User interface language: English | Español

Date May 2016 Marks available 2 Reference code 16M.2.HL.TZ0.6
Level Higher level Paper Paper 2 Time zone Time zone 0
Command term Calculate Question number 6 Adapted from N/A

Question

Two cells of negligible internal resistance are connected in a circuit.

The top cell has electromotive force (emf) 12V. The emf of the lower cell is unknown. The ideal ammeter reads zero current.

Calculate the emf E of the lower cell.

[2]
a.

The diagram shows charge carriers moving with speed v in a metallic conductor of width L. The conductor is exposed to a uniform magnetic field B that is directed into the page.

(i) Show that the potential difference V that is established across the conductor is given by V=vBL.

(ii) On the diagram, label the part of the conductor where negative charge accumulates.

[3]
b.

Markscheme

ALTERNATIVE 1
correct application of Kirchhoff to at least one loop 
E=«4.0×2.0=»8.0V

FOR EXAMPLE
12 = 2.0I1 + 4.0I2 for top loop with loop anticlockwise 
«but I2 = I1 as I3 = 0»
«E =» 8.0 V 

ALTERNATIVE 2
«recognition that situation is simple potential divider arrangement»
pd across 4Ω resistor = \(\frac{{12 \times 4}}{{\left( {2 + 4} \right)}}\)
=8V 

Award [0] for any answer that begins with the treatment as parallel resistors.

a.

(i)
ALTERNATIVE 1
equating electric to magnetic force qE=qvB

substituting \(E = \frac{V}{L}\)

«to get given result»

ALTERNATIVE 2
\(V = \frac{{{\rm{work done}}}}{Q}\) AND work done = force × distance 
work done = qv=Bqv×L 
«to get given result»

(ii)
some mark indicating lower surface of conductor
OR
indication that positive charge accumulates at top of conductor 

Do not allow negative or positive at top and bottom.

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Core » Topic 5: Electricity and magnetism » 5.2 – Heating effect of electric currents
Show 100 related questions

View options