ESAT Physics: Electricity and magnetism, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   Before pumping petrol into a large metal storage tank, a technician connects the tank to the ground using a copper earthing strap. Which one of the following correctly explains why this earthing strap is fitted?

   1. A The strap gives the tank a very high resistance to the ground, so that no current at all can flow into or out of the tank while it is being filled.
   2. B The strap keeps the tank positively charged at all times, so that it repels the approaching fuel and prevents the two from making direct contact.
   3. C The strap gives the tank a low-resistance path to the ground, so any charge that builds up on it from friction between the fuel and the pipe can flow away safely instead of accumulating and producing a spark.
   4. D The strap prevents the friction between the fuel and the pipe walls from generating any charge in the first place, so that no charge is ever produced.
2. 21 mark

   A battery, a lamp and a diode are connected in series, forming a single loop. The diode is initially connected so that the lamp lights. A student then removes the diode and reconnects it the opposite way round, changing nothing else. What happens to the lamp, and why?

   1. A The lamp goes out, because the diode is now connected in reverse: it blocks current in this direction, so no current can flow around the circuit.
   2. B The lamp stays lit just as before, because a diode only affects alternating current and has no effect on the direct current supplied by a battery.
   3. C The lamp becomes dimmer but stays lit, because the diode now offers extra resistance in this direction rather than blocking the current completely.
   4. D The lamp is unaffected either way, because reversing the connections of an ideal diode makes no difference to how it behaves in a circuit.
3. 31 mark

   The graph of current against time for a certain electrical supply is a smooth wave that is above the time axis for half of each cycle and below it for the other half, repeating regularly. Which one of the following statements about this supply, and about direct current, is correct?

   1. A This graph shows direct current, because direct current is still allowed to vary in size over time, and crossing the time axis is simply an example of this variation.
   2. B This graph shows alternating current, and a graph of direct current would be a wave of exactly the same shape, only with smaller peaks.
   3. C This graph shows alternating current, but a graph of direct current could look identical, since crossing or not crossing the time axis are equally valid ways of representing either type of current.
   4. D This graph shows alternating current, because the current's direction reverses regularly, crossing above and below the time axis; a graph of direct current would instead stay on one side of the time axis at all times, never reversing sign.
4. 41 mark

   At a junction in a circuit, three wires meet. A current of 7 A flows into the junction along one wire, and a current of 3 A flows into the junction along a second wire. What current flows out of the junction along the third wire?

   1. A 4 A
   2. B 10 A
   3. C 21 A
   4. D 2.33 A
5. 51 mark

   While setting up a circuit, a student accidentally connects the ammeter directly in parallel with the resistor, instead of in series with it, while the voltmeter remains correctly connected in parallel across the resistor. What is the most significant consequence of this mistake?

   1. A The ammeter will read zero, because no current can flow through a component that is connected in parallel with a resistor rather than in series with it.
   2. B The circuit will behave exactly as intended, because an ammeter's resistance is designed to match the resistance of the component it is connected to, so connecting it in parallel makes no difference.
   3. C The voltmeter reading will become unusually large, because it is now effectively sharing a branch with the ammeter, while the ammeter and resistor are otherwise unaffected.
   4. D Because an ammeter has a very low resistance, most of the current takes this easier path through the ammeter instead of through the resistor; this produces a much larger current through the ammeter than intended, and an ammeter reading that does not represent the true current through the resistor.
6. 61 mark

   In a circuit, a 6 ohm resistor and a 3 ohm resistor are connected in parallel with each other. This parallel combination is then connected in series with a 4 ohm resistor and a battery of emf 12 V (the battery's own resistance can be ignored). What current flows from the battery?

   1. A 6 A
   2. B 1.2 A
   3. C 2 A
   4. D 0.5 A
7. 71 mark

   A student plots current (vertical axis) against potential difference (horizontal axis) for two components in turn: a fixed resistor at constant temperature, and an ideal diode connected the correct way round to conduct. Which one of the following correctly compares the two graphs?

