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What I know is current density is the amount of current passing per unit area perpendicular to the current. But then why do we define it as the rate of change of current with respect to change in area $\left(\dfrac{di}{dS}\right)$.

Aren’t they both two different things?

I’m a high school student. It’s given in the book Concepts Of Physics (volume-2) by H.C. Verma, on page 172.

Proof

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    $\begingroup$ It's not. Can you link to where it is defined that way? $\endgroup$
    – user1247
    Commented Aug 5, 2020 at 16:37
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    Commented Aug 5, 2020 at 16:50

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The current density is not defined as $\mathrm di/\mathrm dS$. The precise definition of current density is the following

$$\mathbf j=\left(\lim _{\Delta S\to 0} \frac{i}{\Delta S}\right) \mathbf{\hat n}$$

where $\mathbf{\hat n}$ is the unit vector in the direction perpendicular to the infinitesimal area ($\Delta S$), and

$$i=\frac{\mathrm d q}{\mathrm dt}$$

where $q$ is the charge passing through/crossing the area $\Delta S$. Note that for any real physical scenario, as $\Delta S$ goes to $0$, so does the current ($i$) passing through it. Thus, when $\Delta S$ is infinitesimal, so is the current, but writing it as $\mathrm di$ is abuse of notation, because $\mathrm d(\rm quantity)$ is primarily used when we are taking an infinitesimal change in the "$\rm quantity$" (not the case here), or taking an infinitesimal element of that "$\rm quantity$" (also, not exactly the case here).

In this light, the definition given in your book is, strictly speaking, incorrect. However, it is highly likely that the author meant to convey the above definition, but abused the notation, thus rendering the definition incorrect.

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  • $\begingroup$ I got to know new thing from your answer that d(quantity) can also be used to represent infinitesimal element of that quantity. But my question is in such cases can we use it in calculus since in calculus d(quantity) has to represent some infinitesimal change. Then only we can use it in calculus terms. Please let me know it fully in case I’m less learned. $\endgroup$
    – abcxyz
    Commented Aug 5, 2020 at 17:11
  • $\begingroup$ Yes. In fact, many of the times, we do use differentials to denote an infinitesimal element of a certain given quantity (implying it's not to be varied, thus changing the quantity doesn't make sense). $\endgroup$
    – user258881
    Commented Aug 5, 2020 at 17:35
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    $\begingroup$ @abcxyz Sorry, I don’t have time to write an answer. I do not consider differentials to necessarily be infinitesimal changes. I consider them simply to be infinitesimal quantities. Sometimes that infinitesimal quantity is an infinitesimal change in something, and sometimes it isn’t. $\endgroup$
    – G. Smith
    Commented Aug 5, 2020 at 19:39
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    $\begingroup$ @abcxyz In your book, $di$ is just an infinitesimal amount of current flowing through an infinitesimal area $dS$. The ratio of these infinitesimal quantities, $di/dS$, is finite and is called the current density. You don’t have to think about anything “changing” here. Every professor I had, and every physicist I ever worked with, was comfortable with this way of thinking, regardless of whether it passes muster with mathematicians. $\endgroup$
    – G. Smith
    Commented Aug 5, 2020 at 19:45
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    $\begingroup$ It is better to say that $\text{j}$ is a current flux. It is expressed via the charge density $\rho$ and the local average charge velocity $\text{v}$: $\text{j}=\rho\cdot \text{v}$. $\endgroup$ Commented Aug 6, 2020 at 18:30

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