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As the title suggests. I was wondering if the Lorentz Force Law equation (or the law in general)for current-carrying wires ($\vec F= I\vec l\times\vec B$) works backward. In other words, if there is a force and magnetic field perpendicular to a wire and each other, do they cause a current to appear?

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To answer your question, imagine a freely moving charged particle with velocity $\vec v$, charge $q$ and mass $m$. As a consequence of this velocity, the Lorentz force law says that in the presence of a magnetic field, $\vec B$, the particle must experience a force: $$ \vec F_\text{B} = q \vec v \times \vec B. $$ Now, if the velocity is initially zero, $\vec v_\text{initial} = 0$, then this force must be zero as well. If at time $t_0$, we start applying a constant external force to this particle, then by Newton's law, $\dot v = F_{\text{ext}} / m$, and thus: $$ \vec v(t) = \frac{1}{m} \int_{t_0}^{t} dt \vec F_{\text{ext}} = \frac{\vec F_{\text{ext}}}{m} (t - t_0), $$ for $t > t_0$. As a consequence of this external force, the particle now experiences a time--dependent Lorentz force: $$ \vec F_\text{B}(t) = (t - t_0)\frac{q}{m} \vec F_\text{ext} \times \vec B. $$ If $\vec F_\text{ext}$ and $\vec B$ are perpendicular, as you asked for, then the resulting force on the charge carrying particle, $\vec F_\text{B}(t)$, will be perpendicular to the two others.

We now put the charge inside a wire and apply a constant force to the wire. Assuming the charged particle stays inside the wire while it is pushed along, then there must be a force on it, which is exactly such as to keep its velocity to match the wire's. Applying a magnetic field perpendicular to the externally applied force and the length of the wire, the charged particle will experience a force in the direction of the wire length, by the arguments above.

From this point of view, you will observe a current.

A follow--up question would then be to ask: but isn't this self--contradictory? What if your frame of reference is following the wire along its acceleration, then from this point of view, the velocity of the charged particle is zero, and there should be no force. The answer is that in such a frame, things get weird and electricity and magnetism start to mix up.

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    $\begingroup$ Great answer, very informative! It helped out a lot and made clear a lot as well. Thanks! $\endgroup$
    – GamrCorps
    Mar 28, 2017 at 6:04
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    $\begingroup$ Very focused and elegant answer, a pleasure to read $\endgroup$ Mar 28, 2017 at 7:54

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