| Date | May 2016 | Marks available | 1 | Reference code | 16M.2.SL.TZ0.5 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | State | Question number | 5 | Adapted from | N/A |
Question
In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.
State what is meant by an ideal voltmeter.
The student adjusts the variable resistor and takes readings from the ammeter and voltmeter. The graph shows the variation of the voltmeter reading V with the ammeter reading I.
Use the graph to determine
(i) the electromotive force (emf) of the cell.
(ii) the internal resistance of the cell.
A connecting wire in the circuit has a radius of 1.2mm and the current in it is 3.5A. The number of electrons per unit volume of the wire is 2.4×1028m−3. Show that the drift speed of the electrons in the wire is 2.0×10−4ms−1.
The diagram shows a cross-sectional view of the connecting wire in (c).
The wire which carries a current of 3.5A into the page, is placed in a region of uniform magnetic field of flux density 0.25T. The field is directed at right angles to the wire.
Determine the magnitude and direction of the magnetic force on one of the charge carriers in the wire.
Markscheme
infinite resistance OR draws no current from circuit/component OR has no effect on the circuit
Do not allow “very high resistance”.
(i)
«vertical intercept = emf» = 8.8 − 9.2 V
(ii)
attempt to evaluate gradient of graph
=0.80Ω
Accept other methods leading to correct answer, eg using individual data points from graph.
Allow a range of 0.78 – 0.82 Ω.
If ε = I(R + r) is used then the origin of the value for R must be clear.
3.5=2.4×1028×π(1.2×10−3)2×1.6×10−19×v«⇒v=2.0×10−4ms−1»
F = «qvB =1.6×10–19 ×2.0×10–4 ×0.25 =»8.1×10–24 N
directed down OR south
