Understand conventional current vs electron flow direction: Exercises

Figure — Understand conventional current vs electron flow direction

Definition One tool we lean on: current density

$J = q n v$ Several problems below ask "which way does the current point?" To answer cleanly we need a small tool. Current density $J$ is a little arrow that shows how much charge flows per second across a patch, and in which direction. Build it from scratch:

Charges of charge $q$ , packed $n$ per cubic metre, drifting with velocity $v$ (a speed and a direction).
The amount of charge sweeping across a patch each second is $q \times n \times v$ — so $J = q n v$ .
The direction is set by the sign of $q$ . If $q$ is positive, $J$ points the same way as $v$ . If $q$ is negative (electrons!), $J$ points opposite to $v$ .

Conventional current is just $J$ 's direction. So "current points along $q v$ " is not a magic rule — it is the definition of current density, and the minus sign in $q = - e$ is exactly why electrons and conventional current disagree. (Full treatment in Current and the Ampere and Drift Velocity.)

Level 1 — Recognition

Recall Solution 1.1

WHAT we ask: the direction of conventional current. WHY: conventional current always leaves the + terminal in the external circuit. Look at the magenta arrow in the master figure: it leaves + and travels along the wire to −. Here + is on the left, so current exits the left terminal, travels through the lamp to the right, and returns to −. Answer: left-to-right through the lamp (+ → lamp → −).

Recall Solution 1.2

Electrons follow the violet arrow in the master figure: they leave the − terminal (right side) and move right-to-left through the lamp toward +. Answer: right-to-left through the lamp (− → lamp → +). It is the opposite of conventional current, exactly as expected.

Recall Solution 1.3

False. In a metal the moving charges are electrons, which are negative. We only draw current as if positive charge moved (Franklin's convention). See Electric Charge and the Coulomb for where the sign of $q$ comes from.

Level 2 — Application

Recall Solution 2.1

Step 1 — total charge. WHY $Δ Q = N e$ : each electron carries the same fixed charge $e$ . If $N$ of them cross the point, the total charge is simply "how many" times "charge each" — you add up $N$ identical parcels, which is a multiplication. $Δ Q = N e = (3.12 \times 1 0^{19}) (1.602 \times 1 0^{- 19} C) \approx 5.0 C .$ Step 2 — current. WHY $I = \frac{Δ Q}{Δ t}$ : current means charge-per-second — a rate. To get "how much per second" you divide the total charge by the time it took. This is the definition, not a trick (see Current and the Ampere). $I = \frac{5.0 C}{1 s} = 5.0 A .$ Answer: about $5.0 A$ .

Recall Solution 2.2

Step 1 — find the charge. WHY $Δ Q = I Δ t$ : if $I$ coulombs flow every second, then in $Δ t$ seconds you get $I \times Δ t$ coulombs total — rate times time is amount, the same way speed times time is distance. This is just $I = Δ Q /Δ t$ rearranged. $Δ Q = I Δ t = (2) (3) = 6 C .$ Step 2 — count electrons. WHY $N = Δ Q / e$ : we know the total charge and the charge of one electron; dividing "total" by "each" tells you how many fit — the inverse of the multiplication in 2.1. $N = \frac{Δ Q}{e} = \frac{6}{1.602 \times 1 0 ^{- 19}} \approx 3.75 \times 1 0^{19} electrons .$ Answer: $6 C$ , about $3.75 \times 1 0^{19}$ electrons.

Recall Solution 2.3

Electrons always move opposite to conventional current (the two arrows in the master figure), so they move west. Direction has nothing to do with the magnitude $2 A$ — only the sign of the carrier flips the arrow.

Level 3 — Analysis

Recall Solution 3.1

Recall from the tool box above: the current direction is the direction of $J = q n v$ , and its sign is set by $q$ .

Electron: $q = - e$ (negative), moving right ( $+ v$ ): $q v = - e (+ v)$ points left (the minus flips the direction).
Proton: $q = + e$ (positive), moving left ( $- v$ ): $q v = + e (- v)$ points left (positive charge keeps its direction). Both currents point left — the same way. A negative charge going right is a positive current going left, so it matches the proton going left.

