Electronic – Should the magnetic field outside a conductor be considered

A simple model to help you visualize: The magnetic flux forms loops, you cannot have a magnetic field line terminate on nothing in free space. That means that all those flux lines inside the solenoid heading (for example) to the right. The flux line must continue outside the solenoid and loop around to join the other end of the flux loop. In doing so it heads to the left. No external flux lines -> no flux.

So what could be the issue?

1) In an infinite solenoid, there is no external magnetic field. One example of this is a Tokamak or toroidal solenoid. Like a big doughnut. The flux lines form loops within the doughnut and have no need to go outside. - of course a mathematically ideal infinite solenoid could have been what they were discussing, in which case the field lines loop around infinity. Hard to test though.

2) The flux density outside of the linear finite solenoid is much less than inside. So while there may be X flux across the throat of the solenoid that same amount of flux (X) is spread out over a much much larger cross sectional area. The density is much less. [here the area is the cross sectional area across the solenoid.]

3) Perhaps they are talking about a shielded solenoid? This would involve having a high permittivity material external to the coils to trap the flux lines.

There are two approaches of the same phenomena. For an observer located inside the wire, the sources of magnetic field are moving, and therefore, an electric field appears according to Faraday's Law, generating an EMF \$\epsilon\$. On the other hand, an observer located outside the wire (in the same inertial frame of the magnetic field source) will only see the charges (electrons and protons) on the wire moving at the velocity \$\overrightarrow{V}\$. Those charges will experiment a Lorentz force \$\overrightarrow{F}=e\left(\overrightarrow{E}+\overrightarrow{V}\times\overrightarrow{B}\right)\$ and will accumulate at the tips of the wire which creates a potential difference \$\epsilon\$ between the tips, same as the EMF calculated by the other approach.

Therefore, since that potential difference \$\epsilon\$ is proportional to the component of the Lorentz force parallel to the wire, the inclination of it will decrease \$\epsilon\$.

In order to obtain an equation, first consider the wire perpendicular to \$\overrightarrow{V}\$. It can be demonstrated that \$\epsilon = \left|\overrightarrow{V}\right|\left|\overrightarrow{B}\right|L \$ (as you said). Now, consider the wire inclinated an angle \$\theta\$ with respect to \$\overrightarrow{V}\$. The component of the Lorentz force in the direction of the wire will be \$\left|\overrightarrow{F}\right|\sin(\theta)\$, so, the resulting potential difference between tips is \$\epsilon = \left|\overrightarrow{V}\right|\left|\overrightarrow{B}\right|L \sin(\theta)\$