   1. A The fixed resistor's graph is a straight line through the origin in both directions; the diode's graph shows zero current while the pd is reversed, and only allows current to flow, rising steeply, once a forward pd is applied.
   2. B Both graphs are straight lines through the origin, since both the resistor and an ideal diode obey Ohm's law equally at every value of potential difference.
   3. C The diode's graph is also a straight line through the origin, like the resistor's, but with a shallower gradient, showing that it simply has a higher resistance in both directions.
   4. D The resistor's graph allows current in one direction only, while the diode's graph is a straight line through the origin in both directions.
8. 81 mark

   An NTC thermistor is connected in series with a fixed resistor and a battery of constant emf. The circuit is initially at room temperature, and the thermistor is then warmed. What happens to the potential difference across the fixed resistor, and why?

   1. A It decreases, because as the thermistor's own resistance falls, its own share of the fixed total potential difference falls too, and since the two components must share this fixed total between them, the fixed resistor's share must fall as well.
   2. B It stays the same, because in a series circuit each component always takes a fixed, unchanging share of the supply's potential difference, regardless of any change in resistance.
   3. C It increases, because the current everywhere in the circuit falls as the thermistor's resistance falls, allowing a larger potential difference to build up across the unchanged fixed resistor.
   4. D It increases, because the thermistor's resistance falls as it warms, reducing the total resistance of the series circuit; since the constant emf now drives a larger current through both components, the potential difference across the unchanged fixed resistor, V=IR, also increases.
9. 91 mark

   A 4 ohm resistor and a 12 ohm resistor are connected in parallel across a 12 V supply. What is (i) the current through the 4 ohm resistor, and (ii) the total current drawn from the supply?

   1. A 1 A through the 4 ohm resistor, and 4 A total from the supply
   2. B 3 A through the 4 ohm resistor, and 4 A total from the supply
   3. C 3 A through the 4 ohm resistor, and 3 A total from the supply
   4. D 2 A through the 4 ohm resistor, and 4 A total from the supply
10. 101 mark

    A charge of 20 coulombs flows through an electrical device in 4 seconds, transferring 100 joules of energy to the device in this time. What power does the device dissipate?

    1. A 25 W
    2. B 5 W
    3. C 80 W
    4. D 100 W
11. 111 mark

    The north pole of a bar magnet is brought close to, but not touching, one end of an unmagnetised soft iron nail. Which one of the following correctly describes what happens to the nail, and why the magnet attracts it?

    1. A The end of the nail nearest the magnet becomes an induced north pole, and the nail is attracted because like poles attract.
    2. B The end of the nail nearest the magnet becomes an induced south pole, and this magnetism is retained permanently because iron is a 'hard' magnetic material.
    3. C The end of the nail nearest the magnet becomes an induced south pole; since unlike poles attract, the nail is pulled towards the magnet. Because iron is a soft magnetic material, the nail loses this induced magnetism again once the magnet is taken away.
    4. D The nail is attracted only because it is a metal, and any metal object would be attracted to a magnet in the same way.
12. 121 mark

    A solenoid has a fixed number of turns and carries a fixed current. Which one of the following changes would NOT increase the strength of the magnetic field produced by the solenoid?

    1. A Increasing the current flowing through the solenoid.
    2. B Placing an iron core inside the solenoid.
    3. C Winding more turns onto the solenoid over the same length.
    4. D Stretching the same solenoid, without changing the number of turns or the current, so its coils are more spread out along a greater length.
13. 131 mark

    A straight wire of length 0.4 m carries a current of 5 A at right angles to a uniform magnetic field of flux density 0.3 T. What is the magnitude of the force on this section of wire?

    1. A 0.06 N
    2. B 0.6 N
    3. C 1.5 N
    4. D 6 N
14. 141 mark

    Two students discuss the motor effect and electromagnetic induction. Student 1 says: 'Fleming's left-hand rule assigns the thumb to the current, the First finger to the field, and the seCond finger to the force.' Student 2 says: 'The size of the voltage induced in a wire moving through a magnetic field can be increased by moving the wire more quickly through the field.' Which student, if either, is correct?

