Electric Current and Current Density

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a copper wire that is thick at one end and thin at the other, like a funnel, and connect it to a cell. Through which part does more current flow, the thick part or the thin part? Most students say the thick part, because it is wider, right? Hold on to that answer; we will settle it today. In JEE, this idea sits under all of current electricity, so let's begin.

0:30Electric current: the rate of flow of charge

So, what is electric current? Imagine a surface cutting across a wire. The current I is the rate at which charge crosses that surface. On screen, I equals d q over d t, where d q is the tiny charge that crosses in a tiny time d t. Okay? The unit is the ampere, written with a capital A on screen, okay? The capital C stands for coulomb, and s to the power minus one means per second. So one ampere is one coulomb of charge crossing every second. Right? Now let me ask you this. In a copper wire, the electrons are moving towards the left. Which way does the current flow? It flows to the right, yes, exactly! Here is the rule on screen. By convention, the direction of current is the direction in which positive charge flows. Clear? And electrons carry negative charge, so the current is opposite to the motion of electrons. Don't mix these two up in the exam. Got it?

1:33From current back to charge

Now, let us turn it around. If I equals d q over d t, then the charge that passes in a time interval is the integral of I, d t. The long S shaped sign is the integral sign. So tell me, what does an integral give you on a graph? The area, yes, very good! So the charge passed is the area under the I versus t graph, that is, current plotted against time. With me? Next, a charge q going round a circle, f times every second. Each time it goes round, the whole charge q passes any fixed point on the circle. So how much charge passes that point every second? Exactly, q times f! So the current I equals q times f. And omega, the Greek letter you see, is the angular speed. Since f equals omega over two pi, the current also equals q omega over two pi. Okay?

2:31Current density

Now, current density. J with an arrow on top is the current density, and the arrow tells you it is a vector. It is the current per unit area, for an area held normal, that is perpendicular, to the flow. Right? To get the current through any surface, we add up J dot d A over the whole surface. Here d A with an arrow is a tiny piece of the area, taken as a vector along its normal, and the dot is the dot product. So the current I equals the integral of J vector dot d A vector. Clear? And when the flow is uniform, this integral becomes a simple product: I equals J times A, nice and easy. But only for a uniform flow. Remember this for example two. Okay? Now a quick question. Current has a direction, we just saw that. So, is current a vector? No, no, it is not, and this is a classic trap! Current is a scalar. At a junction, currents simply add algebraically, with plus and minus signs, and there is no vector addition at all. Got it? But current density, J with the arrow, is a vector. So remember the pair: current is a scalar, and current density is a vector. With me?

3:44Steady state: back to our puzzle

Now, back to our funnel shaped wire, thick at one end and thin at the other. Where is the current larger, in the thick part or in the thin part? Here is the key idea. In a steady state, charge does not pile up anywhere in the wire. Whatever charge enters a section every second must also leave it every second. Okay? So the current is the same at every cross-section of a single wire, whatever its shape. The thick part and the thin part carry the same current! If you said the thick part, that was the trap. Right? What does change is the current density. In the thin part the area is smaller, so J, which equals I over A for a uniform flow, is larger. The same current is squeezed through a smaller area. Clear?

4:33Example 1: 5 C in 2 s, find current and electrons per second

Alright, example one, an easy one. Five coulombs of charge cross a section of a wire in two seconds. Find the current, and the number of electrons crossing per second. Pause the video and try it first. Done? Okay, let's check. Current is charge over time, so I equals q over t, that is five over two, which gives two point five amperes. Okay? Now the electrons. Each electron carries a charge e, of one point six times ten to the power minus nineteen coulomb. So the number crossing per second is the current over e, that is, two point five over one point six times ten to the power minus nineteen. With me? Two point five over one point six gives about one point five six. So the answer is one point five six times ten to the power nineteen, and the s to the power minus one means per second. Clear?

5:27Example 2: current in a wire when J = kr (non-uniform)

Now a harder one, JEE level. The current density in a wire of radius capital R varies with distance from the axis as J equals k times r. Here small r is the distance from the axis, and k is a constant. Find the current. Pause the video and try it first. Alright, many students do this. At the surface, J is k times capital R, and the area is pi times capital R squared. So the current is J times A, which is pi k capital R cubed, right? No, no, careful, that is exactly the trap! I equals J times A only for a uniform flow, and here J changes with r. So I is not equal to J times A, and the slashed equals sign on screen means not equal. Okay? So, cut the cross-section into thin rings. Take a ring of radius small r and width d r. Its area, d A, is its circumference, two pi r, times its width, d r. On this thin ring, J is practically constant, so the ring carries a current of J times d A, simple. With me? Now add up all the rings, from the axis out to the surface. The current I is the integral from zero to capital R of k r times two pi r, d r. Take two pi k outside, and we are left with the integral of r squared, d r. Right? The integral of r squared is r cubed over three. Put in the limits, and I equals two pi k capital R cubed, over three. So whenever J changes across the wire, add up thin rings, and never just multiply J by the area. Got it?

7:10Quick revision: I = dq/dt, area under I-t, J and steady I

So, quick revision. Current is d q over d t, in ampere, in the direction of flow of positive charge. Charge passed is the area under the I versus t graph, and current is the integral of J dot d A, which becomes J times A only for a uniform flow. Current is a scalar, and in a steady state it is the same at every cross-section of a wire. Now go and try the practice problems on this page. You can do them. All the best!