Stand at the end of a platform of Namma Metro in Bengaluru and look down beside the track. A third rail runs there on insulators, kept at 750 V. Each coach touches it with a metal shoe that slides along its top. While a train speeds away from a station, the current through its shoes runs to thousands of amperes. The largest current in your house, through a geyser, is about 10 A.
Charge on the move
Everything so far in this unit has been about charges at rest. Now let charge flow steadily along a wire. Draw an imaginary section across the wire and count the charge that crosses it. The charge that crosses per second is the electric current through that section.
If a net charge \(\Delta q\) crosses a section of a conductor in a time \(\Delta t\), the average current through it is \(I = \Delta q/\Delta t\). The current at an instant is \(i = dq/dt\). The SI unit is the ampere: \(1\,\text{A} = 1\,\text{C/s}\).
When the current does not change with time it is a steady current, and then \(q = It\). The ampere is one of the base units of SI. So the coulomb is defined from it: one coulomb is the charge carried by one ampere in one second.
Fig 1.1 counts a steady \(3.2\,\text{A}\). In \(10\,\text{s}\) the charge through S is \(q = It = 32\,\text{C}\). Each electron carries \(e = 1.6 \times 10^{-19}\,\text{C}\), so each second \(3.2/(1.6 \times 10^{-19})\) \(= 2 \times 10^{19}\) electrons cross S. That is twenty billion billion electrons every second, for an ordinary current.
(a) A steady current of \(3.2\,\text{A}\) flows in a wire. Find the charge that crosses a section in \(10\,\text{s}\). Find also the number of electrons that cross it each second. (b) A charge of \(600\,\text{C}\) passes through a lamp in \(5\) minutes. Find the current, and the time \(1\,\text{C}\) takes to pass.
(a) \(q = It = 3.2 \times 10 = 32\,\text{C}\). Each second \(3.2\,\text{C}\) crosses, which is \(3.2/(1.6 \times 10^{-19})\) \(= 2 \times 10^{19}\) electrons.
(b) \(I = q/t = 600/300 = 2\,\text{A}\). At \(2\,\text{A}\), one coulomb passes in \(1/2 = 0.5\,\text{s}\).
Values used in this lesson: \(e = 1.6 \times 10^{-19}\,\text{C}\).
Which way does the current point?
Charge comes in two signs, so a direction needs a rule. The direction of the current is the direction in which positive charge moves. This is the conventional current. Negative charge moving one way carries the same current as positive charge moving the other way. So electrons that move from A to B make a current from B to A.
In a metal only the free electrons move, so the current always points against their motion. In an electrolyte, such as salt water, positive ions move one way and negative ions the other. Both carry current in the same direction, so their currents add. The rule was fixed long before anyone knew that the carriers in a metal are negative, and nothing measurable depends on it.
In a wire joining two points, \(10^9\) electrons pass from A to B in \(1\,\text{ms}\). What is the current?
Current is a scalar
A current has a direction along its wire, yet it is not a vector. At a junction, the current that leaves is the plain sum of the currents that arrive. The angles between the wires play no part. Charge that arrives each second must leave each second, because it cannot pile up at the junction.
Two wires carrying \(3\,\text{A}\) and \(4\,\text{A}\) meet at a right angle and join a third wire. What current does the third wire carry?
Charge need not flow through a wire to make a current. A long rod carrying charge \(\lambda\) per metre, pushed through a ring at speed \(v\), carries \(\lambda v\) coulombs through the ring each second. So the current is \(I = \lambda v\). A rod with \(2\,\mu\text{C/m}\) moving at \(5\,\text{m/s}\) makes a current of \(10\,\mu\text{A}\).
Now return to the Metro. Take \(2700\,\text{A}\) through the shoes, a fair estimate for a train speeding up; Lesson 5 shows where it comes from. That is \(2700\,\text{C}\) of charge every second. That is \(2700/(1.6 \times 10^{-19})\) \(\approx 1.7 \times 10^{22}\) electrons each second through a few shoes of metal.
