The question asks for the instantaneous speed of a charged particle moving in the magnetic field generated by an almost point-like magnetic dipole at the point of closest approach. The magnetic dipole is located at the origin, and its vector potential is given by \({\rm{A}}\, = \,\frac{{{\rm{\mu }}\,{\rm{sin}}\,{\rm{θ }}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}\widehat ϕ \) in spherical polar coordinates (r, θ, ϕ).

The particle has mass m and charge q and moves in the equatorial plane of the dipole. The equatorial plane corresponds to \({\rm{\theta }} = \frac{{\rm{\pi }}}{{\rm{2}}}\). The particle starts at time t = 0 with an initial speed \(\nu_0\) and an impact parameter b.

In the equatorial plane (\({\rm{\theta }} = \frac{{\rm{\pi }}}{{\rm{2}}}\)), the vector potential becomes:

\[{\rm{A}}\, = \,\frac{{{\rm{\mu }}\,{\rm{sin}}\,({\rm{\pi }}/2)}}{{4{\rm{\pi }}{{\rm{r}}^2}}}\widehat ϕ \, = \,\frac{{{\rm{\mu }}}}{{4{\rm{\pi }}{{\rm{r}}^2}}}\widehat ϕ \]

The force experienced by a charged particle moving in a magnetic field \({\rm{B}}\) is given by the Lorentz force formula:

\[{\rm{F}}\, = \,{\rm{q}}({\rm{E}}\, + \,{\rm{v}} \times {\rm{B}})\]

In this problem, there is no electric field, so \({\rm{E}}\, = \,0\). The force is purely magnetic:

\[{\rm{F}}\, = \,{\rm{q}}({\rm{v}} \times {\rm{B}})\]

A fundamental property of the magnetic force \({\rm{F}}_{\rm{B}} = {\rm{q}}({\rm{v}} \times {\rm{B}})\) is that it is always perpendicular to the velocity vector \({\rm{v}}\). This can be seen from the definition of the cross product.

The work done by a force \({\rm{F}}\) on a particle moving with velocity \({\rm{v}}\) over a small displacement \({\rm{d}}l = {\rm{v}}\,{\rm{dt}}\) is given by \({\rm{dW}} = {\rm{F}} \cdot {\rm{d}}l = {\rm{F}} \cdot ({\rm{v}}\,{\rm{dt}})\). For the magnetic force, the work done is:

\[{\rm{dW}}\, = \,{\rm{F}}_{\rm{B}} \cdot {\rm{v}}\,{\rm{dt}}\, = \,{\rm{q}}({\rm{v}} \times {\rm{B}}) \cdot {\rm{v}}\,{\rm{dt}}\]

Since \({\rm{v}} \times {\rm{B}}\) is perpendicular to \({\rm{v}}\), their dot product is zero: \({\rm{(v}} \times {\rm{B}}) \cdot {\rm{v}} = 0\).

Therefore, the work done by the magnetic force is always zero:

\[{\rm{dW}}\, = \,0\]

The work-energy theorem states that the total work done on a particle is equal to the change in its kinetic energy. Since the magnetic force is the only force doing work, and the work done is zero, the change in kinetic energy is zero.

\[\Delta {\rm{KE}}\, = \,0\]

This means the kinetic energy of the particle remains constant throughout its motion in the magnetic field. Kinetic energy is given by \({\rm{KE}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}\). Since m is constant and KE is constant, the speed \({\rm{v}}\) of the particle must also remain constant.

The initial speed of the particle is given as \({\rm{\nu }}_0\). Since the speed remains constant throughout the motion, the speed at any point in the path, including the point of closest approach, will be the same as the initial speed.

Thus, the instantaneous speed at the point of closest approach is \({\rm{\nu }}_0\).

Magnetic Force and Particle Speed

• The force on a charged particle moving in a magnetic field is \({\rm{F}}_{\rm{B}} = {\rm{q}}({\rm{v}} \times {\rm{B}})\).
• This force is always perpendicular to the velocity \({\rm{v}}\).
• Work done by a force is \({\rm{W}} = \int {\rm{F}} \cdot {\rm{d}}l\).
• Since \({\rm{F}}_{\rm{B}}\) is perpendicular to \({\rm{v}}\) (and thus \({\rm{d}}l\)), the dot product \({\rm{F}}_{\rm{B}} \cdot {\rm{v}} = 0\), and the work done by the magnetic force is zero.
• By the work-energy theorem, the change in kinetic energy is zero if only magnetic forces are acting.
• Kinetic energy \({\rm{KE}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}\) is constant.
• Therefore, the speed \({\rm{v}}\) must be constant.

Constant Speed Motion

The particle starts with speed \({\rm{\nu }}_0\). As the speed remains constant throughout the motion due to the nature of the magnetic force, the speed at the point of closest approach will be equal to the initial speed.

Initial speed = \({\rm{\nu }}_0\)

Speed at any point = Initial speed

Speed at the point of closest approach = \({\rm{\nu }}_0\)

This conclusion relies on the fact that only the magnetic force acts on the particle, and magnetic forces do no work, hence conserving kinetic energy and speed.

The final answer is \({\rm{\nu }}_0\).