated, with the magnetic force being a relativistic effect. The direction of the unit vector in relation to the direction of the position vector is also needed. It is suggested to solve for the electrostatic problem in the frame where the rod is at rest and then transform the electromagnetic field with a Lorentz transformation to get the magnetostatic solution in the reference frame where the rod is moving. However, the OP's approach in the first post only requires the correction for length contraction to accurately calculate the force without needing to know the
  • #1
ARoyC
56
11
Homework Statement
An infinite rod of charge density λ lies parallel to the x-axis. A point charge q is stationary at a distance r from the rod. An observer is moving with a velocity -vx̂. What is the force on the point charge from that observer's frame?
Relevant Equations
F = F_E + F_B
B of an Infinitely Long Current carrying Wire = µI/2πr
F_B = q(vxB)
F_E = qE
E of an Infinitely Long Uniformly Charged Rod = λ/2πεr r̂
As the observer is moving, there will be a magnetic force.

Electric Field of the Rod = λ/2πεr
Electric Force on the Point Charge = qλ/2πεr
Magnetic Force on the Point Charge = q(vxB) = qvB = qv(µI/2πr) = qv(µλv/2πr)
= µqλv²/2πr

Total Force = Electric Force + Magnetic Force
 
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  • #2
I think that for this problem you need to use the relativistic transformation equations. Find what the fields that might exert a force on the particle look like in the rest frame of the particle and transform to the rest frame of the observer.
 
  • #3
kuruman said:
I think that for this problem you need to use the relativistic transformation equations. Find what the fields that might exert a force on the particle look like in the rest frame of the particle and transform to the rest frame of the observer.
Isn't Magnetic Force just a Relativistic Effect?
 
  • #4
ARoyC said:
Isn't Magnetic Force just a Relativistic Effect?
It is, but you are confusing the two frames. When you say
ARoyC said:
Electric Field of the Rod = λ/2πεr
that's the electric field in the frame of charged wire. Then you say
ARoyC said:
Magnetic Force on the Point Charge = q(vxB) = qvB = qv(µI/2πr) = qv(µλv/2πr)
= µqλv²/2πr
There is no magnetic force in the frame of the wire because the particle is at rest in that frame.

The problem clearly asks you to find the force on the point charge in the observer's frame. That is why you need to transform the field(s) to that frame.
 
  • #5
ARoyC said:
As the observer is moving, there will be a magnetic force.

Electric Field of the Rod = λ/2πεr
Electric Force on the Point Charge = qλ/2πεr
Magnetic Force on the Point Charge = q(vxB) = qvB = qv(µI/2πr) = qv(µλv/2πr)
= µqλv²/2πr

Total Force = Electric Force + Magnetic Force
Your work looks good except you haven't taken into account the effect of length contraction on ##\lambda##.

Also, you need to give the direction of the unit vector ##\hat n##.

With your approach, you will not need to know the relativistic transformation equations for the fields.
 
  • #6
TSny said:
Your work looks good except you haven't taken into account the effect of length contraction on ##\lambda##
If I multiply the Electric Field expression by Gamma, will the expression of the Magnetic Force remain the same?
TSny said:
Also, you need to give the direction of the unit vector ##\hat n##.
How would I find that as no position of q has been given?
 
  • #7
ARoyC said:
If I multiply the Electric Field expression by Gamma, will the expression of the Magnetic Force remain the same?
What is the justification for multiplying the electric field by gamma?

I'm not sure what you mean when you ask if the expression for B will remain the same.
ARoyC said:
How would I find that as no position of q has been given?
You should be able to specify the direction of ##\hat n## in relation to the direction of ##\hat r##.
 
  • #8
TSny said:
What is the justification for multiplying the electric field by gamma?
Transformation of the Electric Field to the Moving Frame?
TSny said:
I'm not sure what you mean when you ask if the expression for B will remain the same
Isn't F = qE + qvB, where is the Electric Field from the rest frame?
TSny said:
You should be able to specify the direction of ##\hat n## in relation to the direction of ##\hat r##.
How to do that?
 
  • #9
Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.
 
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  • #10
vanhees71 said:
Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.
Yes, I can do it. But I want to understand the mistake in my solution.
 
  • #11
vanhees71 said:
Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.
Yes, you can certainly solve it this way. But, the OP's approach in the first post will give the correct answer without having to know the general transformation equations for the fields. The only correction needed is to use length contraction to get an expression for ##\lambda## in the observer's frame. I think this approach has pedagogical advantages.
 
  • #12
Well, then you have to know the transformation laws for line-charge densities...
 
  • #13
vanhees71 said:
Well, then you have to know the transformation laws for line-charge densities...
Yes, but that is just elementary relativity (length contraction) along with invariance of charge.
 
  • #14
ARoyC said:
Transformation of the Electric Field to the Moving Frame?
If you want to solve the problem by transforming the fields to the moving frame as suggested by @kuruman and @vanhees71, then that is a good approach. But, the method in your first post will also work, and it does not require knowing the general transformation equations of the fields.