A current that changes
Read \(i = dq/dt\) the other way round. In a short time \(dt\) the charge \(dq = i\,dt\) crosses. Add these small pieces from \(t_1\) to \(t_2\) and you get \(q = \int i\,dt\). On a graph of \(i\) against \(t\) this is the area under the graph, just as the area under a v-t graph is the displacement.
Set the first chip, \(i = 2t + 3\), and end the interval at \(2\,\text{s}\). The area is \(\int_0^2 (2t + 3)\,dt\) \(= [t^2 + 3t]_0^2 = 10\,\text{C}\). The average current is \(10/2 = 5\,\text{A}\). Because \(i\) grows in a straight line, the average equals the current half-way through, at \(t = 1\,\text{s}\).
The current in a wire rises steadily from zero to \(4\,\text{A}\) in the first \(2\,\text{s}\), then stays at \(4\,\text{A}\). Find the charge that passes from \(t = 0\) to \(t = 5\,\text{s}\), and the average current.
The i-t graph is a triangle from 0 to 2 s, then a rectangle from 2 s to 5 s, both of height \(4\,\text{A}\).
Charge = area of the triangle + area of the rectangle.
Triangle: \(\tfrac12 \times 2 \times 4 = 4\,\text{C}\). Rectangle: \(3 \times 4 = 12\,\text{C}\). Total \(16\,\text{C}\).
The charge is \(16\,\text{C}\), and the average current is \(16/5 = 3.2\,\text{A}\).
The charge that has passed a section of a wire is \(q = 4t^2 + 2t\), with \(q\) in coulombs and \(t\) in seconds. Find the current at \(t = 2\,\text{s}\), in amperes.
A current of \(i = 2t + 3\) amperes flows from \(t = 0\) to \(t = 4\,\text{s}\). How much charge passes, in coulombs?
How crowded is the flow?
A current is a number for a whole section. A thick copper bar and a hair-thin wire can carry the same current, but the flow is crowded in the thin wire. The current density \(J\) describes this. It is the current per unit area of a section drawn at right angles to the flow.
The current density at a point is the current per unit area through a small area at right angles to the flow there, \(J = dI/dA\). It is a vector that points along the conventional current. For a current spread evenly over a section of area \(A\), \(J = I/A\).
A wire of section \(1\,\text{mm}^2\) carrying \(1.7\,\text{A}\) has \(J = 1.7/10^{-6}\) \(= 1.7 \times 10^6\,\text{A/m}^2\). A wire of the same metal with section \(2\,\text{mm}^2\), at the same current density, carries \(1.7 \times 10^6 \times 2 \times 10^{-6}\) \(= 3.4\,\text{A}\). Twice the area at the same crowding carries twice the current.
If a section is tilted, fewer charges cross it. Only the area \(A\cos\theta\) faces the flow, where \(\theta\) is the angle between \(\vec J\) and the normal to the section. So \(I = JA\cos\theta = \vec J \cdot \vec A\), the same dot product as the flux of the chapter on Electric Flux and Gauss's Law.
For a section of any shape, add up \(\vec J \cdot d\vec A\) over it: \(I = \int \vec J \cdot d\vec A\). The current through a surface is the flux of the current density through it, just as the electric flux is the flux of \(\vec E\).
A current of \(1.6 \times 10^{-7}\,\text{A}\) is spread evenly over a section of area \(10\,\text{mm}^2\). Find the current density, in A/m².
A power bank marked 5000 mAh states a charge, not an energy or a current. One milliampere-hour is \(10^{-3}\,\text{A} \times 3600\,\text{s} = 3.6\,\text{C}\). So 5000 mAh is \(18\,000\,\text{C}\). Delivered at \(2\,\text{A}\), that charge lasts \(18\,000/2 = 9000\,\text{s}\), two and a half hours.
- Giving the current the direction in which the electrons move.Conventional current follows positive charge; in a metal it points against the electrons.
- Adding currents at a junction like vectors.Current is a scalar. The currents into a junction add as numbers.
- Using \(q = It\) for a current that changes.Use \(q = \int i\,dt\), the area under the i-t graph.
- Leaving the area in mm² in \(J = I/A\).\(1\,\text{mm}^2 = 10^{-6}\,\text{m}^2\); convert before dividing.