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I am reading chapter "Moving Charges and Magnetism" in NCERT book. In the part talking about the force on a current-carrying wire in a uniform magnetic field, I read this strange transformation of vectors. I have highlighted the relevant parts.

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Can someone explain how and why did that happen?

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  • $\begingroup$ Does this answer your question? How can length be a vector? $\endgroup$
    – ZeroTheHero
    Commented Oct 2, 2021 at 17:38
  • $\begingroup$ @ZeroTheHero isn't my question more about the transfer of vectors ? $\endgroup$
    – Ankit
    Commented Oct 2, 2021 at 17:46
  • $\begingroup$ There is no such thing as “transfer of vectors” and that phraseology is misleading. It so happens that $I d\vec \ell$ is defined to have the vector part over the line element rather than the current element, whereas for other geometries the vector part is on the current element, as per physics.stackexchange.com/a/325898/36194 $\endgroup$
    – ZeroTheHero
    Commented Oct 2, 2021 at 17:50
  • $\begingroup$ A very poorly worded presentation. $\endgroup$
    – garyp
    Commented Oct 2, 2021 at 19:48

2 Answers 2

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$nq$ is the density of moving charges $\rho$. So, $$\mathbf j = \rho \frac{\mathbf{dx}}{dt}$$ If the density of current is constant over a straight wire: $$\mathbf j = \rho \frac{\mathbf{\Delta x}}{\Delta t}$$
Naming I as $$I = \frac{\Delta Q}{\Delta t} = \frac{\rho lA}{\Delta t}$$

and $\mathbf l = \mathbf{\Delta x}$, we get the final expression.

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The (j) is a unit vector indicating a direction (in an xyz coordinate system). The (l) is a length which has a magnitude. Assuming (l) is in the direction of (j), when you put them together you get a vector with a magnitude and direction.

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  • $\begingroup$ how is j a unit vector ? Isn't it the current density vector ? $\endgroup$
    – Ankit
    Commented Oct 3, 2021 at 4:22
  • $\begingroup$ My mistake, but (j) is parallel to (l), so they do share the same direction. $\endgroup$
    – R.W. Bird
    Commented Oct 3, 2021 at 15:47

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