CAIE A-Level Physics 9.1 Electric Current

Practise relating current to charge flow, Q = It, charge quantisation and I = Anvq, checking carriers, units and conductor data.

Exam points

interpret electric current as flow of charge carriers and use Q = It to calculate charge, current, time or carrier count
apply I = Anvq to analyse carrier density, drift speed, cross-sectional area and charge in a current-carrying conductor
use quantisation of charge to judge possible charge values and relate charge to elementary carriers

Covered syllabus points

9.1.1—An electric current is a flow of charge carriers 9.1.2—The charge on charge carriers is quantised 9.1.3—Q = It 9.1.4—Use, for a current-carrying conductor, the expression I = Anvq

[Maximum number: 6]

(a)

State what is meant by an electric current.

[ 1 ]

(b)

A metal wire has length L and cross-sectional area A, as shown in Fig. 6.1.

Fig. 6.1

I is the current in the wire,
n is the number of free electrons per unit volume in the wire,
v is the average drift speed of a free electron and
e is the charge on an electron.

[ 3 ]

(i)

State, in terms of A, e, L and n, an expression for the total charge of the free electrons in the wire.

[ 1 ]

(ii)

Use your answer in (i) to show that the current I is given by the equation

I = nAve.

[ 2 ]

( t is time taken for electrons to move length L )
I=Q / t

B1

I=n A L e / t

or
I=n A L e /(L / v)

or
$I=n A \text { vte } / t \text { and } I=n A \text { ve }$

B1

(c)

A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross

sectional area of the wire is reduced over a length of 3.0 mm , as shown in Fig. 6.2.

Fig. 6.2

The wire has diameter d at cross-section X and diameter 0.69 d at cross-section Y . The current in the wire is 0.50 A .

[ 2 ]

(i)

Determine the ratio
$\frac{\text { average drift speed of free electrons at cross-section } \mathrm{Y}}{\text { average drift speed of free electrons at cross-section } \mathrm{X}}$ .
ratio =

[ 2 ]

ratio = area at X/area at Y
$=\left[\pi d^{2} / 4\right] /\left[\pi(0.69 d)^{2} / 4\right] \text { or } d^{2} /(0.69 d)^{2} \text { or } 1 / 0.69^{2}$

C1

= 2.1

A1

[Maximum number: 1]

A spherical oil drop has a radius of $1.2 \times 10^{-6} \mathrm{~m}$ . The density of the oil is $940 \mathrm{~kg} \mathrm{~m}^{-3}$ .

The oil drop is charged. Explain why it is impossible for the magnitude of the charge to be $8.0 \times 10^{-20} \mathrm{C}$ .

minimum charge (on drop) is $1.6 \times 10^{-19} \mathrm{C}$

B1

[Maximum number: 2]

A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 6.1.

Fig. 6.1

The current in the circuit is I and the potential difference across the variable resistor is V.

The battery stores 9.2 kJ of energy. The variable resistor is adjusted so that $V=2.1 \mathrm{~V}$ . Use Fig. 6.2 to:

calculate the number of conduction electrons moving through the battery in a time of 1.0 s

number =

$\begin{aligned} \text { number } =0.50 / 1.60 \times 10^{-19} =3.1 \times 10^{18} \end{aligned}$

A1

All question bank results loaded

Covered syllabus points

9.1 Electric current question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (i)

9.1 Electric current question 2

9.1 Electric current question 3