How to write Magnetic field in terms of Force, charge and velocity:

Found this relationship in an old notebook:

$$\vec{B} = \frac{\vec{F}}{q\vec{v}}~~~[\text{Tesla}]$$

wasn't sure if its technically true, since:

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

If I wanted to rearrange the cross product to put $\vec{B}$ on one side of equation, how would you do that?

  • $\begingroup$ The division of two vectors has not been defined, as far as I know.But it could I guess. $\endgroup$
    – user65081
    Jan 27, 2020 at 0:49

2 Answers 2


The magnitude of the cross product can be written as:

$|\overrightarrow{v} \times \overrightarrow{B}| = vB \sin{\theta}$

However, once you do this, you only get the magnitude of the vector. You can't really "undo" a cross product. You can look at this math.se thread for more information.

  • $\begingroup$ thanks...so really its: $\vec{B} = \frac{|\vec{F}|}{q|\vec{v}|\sin \theta} \hat{B}$......at a point in the field.. $\endgroup$
    – pico
    Jan 27, 2020 at 0:59
  • 1
    $\begingroup$ Yes, that correct. Notice that if when $\theta$ goes to $\pi /2$, the $\sin {\theta}$ goes to $1$. That is probably why in your notes you have that equation: you've assumed that the force vector is orthogonal to the velocity vector. $\endgroup$ Jan 27, 2020 at 1:03
  • $\begingroup$ @MrMineHeads They've assumed that the magnetic field vector is orthogonal to the velocity vector. The force vector is always orthogonal to the other two. $\endgroup$
    – J. Murray
    Jan 27, 2020 at 1:45
  • $\begingroup$ @J.Murray you are correct, but for some reason I cannot edit my comment to fix that. $\endgroup$ Jan 27, 2020 at 1:58
  • $\begingroup$ @MrMineHeads No worries. Comments can be edited for 5 mins after they are posted, at which point they become permanent (unless they are deleted). It's a good policy for keeping comment threads honest, but a bit annoying when it comes to typos. $\endgroup$
    – J. Murray
    Jan 27, 2020 at 2:00

I guess that the first equation you wrote isn't correct as we can't divide two vectors as well as even if the first equation is written in terms of magnitude then it represents the special case where the charged particle performs uniform circular motion.

More precisely we can write, $$q|\overrightarrow{v} \times \overrightarrow{B}| = qvB \sin{\theta}$$ $\theta$ is the angle between velocity and magnetic field.


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