    1. A Student 2 only
    2. B Student 1 only
    3. C Both students
    4. D Neither student
15. 151 mark

    A step-up transformer at a power station has 200 turns on its primary coil and 5000 turns on its secondary coil, and is connected to a 400 V primary supply. Assume the transformer is 100 per cent efficient. Which one of the following correctly gives the secondary voltage, and correctly explains why power is transmitted through the National Grid at such a high voltage?

    1. A 16 V; a step-up transformer reduces the voltage for safer transmission, since Vs = Vp x np/ns = 400 x 200/5000 = 16 V.
    2. B 10000 V (10 kV); transmitting at high voltage reduces power losses because the resistance of the cables themselves falls as the voltage across them rises.
    3. C 10000 V (10 kV); transmitting at this high voltage allows a much smaller current to deliver the same power (since power = voltage x current), and because the power lost as heat in the cables is proportional to the square of the current, a smaller current greatly reduces the power lost during transmission.
    4. D 10000 V (10 kV); transmitting at high voltage increases the current needed to deliver the same power, but this is not a problem because high-voltage cables are made from a special lower-resistance material.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: C

   1. Friction between the fuel and the pipe walls as it flows can transfer charge, building up a static charge on the tank.
   2. If this charge is allowed to build up, a spark could eventually jump between the tank and a nearby earthed object, which is highly dangerous near flammable fuel vapour.
   3. The earthing strap provides a low-resistance path to the ground, so any charge that builds up flows away safely as it forms, rather than accumulating.
   4. Therefore the answer is C.
   • Why not A: This has earthing working by high resistance, but an earthing connection must have a very low resistance to allow charge to flow away easily; a high-resistance path would not remove built-up charge effectively at all.
   • Why not B: This invents an unrelated mechanism. Earthing does not keep an object charged in a particular way; it provides a path for any charge that builds up to flow away, keeping the object close to zero net charge, not deliberately charging it to repel anything.
   • Why not D: This claims earthing stops friction charging from happening at all, but friction charging is a mechanical process caused by the rubbing of the fuel against the pipe; earthing cannot prevent this charge from being generated, it only provides a safe route for the charge to leave once it has been generated.

2. Question 2Answer: A

   1. An ideal diode has a very low resistance to current flowing in one direction (forward biased) and a very high resistance to current flowing the other way (reverse biased).
   2. Originally the diode was connected so current could flow and light the lamp; reversing the diode's connections switches it from forward biased to reverse biased.
   3. With the diode now blocking current in this direction, no current can flow around the single loop, so the lamp goes out.
   4. Therefore the answer is A.
   • Why not B: This claims a diode only affects ac, but a diode's one-way behaviour applies to direct current from a battery just as much as to any other supply; in fact, the classic demonstration of a diode's action uses exactly this kind of simple dc circuit.
   • Why not C: This softens the diode's reverse behaviour into 'extra resistance', but an ideal diode does not just increase its resistance in reverse: it blocks current almost completely, so the lamp should go out, not merely dim.
   • Why not D: This denies that direction matters to a diode at all, but the defining property of a diode is that it conducts in one direction and blocks the other; reversing its connections must change whether current can flow.

3. Question 3Answer: D

   1. Alternating current periodically reverses the direction in which it flows, while direct current flows in one direction only.
   2. On a current-time graph, a reversal of direction shows up as the graph crossing the time axis and becoming negative for part of each cycle, exactly as described here.
   3. A graph of direct current against time never crosses the time axis in this way, since its direction never reverses.
   4. Therefore the answer is D.
   • Why not A: This treats crossing the time axis as just another kind of 'variation in size', but crossing the axis means the current's direction has reversed, which is precisely what distinguishes alternating current from direct current; direct current may vary in size but never reverses direction.
   • Why not B: This assumes a direct current graph must be a wave of reduced height, but a steady direct current gives a straight horizontal line, not a wave shape at all, since its direction and, for a steady supply, its size do not change over time.
   • Why not C: This denies that the two types of current can be told apart from their graphs, but the reversing, sign-changing wave described here is a graph only alternating current can produce; a genuine direct current graph would never cross to the opposite side of the time axis.

