Particle/dipole acceleration in non-uniform magnetic field

  • #1
kmm
170
10
I've come across a few places that mention that dipoles and charged particles accelerate in non-uniform magnetic fields. Is this true? If the Magnetic force is always perpendicular to the velocity of a charge, I don't see why it would accelerate. I see it having centripetal acceleration with constant kinetic energy unless the centripetal acceleration is all that's needed for the charge to emit a photon. I also don't see a dipole would accelerate in a non-uniform magnetic field.
 

Answers and Replies

  • #2
WannabeNewton
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A charged particle with some non-zero velocity will accelerate in any magnetic field as per the Lorentz force law, given that the velocity and magnetic field are not parallel. Simple as that.

As for magnetic dipoles, if we have an infinitesimal current loop with magnetic dipole moment ##\vec{m}## in a magnetic field ##\vec{B}## then it can be shown that it experiences a force ##\vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B})##. See problem 6.4 in Griffiths.
 
  • #3
kmm
170
10
A charged particle with some non-zero velocity will accelerate in any magnetic field as per the Lorentz force law, given that the velocity and magnetic field are not parallel. Simple as that..
Except the Lorentz force law says that the magnetic force is [itex] \mathbf{F_{magnetic}} = q (\mathbf{v} \times \mathbf{B}) [/itex] so the force is perpendicular and therefore no acceleration, unless you're simply referring to centripetal acceleration.

As for magnetic dipoles, if we have an infinitesimal current loop with magnetic dipole moment ##\vec{m}## in a magnetic field ##\vec{B}## then it can be shown that it experiences a force ##\vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B})##. See problem 6.4 in Griffiths.
I will check Griffiths. Also, what then is the advantage of a non-uniform magnetic field in particle accelerators or like that used in the Stern-Gerlach experiment? Simply to control the path better?
 
  • #4
WannabeNewton
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Except the Lorentz force law says that the magnetic force is [itex] \mathbf{F_{magnetic}} = q (\mathbf{v} \times \mathbf{B}) [/itex] so the force is perpendicular and therefore no acceleration, unless you're simply referring to centripetal acceleration.
The force being perpendicular to the velocity doesn't imply that there is no acceleration. A force always imparts an acceleration as per Newton's 2nd law. What makes you think centripetal acceleration is not an acceleration?

I will check Griffiths. Also, what then is the advantage of a non-uniform magnetic field in particle accelerators or like that used in the Stern-Gerlach experiment? Simply to control the path better?
Well in the Stern-Gerlach experiment we need a non-uniform magnetic field in order for an interacting magnetic dipole to get deflected in a way that distinguishes between different dipole moments. Classically this goes back to that equation I wrote above: ##\vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B})## specialized to a constant dipole moment. See section 1.1 of Sakurai's QM text.
 
  • #5
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170
10
The force being perpendicular to the velocity doesn't imply that there is no acceleration. A force always imparts an acceleration as per Newton's 2nd law. What makes you think centripetal acceleration is not an acceleration?
I don't think my original question was asked well and I did imply that centripetal acceleration isn't acceleration. What I meant was that I wasn't sure if it was the magnitude of the velocity that changed since that's the most common context I've heard it in. OK, so yes there is an acceleration, just no change in speed.


Well in the Stern-Gerlach experiment we need a non-uniform magnetic field in order for an interacting magnetic dipole to get deflected. Classically this goes back to that equation I wrote above: ##\vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B})## specialized to a constant dipole moment. See section 1.1 of Sakurai.
Unfortunately, I don't have Sakurai. Although, if the force on the dipole is [itex] \vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B}) [/itex] then it seems the dipole would be deflected whether the magnetic field was uniform or not.
 
  • #6
WannabeNewton
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Unfortunately, I don't have Sakurai. Although, if the force on the dipole is [itex] \vec{F} = \vec{\nabla}(\vec{m}\cdot \vec{B}) [/itex] then it seems the dipole would be deflected whether the magnetic field was uniform or not.
If the magnetic field is uniform then the gradient will vanish and there won't be a force.
 
  • #7
kmm
170
10
If the magnetic field is uniform then the gradient will vanish and there won't be a force.
Oh of course, thanks for helping me clear that up.
 

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