2023 A/L Physics Past Paper · Question 48 (English Medium)

Answer: (4) Graph (4)

What the question asks

We want: the graph of $v_d$ against $x$.

On the horizontal axis is $x$, the distance from the left end. On the vertical axis is $v_d$, the drift velocity of the electrons at that place.

What each letter means:

$I$ ($\text{A}$): the current. It is the same at each cross-section of the wire.

$r$ ($\text{m}$): the radius of the wire at a place. It gets smaller from the left end to the neck in the middle, then larger again.

$A$ ($\text{m}^2$): the area of the cross-section, $A = \pi r^2$.

$n$, $e$: the number of electrons in unit volume and the charge of an electron. They are the same along the wire.

Relevant theory

The velocity of the electrons of an electric current flowing through a conductor is called the "drift velocity" of the electrons.

per second, current $I = \frac{Q}{t} = \frac{vAne}{1} = vAne$

The first quote names the velocity of the electrons in a current: the drift velocity.

The second quote gives the current through a cross-section of area $A$: $I = vAne$, where $n$ is the number of electrons in unit volume and $e$ is the charge of an electron.

Turn it round: $v = \dfrac{I}{Ane}$. With the same $I$, $n$ and $e$, the drift velocity is large where the area is small.

For example, think of a crowd leaving a stadium. The same number of people pass each point each minute. Where the corridor is wide they walk slowly. At a narrow gate they must move fast to keep the same flow.

Required equations

$I = v_d A n e$

$A = \pi r^2$

$v_d = \dfrac{I}{\pi r^2 n e}$

Reasoning and calculation

At the left end, $x = 0$:

The wire is widest, so the area is largest. The drift velocity is small. It is not zero, because the current $I$ flows there too.

Split the wire at the neck.

From the left end to the neck, the radius gets smaller, so $\dfrac{1}{r^2}$ grows, and $v_d$ rises.

From the neck to the right end, the radius gets larger, so $v_d$ falls.

The wire looks the same on the two sides of the neck, so the two parts of the graph mirror each other.

At the neck:

The area is smallest, so $v_d$ is largest. The neck is narrow but not cut, so its area is not zero. The peak is high but finite: the graph comes to a closed point.

The shape of each part:

The radius changes in a straight line, but $v_d$ goes as $\dfrac{1}{r^2}$. Halving the radius makes $v_d$ four times larger.

So the rise is gentle near the ends and steep near the neck. Each part is a curve that bends upward, not a straight line.

$4$

$\dfrac{v_{neck}}{v_{end}} = \left(\dfrac{r_{end}}{r_{neck}}\right)^2 = 2^2$

$(R / r)^2$

Reading the options

Now match it with the five graphs.

Graph (4) starts above zero, curves up to a closed peak at the neck, and falls the same way. It matches.

Graph (1) is made of straight lines. Graph (2) has an open peak that goes up without end. Graph (3) starts and ends at zero. Graph (5) is flat. None of these match.

Why the other options fail

Graph (1) is straight, but $v_d$ goes as $\dfrac{1}{r^2}$, so it curves up steeply near the neck.

Graph (2) goes up without end, but the neck has a non-zero area, so the peak is finite.

Graph (3) is zero at the ends, but the current flows there too, so $v_d$ is small but not zero.

Graph (5) is flat, but the area changes along the wire while $I$ stays the same.

Conclusion

$v_d = \dfrac{I}{\pi r^2 n e}$

The drift velocity $v_d = \dfrac{I}{\pi r^2 n e}$ is small but not zero at the wide ends, curves up to a finite peak at the narrow neck, and falls again. That is graph (4).

The answer is (4).

The quotes give the drift velocity and $I = vAne$. That the radius changes in a straight line, and that the neck has a small but non-zero area, are our reading of the figure.

Teacher's note

In a wire carrying a steady current, $I$ is the same at each cross-section. So $v_d$ goes as $\dfrac{1}{A}$, or $\dfrac{1}{r^2}$. Check the ends (small, not zero) and the narrowest point (largest, finite).

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