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.

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.

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.

(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.