4. Question 4Answer: B

   1. At any junction in a circuit, charge is conserved, so the total current flowing into the junction must equal the total current flowing out of it.
   2. Two currents flow into this junction, 7 A and 3 A, so the total current flowing in is 7+3=10 A.
   3. Since only one wire carries current away from the junction, that wire must carry the full current leaving, 10 A.
   4. Therefore the answer is B.
   • Why not A: This subtracts the two currents (7-3=4) instead of adding them; at a junction the total current flowing in must equal the total current flowing out, so the two currents entering the junction must be added together, not subtracted.
   • Why not C: This multiplies the two currents (7x3=21) instead of adding them; conservation of charge at a junction requires the incoming currents to be summed, not multiplied.
   • Why not D: This divides one current by the other (7/3=2.33) instead of adding them; dividing the two currents like this has no physical meaning at a junction, where charge conservation requires them to be added.

5. Question 5Answer: D

   1. An ammeter is built to have a very low resistance so that, connected correctly in series, it does not significantly change the current it is measuring.
   2. Placed in parallel with the resistor instead, this very low resistance offers current an easy alternative path, so most of the current bypasses the resistor and flows through the ammeter.
   3. This produces an unexpectedly large current through the ammeter, and a reading that does not represent the current actually flowing through the resistor.
   4. Therefore the answer is D.
   • Why not A: This claims no current flows through a parallel branch, but a parallel connection gives current an additional path to flow through, not a blocked one; with the ammeter's very low resistance, in fact most of the current will flow this way, not none of it.
   • Why not B: An ammeter is designed to have a very low resistance (ideally close to zero) precisely so that it disturbs a circuit as little as possible when placed correctly in series; its resistance is not matched to the components around it, so placing it in parallel does disturb the circuit significantly.
   • Why not C: The voltmeter remains correctly connected across the resistor and is not directly affected by the ammeter's mistaken placement; if anything, the voltmeter's reading would fall towards zero, since the resistor is now almost short-circuited by the low-resistance ammeter, not rise.

6. Question 6Answer: C

   1. The 6 ohm and 3 ohm resistors in parallel combine to (6x3)/(6+3)=18/9=2 ohm.
   2. Adding the 4 ohm resistor in series gives a total resistance of 2+4=6 ohm.
   3. The current from the battery is I=V/R=12/6=2A.
   4. Therefore the answer is C.
   • Why not A: This uses only the parallel combination's resistance (2 ohm) and ignores the 4 ohm resistor in series entirely, giving 12/2=6A instead of using the full circuit's total resistance.
   • Why not B: This assumes the parallel combination's resistance equals the larger of the two resistors (6 ohm) rather than combining them properly, giving a total of 6+4=10 ohm and 12/10=1.2A.
   • Why not D: This correctly finds the total resistance as 6 ohm, but then inverts the final step, calculating resistance divided by voltage (6/12=0.5) instead of voltage divided by resistance.

7. Question 7Answer: A

   1. A fixed resistor at constant temperature obeys Ohm's law, giving a straight-line current-pd graph through the origin for either direction of pd.
   2. An ideal diode conducts current in only one direction: when reverse biased it allows (ideally) no current at all, and when forward biased it allows current to flow freely once conducting.
   3. So the diode's graph stays at zero for reversed pd and then rises steeply once a forward pd is applied, unlike the resistor's straight line in both directions.
   4. Therefore the answer is A.
   • Why not B: This claims an ideal diode obeys Ohm's law like a resistor does, but a diode is specifically a non-ohmic component: its current does not vary in the same way with potential difference in both directions, since it conducts in only one of them.
   • Why not C: This softens the diode's one-way behaviour into merely 'a higher resistance', but an ideal diode does not conduct at all when reverse biased; it is not simply a higher-resistance version of a resistor in that direction, it is effectively a break in the circuit.
   • Why not D: This swaps the two components round: it is the fixed resistor that conducts equally well in either direction and gives the straight line through the origin, while it is the diode that only conducts in one direction.

