I know that the Lorentz Force law states that:

$\vec{F} = q\vec{E}+q\vec{v} \times \vec{B}$

And then for the magnitude of the force, where $q_2$ is the moving charge:

$E = F/q_2$, and therefore $E = kq_1q_2/r^2 * q_2$, which simplifies to $E=kq_1/r^2$

And continuing to find the magnitude, we can replace $\vec{v}\vec{B}$ with $vBsin(\theta)$

Meaning that the magnitude of the force is:

$F = kq_1/r^2 + q_2vBsin(\theta)$

My question is: Is my thought process here correct? Have I arrived at an equation that will accurately describe the magnitude of the force exerted on a moving charge by a magnetic field? If not, what did I do wrong? I'm still trying to get familiar with all of these equations, so anything would help.

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    $\begingroup$ No, that isn't how you add vectors and your last equation is dimensionally incorrect. $\endgroup$ – Rob Jeffries Sep 2 at 17:54
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    $\begingroup$ @OP -- I am sure you meant $q\vec{v}\times\vec{B}$. $\endgroup$ – Sayan Mandal Sep 2 at 17:55
  • $\begingroup$ E is not necessarily electrostatic in origin. Your last equation is incorrect. Why are you deriving equations? You should read more until you are in full control of the theory. $\endgroup$ – my2cts Sep 2 at 19:28
  • $\begingroup$ You can’t understand electromagnetism until you understand vectors, which you normally learn about in a Newtonian mechanics course. $\endgroup$ – G. Smith Sep 2 at 21:44

No. Firstly the magnetic force is the vector product $q{\bf v} \times {\bf B}$ and not the undefined notion "${\bf v} {\bf B}$". It's true that $|{\bf v} \times {\bf B}| = |{\bf V|}|{\bf B}| \sin \theta$, but the vector $q {\bf v}\times {\bf B}$ is perpendicular to both ${\bf v}$ and ${\bf B}$. You have not given enough information to know its direction compared to that of ${\bf E}$ and you need to know all three directions and use vector addition to get the total force. Its magnitude is only the sum of the magnitudes of $q{\bf v} \times {\bf B}$ and $q{\bf E}$ when the vector ${\bf v} \times {\bf B}$ is pointing in the same direction as ${\bf E}$.


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