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My textbook lays out the following explanation of Force experienced by a current-carrying wire:

My question is where did the negative sign of the charge on electron go? How did Drift velocity go from vector to scalar and how did dl enter the vector form? I know that this formula can be derived using current density or taking q through wire to be positive but I would like to know how the author came to such conclusions?

derivation of force on a current-carrying conductor

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Where did the negative sign go? It's simply a matter of notation: your textbook starts talking about electron because indeed they are the particles that move in a wire; but as you probably already know the convention is to think about currents of positive charges and not negative ones. So when the book starts talking about currents it readily flips the sign, to be consistent with the convention. I can well understand why you have been confused by it. Hope it's clear now. (Another way to see why we get the disappearance of the minus sign is indeed to think about the direction of the vector $d\vec{l}$, that is conventionally in the direction of the current of positive charges)

Why $v$ stops being a vector and $dl$ starts? You see: if you think about electron moving in a wire their velocity in a specific position must be in the direction of the infinitesimal element of the wire in that position; if that wasn't the case then charges would escape the wire! So if we think about the two vectors $\vec{v},d\vec{l}$ they must be aligned. But we want only one vector, not two! The other quantity serves the purpose of specifying the module of the vector in question. This means that we can swap the vector status between $v$ and $dl$ as we please.

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  • $\begingroup$ But isn't dl vector taken in the direction of current that is opposite of Vd? That's what I was taught it class, or is that also a matter of convention? So you mean to say that Vd and dl vectors were switched because we needed Vd to formulate i which made the formula simpler? $\endgroup$ Commented Jul 10, 2020 at 13:29
  • $\begingroup$ The answer to your first question is yes, and this indeed explains the disappearance of the minus sign. The answer to the second question is also yes. And my answer tries to explain why we can do this switch. $\endgroup$
    – Noumeno
    Commented Jul 10, 2020 at 14:09
  • $\begingroup$ okay, so if i am getting this right-dl is in the direction of the current, but as the negative sign got canceled, even Vd was in the direction of current. Now as Vd and dl are in the same direction, we switched their unit vectors? Thank you so much! $\endgroup$ Commented Jul 10, 2020 at 15:24

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