8. Question 8Answer: D

   1. An NTC thermistor's resistance falls as its temperature rises, so warming it reduces the total resistance of the series circuit.
   2. With the emf unchanged, a smaller total resistance means a larger current now flows through both components, since they are in series and share the same current.
   3. The potential difference across the fixed resistor is V=IR; since its own resistance R is unchanged but the current I has increased, this potential difference increases.
   4. Therefore the answer is D.
   • Why not A: This assumes the thermistor's own falling share must drag the fixed resistor's share down with it, but the two shares are not tied together like this; because the circuit's total resistance has fallen, the current everywhere increases, and it is this larger current, multiplied by the fixed resistor's unchanged resistance, that raises its share of the potential difference, not lowers it.
   • Why not B: This claims the potential difference splits in fixed, unchanging shares, but in a series circuit the shares depend on the resistances of the components at that moment; changing one resistance, as the thermistor's does here, necessarily changes how the total potential difference is shared between the components.
   • Why not C: This reaches the correct final answer (an increase) but for the wrong reason: current does not fall when the thermistor's resistance falls. For a constant emf, a fall in total resistance means the current everywhere in the series circuit rises, not falls, and it is this larger current that raises the potential difference across the fixed resistor.

9. Question 9Answer: B

   1. Both resistors are connected across the same 12V supply, so each has the full 12V across it.
   2. The current through the 4 ohm resistor is I=V/R=12/4=3A, and the current through the 12 ohm resistor is I=V/R=12/12=1A.
   3. The total current drawn from the supply is the sum of the branch currents, 3+1=4A.
   4. Therefore the answer is B.
   • Why not A: This swaps which resistor carries the larger current: since both resistors share the same 12V potential difference, the smaller resistance (4 ohm) must carry the larger current (I=V/R), so the 4 ohm resistor carries 3A, not 1A.
   • Why not C: This correctly finds the current through the 4 ohm resistor but then forgets to add the current through the 12 ohm resistor (1A) to find the total current drawn from the supply, simply repeating the 4 ohm branch's own current as the total.
   • Why not D: This correctly finds the total resistance and total current, but then wrongly assumes the total current splits equally between the two branches; a parallel circuit shares the same potential difference across each branch, not the same current, so the branches carry different currents unless their resistances are equal.

10. Question 10Answer: A

    1. The current is I=Q/t=20/4=5A.
    2. The potential difference is V=E/Q=100/20=5V.
    3. The power is P=IV=5x5=25W, which agrees with P=E/t=100/4=25W as a check.
    4. Therefore the answer is A.
    • Why not B: This calculates E/Q = 100/20 = 5, but that calculation gives the potential difference V=E/Q, not the power; 5 V has been mislabelled as a power in watts.
    • Why not C: This multiplies the charge by the time (20x4=80), which does not correspond to any formula relating these quantities to power; two unrelated numbers have simply been multiplied together.
    • Why not D: This uses P = current x charge (5x20=100) instead of P = current x potential difference (5x5=25); the charge has been substituted for the voltage in the power formula.

11. Question 11Answer: C

    1. Bringing a magnetic pole close to an unmagnetised piece of iron induces magnetism in it, with the nearer end always taking on the opposite pole to the one approaching it.
    2. Since unlike poles attract, the nail's induced south pole (nearest the magnet's north pole) is attracted towards the magnet.
    3. Iron is a 'soft' magnetic material: it magnetises very easily but also loses its induced magnetism again as soon as the external magnet is taken away.
    4. Therefore the answer is C.
    • Why not A: This gets two things backwards at once: the end of the nail nearest an approaching magnetic pole is always induced with the opposite pole, not the same one, and it is unlike poles, not like poles, that attract each other.
    • Why not B: This correctly identifies the induced pole but wrongly calls iron a 'hard' magnetic material that keeps its magnetism permanently; iron is in fact the classic 'soft' magnetic material, which magnetises easily but loses its magnetism again once the external magnet is removed. Steel is the material that behaves as described here.
    • Why not D: This claims any metal would be attracted in the same way, but magnetic attraction to a permanent magnet is a property of specific magnetic materials such as iron, steel, nickel and cobalt; most metals, including copper and aluminium, are not attracted to a magnet at all.

