# Magnetic force acting on a current carrying wire

I am a high school student and I would like to know why in a magnetic field the Force, $F$, is equal to $BIL\sin(\theta)$, where $\theta$ represents the angle between the magnetic field and the current. I understand that $F$ is proportional to $B$, $I$ and $L$, but I do not understand the inclusion of $\sin (\theta)$. In other words, I would like to see a simple proof or explanation of why the effective length of the wire is $L \sin(\theta)$

I wouldn't think of it in terms of effective length as you have stated. I think you have to go back to a more fundamental equation, and that is the equation of force on a moving charged particle in a magnetic field which is given by $$F=|q(\mathbf{v}\times\mathbf{B})| = qvB\sin(\theta)$$

The magnetic force only acts on the component of the velocity that is perpendicular to the field. In this case, the moving charges in the wire make up the current, and the force will only act on the component of the current that is perpendicular to the field as well. The $\sin(\theta)$ term comes from the cross product in the above equation, and your force on the wire should be more strictly written as $$\mathbf{F}=I\mathbf{L}\times\mathbf{B}$$ where $\mathbf{L}$ points in the direction of the current.

EDIT: After thinking about it a bit more, effective length might be a fine way to think about, but in the sense that it is the effective length that is perpendicular to the field (in fact, this is probably exactly how it should be thought of).

• Yes, that is a better to think about it. Now my question would be "Why the magnetic force acts only on the component of the velocity that is perpendicular to the field?" – daljit97 Feb 23 '16 at 23:19
• @daljit97 This has been asked before in this question: physics.stackexchange.com/questions/29133/… The top voted answer is a bit math-heavy, but there are other explanations as well in the other answers that may be helpful. – tmwilson26 Feb 24 '16 at 13:17