In the observer's frame (primed frame), there is an infinitely long line charge moving parallel to itself. The charge density in this frame is ##\lambda'##, which differs from ##\lambda##. Gauss' law is still valid in this frame. So, the expression for the E-field in this frame is the same as the expression for the E-field of an infinite line charge at rest except that you need to use ##\lambda'## instead of ##\lambda##.

In the observer's frame, ##B'## is just the field of a an infinitely long straight current where the current, ##I'##, can be expressed in terms of ##\lambda'## and ##v##.
 
  • #15
I think the quickest way to the answer would be to use the relativistic transformation equations for force. :oldsmile:
 
  • #16
TSny said:
But, the OP's approach in the first post will give the correct answer without having to know the general transformation equations for the fields.
I would like to see how that is done specifically the equation for the magnetic field in the observer's frame. It does not match what one gets with the field transformation equation.
TSny said:
I think the quickest way to the answer would be to use the relativistic transformation equations for force. :oldsmile:
Tsk, tsk, tsk. What happened to solutions with pedagogical advantages? :wink:
 
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  • #17
kuruman said:
Tsk, tsk, tsk. What happened to solutions with pedagogical advantages? :wink:
Exactly!
 
  • #18
TSny said:
If you want to solve the problem by transforming the fields to the moving frame as suggested by @kuruman and @vanhees71, then that is a good approach. But, the method in your first post will also work, and it does not require knowing the general transformation equations of the fields.

In the observer's frame (primed frame), there is an infinitely long line charge moving parallel to itself. The charge density in this frame is ##\lambda'##, which differs from ##\lambda##. Gauss' law is still valid in this frame. So, the expression for the E-field in this frame is the same as the expression for the E-field of an infinite line charge at rest except that you need to use ##\lambda'## instead of ##\lambda##.

In the observer's frame, ##B'## is just the field of a an infinitely long straight current where the current, ##I'##, can be expressed in terms of ##\lambda'## and ##v##.
Got it. Thank you.
 
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  • #19
It's even simpler. You don't need any Lorentz transformations.

In the restframe of the wire you have an electrostatic situation, i.e.,
$$A^{*\mu}(x^*)=\begin{pmatrix} \Phi(x^*) \\0 \\0 \\0 \end{pmatrix}.$$
This you can write in a manifestly covariant way as
$$A^{*\mu}(x^*)=u^{*\mu} \Phi(x^*).$$
Now you only need to write this in a general frame, where ##u## is an arbitrary four-velocity of the wire:
$$A^{\mu}(x)=u^{\mu} \Phi(x^*),$$
and then you simply need to express the ##x^*## by ##x##, using the Lorentz transformation ##x=\Lambda^{-1}(\vec{v})## with ##\vec{v}=\vec{u}/u^0##.

The other way, is also not as complicated as I first thought. By the same reasoning you have
$$\lambda^{*\mu}=\begin{pmatrix}\lambda^* \\ 0 \\0 \\ 0 \end{pmatrix},$$
where ##\lambda^*## is the charge-line-density in the restframe of the wire. In the general frame again you have
$$\lambda^{\mu}=u^{\mu} \lambda^*.$$
Then you can solve the corresponding magnetostatic problem with this given four-current-line density.
 

FAQ: Total Force on a Point Charge in Motion

What is the total force on a point charge in motion?

The total force on a point charge in motion is the sum of the electric force and the magnetic force acting on the charge. This is often described by the Lorentz force law, which states that the total force \(\mathbf{F}\) is given by \(\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\), where \(q\) is the charge, \(\mathbf{E}\) is the electric field, \(\mathbf{v}\) is the velocity of the charge, and \(\mathbf{B}\) is the magnetic field.

How does the velocity of a point charge affect the total force?

The velocity of a point charge affects the magnetic component of the total force. According to the Lorentz force law, the magnetic force is given by \(q(\mathbf{v} \times \mathbf{B})\). This means that the force depends on the cross product of the velocity vector \(\mathbf{v}\) and the magnetic field vector \(\mathbf{B}\). Therefore, the direction and magnitude of the magnetic force change with the velocity of the charge.

What role does the electric field play in the total force on a point charge in motion?

The electric field exerts a force on the point charge that is independent of its motion. This force is given by \(q\mathbf{E}\), where \(q\) is the charge and \(\mathbf{E}\) is the electric field. This component of the force acts in the direction of the electric field and is proportional to the magnitude of the charge and the electric field.

How does the magnetic field influence the total force on a moving point charge?

The magnetic field influences the total force by contributing a magnetic force component that depends on the velocity of the charge. This magnetic force is given by \(q(\mathbf{v} \times \mathbf{B})\). The direction of this force is perpendicular to both the velocity of the charge and the magnetic field, and its magnitude depends on the charge, the velocity, and the strength of the magnetic field.

Can a point charge in motion experience a force if only a magnetic field is present?

Yes, a point charge in motion can experience a force even if only a magnetic field is present. This force is the magnetic component of the Lorentz force, given by \(q(\mathbf{v} \times \mathbf{B})\). If the charge is moving and there is a magnetic field, the charge will experience a force that is perpendicular to both its velocity and the magnetic field.

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