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We have a wire with cross-sectional area $A$, length $L$ and current $I$.

If the wire is in a magnetic field $\vec B$, the magnetic force on each charge is $\vec F =q\vec v_d \times \vec B$.

$\vec v_d$ is the drift velocity or average velocity

The number of charges in the wire segment is the number $n$ per unit volume, multiplied by the volume $AL$.

Thus the total force in the wire segments is $$(1)\: \:\vec F =(q\vec v_d \times \vec B)nAL$$

We know that the current in the wire is $(2)$ $I=nqv_d A$

Combining (2) and (1) the force can be written as $(3)$ $\vec F=I\vec L \times \vec B$

This is a paragraph from my textbook.

I know that the cross product has the property $(rA \times B)=(A \times rB) = r(A \times B)$

What I don't understand is why $\vec v_d$, that is a vector in (1) becomes a scalar in (3) as a part of $I$. And why, on the other hand, $L$ that is an scalar in (1) becomes a vector in (3). What property allows you to do that?

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  • $\begingroup$ (3) should probably be something like $\vec{F}=I\hat{\ell}\times\vec{B}$, where $\hat{\ell}$ is a unit vector in the direction of the axis of the wire. The reason being that $I$ is a scalar, but you want to have a vector pointing in the direction the current is flowing. $\endgroup$ – The Photon Aug 22 at 22:37

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