Recall Solution 3.2

Magnitude is unchanged: still $1.0 A$ — flipping the battery swaps which terminal is +, not how many charges move per second. Direction: the + terminal is now on the other side, so both arrows in the master figure spin around — conventional current and electron flow reverse together. They remain antiparallel (opposite). The relationship never breaks — see Example 3 in the parent note.

Recall Solution 3.3

In the salt water (b). There the carriers are positive ions ( $q > 0$ ), so in $J = q n v$ the current points the same way as the carriers — the same way as the conventional-current arrow. In copper (a) the carriers are negative electrons ( $q < 0$ ), so they move opposite to conventional current. Both give a valid $2.0 A$ — the convention was designed so the math is identical regardless of carrier sign. (Positive carriers become physically important in Semiconductors and Holes.)

Level 4 — Synthesis

Recall Solution 4.1

WHY this formula: it is the current-density tool $J = q n v$ multiplied by the wire's area $A$ . $n A v$ counts how many electrons cross the wire's face each second (density × area × speed = electrons/second), and $\times e$ turns that count into coulombs/second = amperes. This is Drift Velocity feeding into current. (a) $I = n e A v = (8.5 \times 1 0^{28}) (1.602 \times 1 0^{- 19}) (1.0 \times 1 0^{- 6}) (1.0 \times 1 0^{- 4}) .$ $I \approx 1.36 A .$ (b) Electrons drift right, so conventional current points left (opposite the electrons — the minus sign in $q = - e$ from the tool box).

Recall Solution 4.2

From $I = n e A v$ , with $I$ , $e$ , $A$ fixed, $v \propto \frac{1}{n}$ . Halving $n$ doubles $v$ . Fewer carriers must each move twice as fast to deliver the same charge-per-second. Answer: $v$ goes up, by a factor of 2.

Level 5 — Mastery

Recall Solution 5.1

(a) Convert time: $2.0 min = 120 s$ (WHY convert: the formula's time must be in seconds to match amperes = coulombs per second). Then $Δ Q = I Δ t = (0.50) (120) = 60 C$ . (b) $N = \frac{Δ Q}{e} = \frac{60}{1.602 \times 1 0 ^{- 19}} \approx 3.75 \times 1 0^{20}$ electrons. (c) Conventional current points north, so electrons flow south. Reason: current points along $J = q n v$ . For electrons $q = - e < 0$ , so $J$ points opposite to the electrons' velocity $v$ . For $J$ (and hence the conventional current) to point north, $v$ must point south.

Recall Solution 5.2

Verify: TRUE.

Conventional current is defined as the direction positive charge moves. In the tool $J = q n v$ with $q > 0$ , the current points the same way as the carriers, so the carrier arrow and the conventional arrow coincide — no mismatch. This is exactly the hole picture in Semiconductors and Holes.
$I = Δ Q /Δ t$ never cared about the carrier's sign; it just counts coulombs per second (used in Ohm's Law). So the formula still holds. The parent note's key insight — "a positive charge right = a negative charge left" — is precisely why both pictures give the same $I$ .

Recall Solution 5.3

Same current, $0.30 A$ , through Y. In a single series loop the same charge-per-second passes every point — charge is not created or lost along the way (conservation of charge). Electron flow: electrons travel the whole loop from − back to +, so they pass through both bulbs in the same order (opposite to conventional current in each). Through both bulbs the electrons move − → +, and conventional current moves + → −.

Recall Self-test recap (fill in the blank)

Try to answer each before reading the part after the dash.

Conventional current flows from ? to ? in the external circuit — from + to −.
Electron flow goes from ? to ? — from − to +.
Current from charge: $I = \frac{?}{?}$ — $Δ Q /Δ t$ .
Reversing the battery keeps the magnitude but does what to the two directions? — flips both together, still opposite.
In a series loop, the current at every point is — the same.

Connections

Current and the Ampere — the definition $I = Δ Q /Δ t$ used throughout.
Electric Charge and the Coulomb — where $e$ and the sign of $q$ come from.
Drift Velocity — feeds the $I = n e A v$ synthesis problems.
Voltage and EMF — why current has the same magnitude when a battery is reversed.
Ohm's Law — uses this conventional $I$ in every equation.
Semiconductors and Holes — the L5 case where positive carriers are real.