Question 23M.2.HL.TZ1.7
| Date | May 2023 | Marks available | [Maximum mark: 12] | Reference code | 23M.2.HL.TZ1.7 |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Determine, Explain | Question number | 7 | Adapted from | N/A |
A vertical rectangular loop of conducting wire is dropped in a region of horizontal magnetic field. The diagram shows the loop as it leaves the region of the magnetic field.
Explain, by reference to Faraday’s law of electromagnetic induction, why there is an electromotive force (emf) induced in the loop as it leaves the region of magnetic field.
[2]
The induced emf is equal/proportional/related to the «rate of» change of «magnetic» flux/flux linkage ✓
Flux is changing because the area pierced/enclosed by magnetic field lines changes «decreases»
OR
Flux is changing because the loop is leaving/moving out of the «magnetic» field. ✓
Need to see a connection between the EMF and change in flux for MP1.
Need to see a connection between the area changing or leaving the field and the change in flux for MP2
Just before the loop is about to completely exit the region of magnetic field, the loop moves with constant terminal speed v.
The following data is available:
| Mass of loop | m = 4.0 g |
| Resistance of loop | R = 25 mΩ |
| Width of loop | L = 15 cm |
| Magnetic flux density | B = 0.80 T |
Determine, in m s−1 the terminal speed v.
[4]
mg = BIL OR I = 0.33 «A» ✓
BvL = IR OR ℰ = 8.25 × 10−3 «V» OR ℰ = 0.12v ✓
Combining results to get v = ✓
v = « =» 0.068 «m s−1»
Allow ECF between steps if clear work is shown.
Three capacitors C1 = 3.0 μF, C2 = 2.0 μF and C3 = 4.0 μF are connected to a cell of emf 12 V and negligible internal resistance. The capacitors are initially uncharged.
Calculate
the total capacitance of the circuit.
[2]
The 2 in parallel give a total of 6.0 «μF»✓
The total is «» = 2.0 «μF»✓
Allow ECF from MP1
Accept other powers of 10 for capacitances with proper unit included.
the total energy stored in the three capacitors.
[1]
E = «CV2 = × 2.0 × 10−6 × 122» 1.44 × 10−4 «J» ✓
Allow ECF from c(i) (=72 × c(i))
the charge on C2.
[3]
ALTERNATE 1
Voltage across C2 is half that across C1 ✓
So voltage across C2 is 4.0 V ✓
Charge is «C2V2 = 2.0 × 10−6 × 4.0» 8.0 × 10−6 «C» ✓
ALTERNATE 2
Charge on C1 is «CTVT = 2.0 × 10−6 × 12» 24 «µC» ✓
So voltage across C1 is «» 8.0 «V» ✓
Charge on C2 is «C2V2 = 2.0 × 10−6 × 4.0» 8.0 × 10−6 «C» ✓
ALTERNATE 3
«C3 = 2C2 leading to » q3 = 2q2 ✓
Total charge in parallel = «q2 + q3 = q2 + 2q2 =» 3q2 ✓
3q2 = 24 leading to q2 = 8 × 10−6 «C» ✓
ECF for MP3 allowed in ALT 1 and ALT 2