12. Question 12Answer: D

    1. A solenoid's magnetic field strength increases with the current flowing through it, with the number of turns per unit length, and with the presence of a suitable core.
    2. Increasing the current, adding an iron core, or winding more turns over the same length would each increase the field strength, so none of these is the change described in the question.
    3. Stretching the solenoid without changing the number of turns or the current spreads the same turns over a greater length, reducing the number of turns per unit length; this reduces, rather than increases, the field strength.
    4. Therefore the answer is D.
    • Why not A: Increasing the current is one of the direct factors that increases a solenoid's magnetic field strength, so this change would increase it, making this an incorrect choice for a change that does NOT increase the field.
    • Why not B: Placing an iron core inside the solenoid is a standard way of substantially increasing its magnetic field strength (this is how an electromagnet is strengthened), so this change would increase the field, not leave it the same or reduce it.
    • Why not C: Winding more turns over the same length increases the number of turns per unit length, which is one of the direct factors that increases a solenoid's field strength, so this change would increase it.

13. Question 13Answer: B

    1. The force on a current-carrying wire at right angles to a magnetic field is F=BIL.
    2. Substituting the given values: F=0.3x5x0.4.
    3. 0.3x5=1.5, and 1.5x0.4=0.6, so F=0.6N.
    4. Therefore the answer is B.
    • Why not A: This uses a current of 0.5A instead of the given 5A somewhere in the multiplication (0.3x0.5x0.4=0.06), a factor-of-ten misreading of the current.
    • Why not C: This calculates B x I (0.3x5=1.5) but leaves out the length of the wire entirely; F=BIL requires all three quantities multiplied together, not just two of them.
    • Why not D: This uses a flux density of 3T instead of the given 0.3T, a factor-of-ten misplacement of the decimal point.

14. Question 14Answer: A

    1. Fleming's left-hand rule uses the First finger for the magnetic Field, the seCond finger for the Current, and the thuMb for the resulting Motion (the force); Student 1 has assigned current to the thumb rather than to the second finger, so Student 1 is incorrect.
    2. Increasing the speed at which a wire cuts through magnetic field lines is one of the factors that increases the size of the induced voltage, so Student 2's statement is correct.
    3. Only Student 2 is correct.
    4. Therefore the answer is A.
    • Why not B: This accepts Student 1's rule, but Fleming's left-hand rule assigns the First finger to the magnetic Field, the seCond finger to the Current, and the thuMb to the resulting Motion (force); Student 1 has swapped the roles of the thumb and second finger, wrongly giving the thumb to current.
    • Why not C: This wrongly accepts Student 1 alongside Student 2; Student 1's assignment of the rule's fingers and thumb is incorrect, as explained above.
    • Why not D: This wrongly rejects Student 2's statement, which correctly identifies increasing speed as one of the factors that increases the size of an induced voltage.

15. Question 15Answer: C

    1. For an ideal transformer, Vp/Vs = np/ns, so Vs = Vp x ns/np = 400 x 5000/200 = 10000 V.
    2. For a fixed power transmitted, P=VI means that using a higher voltage allows the same power to be delivered by a smaller current.
    3. The power lost as heat in the transmission cables is given by P_loss = I^2 R, so a smaller current greatly reduces this loss, since it depends on the square of the current.
    4. Therefore the answer is C.
    • Why not A: This inverts the turns-ratio relationship, using np/ns instead of ns/np, and also misdescribes a step-up transformer as reducing voltage; a step-up transformer, having more turns on its secondary coil than its primary, always increases the voltage from primary to secondary.
    • Why not B: This reaches the correct secondary voltage but gives the wrong mechanism for reduced losses: the resistance of the transmission cables does not change with the voltage used; it is the current that falls as the voltage rises for a fixed power, and it is this smaller current, through P=I^2R, that reduces the power lost as heat.
    • Why not D: This reaches the correct secondary voltage but wrongly claims transmitting at high voltage requires a larger current for the same power. Since power = voltage x current, increasing the voltage for a fixed power actually decreases the current required, and there is no need for cables of a special lower-resistance material for this reason.