DP Physics (last assessment 2024)

Question 22N.2.SL.TZ0.5

Date	November 2022	Marks available	[Maximum mark: 8]	Reference code	22N.2.SL.TZ0.5
Level	SL	Paper	2	Time zone	TZ0
Command term	Describe, Determine, Explain, Show that, State	Question number	5	Adapted from	N/A

[Maximum mark: 8]

22N.2.SL.TZ0.5

Resistor R is connected in a circuit with a cell that has internal resistance.

The ammeter and the voltmeter are ideal.

(a)

State what is meant by an ideal voltmeter.

[1]

Markscheme

infinite resistance

no current is flowing through it ✓

Examiners report

A majority of candidates scored a mark by simply stating infinite resistance. Several answers went the other way round, stating a resistance of zero.

The cell has an emf of 1.49 V. The resistance of R is 50.0 Ω. The voltmeter reads 1.47 V.

(b.i)

Show that the internal resistance of the cell is about 0.7 Ω.

[2]

Markscheme

current $= \frac{1.47}{50.0} = 2.94 \times 10^{- 2}$ «A» ✓

r $= \frac{1.49}{2.94 \times 10^{- 2}} - 50.0$ OR $0.68$ «Ω» ✓

For MP2, allow any other correctly substituted expression for r.

Examiners report

Many answers here produced a number that did not round to 0.7 but students claimed it did. The simultaneous equation approach was seen in the best candidates, getting the right answer. It is worthy of reminding about the need of showing one more decimal place when calculating a show that value type of question.

(b.ii)

Determine the total power dissipated in the circuit.

[2]

Markscheme

$1.49 \times 2.94 \times 10^{- 2}$

$\frac{1 . 49^{2}}{50.0 + 0.68}$

${(2.94 \times 10^{- 2})}^{2} (50.0 + 0.68)$ ✓

$0.0438$ «W» ✓

Accept use of 0.7 Ω in MP1.

Examiners report

The calculation of total power dissipated was not usually well done, as it often failed to include the internal resistance therefore calculating the power dissipated only in the resistor.

One of the connecting wires is placed in a magnetic field. The direction of the current in the wire is shown.

(c.i)

Explain, by reference to charge carriers in the wire, how the magnetic force on the wire arises.

[2]

Markscheme

charge/carriers are moving in a magnetic field ✓

there is a magnetic force on them / quote F = qvB

this creates a magnetic field that interacts with the external magnetic field ✓

Accept electrons.

For MP2, the force must be identified as acting on charge / carriers.

Examiners report

Many scored MP1 here but did not get MP2 as they jumped straight to the wire rather than continuing with the explanation of what was going on with the charge carriers.

(c.ii)

Every current-carrying wire produces a magnetic field.

Describe one piece of evidence that supports this statement.

[1]

Markscheme

magnetic needle is deflected by nearby currents

two «parallel» current-carrying wires exert a force on each other

magnetic field due to a current can be measured directly with a probe ✓

Only accept argument that refers to an observation or experiment.

Examiners report

Students usually failed to identify appropriate evidence.