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I don't understand how to explain why kinetic energy increases using Faraday's Law as a charged particle reaches the end of a magnetic bottle.

I know that along the field line in the bottle, the magnetic moment of the particle is conserved (first adiabatic invariant) which leads to $v_{\perp}$ icreasing as $v_{\parallel}$ decreases, and the kinetic energy of the particle is just equal to $K.E.=\frac{1}{2}v_{\perp}^{2}$. This makes sense, but I have no idea how this relates to Faraday's Law.

I am tempted to invoke the integral form of Faraday's Law:

$\oint_C {E \cdot d\ell = - \frac{d}{{dt}}} \int_S {B_n dA}$

where the closed loop is the path the particle is gyrating around the magnetic field, but I am not sure that is right. Obviously, as the field line density increases towards the end of a magnetic bottle, the flux through a given area increases.

Any help here would be appreciated. I am just having a hard time contextualizing these principles for a magnetic bottle.

honeste_vivere
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plodot145
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    Why do you think kinetic energy of the charged particle increases? If the magnetic field is static, and there is no electric field, then kinetic energy has to be constant in time. – Ján Lalinský Sep 13 '23 at 01:54
  • Or are you assuming a moving magnetic bottle that accelerates the charged particle? – Ján Lalinský Sep 13 '23 at 01:55
  • Magnetic flux increases as the bottle gets smaller, therefore the field strength increases and the perpendicular component (compared to the field line) of the particle's velocity increases. – plodot145 Sep 13 '23 at 03:04
  • So you're assuming the bottle shrinks, that is, the magnetic field changes in time? – Ján Lalinský Sep 13 '23 at 03:44
  • The total kinetic energy remains constant in the rest frame of the magnetic bottle but the components of the particles speed will change (i.e., as $v_{\parallel}$ decreases, $v_{\perp}$ will increase). See also https://physics.stackexchange.com/a/252885/59023 and https://physics.stackexchange.com/a/670591/59023 – honeste_vivere Sep 13 '23 at 20:19

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I suspect you know that the total energy doesn't change; the kinetic energy due to the velocity perpendicular to the axis of the magnetic bottle, $1/2 m v_\perp^2$ increases at the expense of the remaining kinetic energy term $1/2 m v_{||}^2$. As the particle moves into a region of higher magnetic field, the latter term ideally drops to zero, at which point the particle stops moving along the bottle axis, and starts changing direction. So energy is conserved over all.

That all sounds good from an energy-conservation perspective, but you ask the excellent question about how to make sense of this via Faraday's Law. Indeed, the energy doesn't just swap from $1/2 m v_{||}^2$ to $1/2 m v_\perp^2$ for no reason; there's a physical mechanism. And the reason that $v_\perp^2$ increases in the first place is consistent with an induced emf from Faraday's Law.

The easiest way to visualize this is to imagine going into the reference frame of $v_{||}$, and to draw an imaginary loop which happens to coincide with the Larmor orbit of the charged particle at some moment. In this reference frame, the magnetic flux through your imaginary loop is increasing (as the end of the magnetic bottle approaches). According to Faraday's law, this change in flux generates an emf. In particular, it generates an electric field which increases $v_\perp$. That's the physical mechanism which increases the perpendicular component of the kinetic energy ( at least in that reference frame).

The physical mechanism for the corresponding change in $v_{||}$ is harder to see, but it's a lot easier if you think of the Larmor orbit as a little magnetic dipole. Then the force on a magnetic dipole is proportional to the gradient of the magnetic field, in that geometry.

Ken Wharton
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