# Does the Lorentz (Laplace) Force Law work backwards?

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?

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.