Question 1
1
A light spring is suspended from a fixed point. A bar magnet is attached to the end of the spring, as shown in Fig. 1.1. In order to shield the magnet from draughts, a cardboard cup is placed around the magnet but does not touch it. The magnet is displaced vertically and then released. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 1.2. The mass of the magnet is 130 g .
structured3 marks
Question 1(b)
1(b)
The cardboard cup is now replaced with a cup made of aluminium foil. During 10 complete oscillations of the magnet, the amplitude of vibration is seen to decrease to 0.75 cm from that shown in Fig. 1.2. The change in angular frequency is negligible.
structured3 marks
Question 1(b)(i)
1(b)(i)
Use Faraday's law of electromagnetic induction to explain why the amplitude of the oscillations decreases.
Mediumstructured3 marks
Answer
as magnet moves, flux is cut by cup/aluminium giving rise to induced e.m.f. (in cup) B1 induced e.m.f. gives rise to currents and heating of the cup B1 thermal energy derived from oscillations of magnet so amplitude decreases B1 or induced e.m.f. gives rise to currents which generate a magnetic field (B1) the magnetic field opposes the motion of the magnet so amplitude decreases (B1)
Question 3
3
A bar magnet is suspended from the free end of a helical spring, as illustrated in Fig. 3.1. One pole of the magnet is situated in a coil of wire. The coil is connected in series with a switch and a resistor. The switch is open. The magnet is displaced vertically and then released. As the magnet passes through its rest position, a timer is started. The variation with time t of the vertical displacement y of the magnet from its rest position is shown in Fig. 3.2. At time \(t=4.0 \mathrm{~s}\), the switch is closed.
structured6 marks
Question 3(b)
3(b)
6 marks
Question 3(b)(i)
3(b)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. is proportional to rate of change/cutting of (magnetic) flux (linkage) M1 A1 [2]
Question 3(b)(ii)
3(b)(ii)
Explain why, after time \(t=4.0 \mathrm{~s}\), the amplitude of vibration of the magnet is seen to decrease.
Mediumstructured4 marks
Answer
a current is induced in the coil as magnet moves in coil current in resistor gives rise to a heating effect thermal energy is derived from energy of oscillation of the magnet M1 A1 M1 A1 [4]
Question 5
5
A bar magnet is suspended vertically from the free end of a helical spring, as shown in Fig. 5.1. One pole of the magnet is situated in a coil. The coil is connected in series with a high-resistance voltmeter. The magnet is displaced vertically and then released. The variation with time t of the reading V of the voltmeter is shown in Fig. 5.2.
structured3 marks
Question 5(a)
5(a)
3 marks
Question 5(a)(i)
5(a)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2]
Question 5(a)(ii)
5(a)(ii)
Use Faraday's law to explain why 1. there is a reading on the voltmeter, 2. this reading varies in magnitude, 3. the reading has both positive and negative values.
Mediumstructured1 marks
Answer
1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1]
Question 5
5
A Hall probe is placed a distance d from a long straight current-carrying wire, as illustrated in Fig.5.1. The direct current in the wire is 4.0 A . Line XY is normal to the wire. The Hall probe is rotated about the line X Y to the position where the reading \(V_{H}\) of the Hall probe is maximum.
structured2 marks
Question 5(c)
5(c)
A student suggests that the Hall probe in (a) is replaced with a small coil connected in series with a millivoltmeter. The constant current in the wire is 4.0 A . In order to obtain data to plot a graph showing the variation with distance x of the magnetic flux density, the student suggests that readings of the millivoltmeter are taken when the coil is held in position at different values of x. Comment on this suggestion.
Hardstructured2 marks
Answer
e.m.f. induced in coil when magnetic field/flux is changing/cutting either at each position, magnetic field does not vary so no e.m.f. is induced in the coil/no reading on the millivoltmeter or at each position, switch off current and take millivoltmeter reading or at each position, rapidly remove coil from field and take meter reading
Question 5
5
5 marks
Question 5(a)
5(a)
State the relation between magnetic flux density B and magnetic flux \(\Phi\), explaining any other symbols you use.
Easystructured0 marks
Answer
\[ \phi=B A \] (a) \[ \text { either } \phi=B A \sin \theta \] where A is the area (through which flux passes) \(\theta\) is the angle between B and (plane of) A or where A is area normal to B M1 A1 (M1) (A1) [2]
Question 5(c)
5(c)
5 marks
Question 5(c)(i)
5(c)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to
Question 5(c)(ii)
5(c)(ii)
The Hall probe in (b) is replaced by a small flat coil of wire. The coil is moved at constant speed along the line XY . The plane of the coil is parallel to the faces of the poles of the magnet. On the axes of Fig. 5.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil.
Mediumstructured3 marks
Answer
short pulse on entering and on leaving region between poles pulses approximately the same shape but opposite polarities e.m.f. zero between poles and outside
Question 5
5
6 marks
Question 5(b)
5(b)
A long solenoid has an area of cross-section of \(28 \mathrm{~cm}^{2}\), as shown in Fig. 5.1. A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of the solenoid. The magnetic flux density B at the centre of the solenoid is given by the expression where I is the current in the solenoid, n is a constant equal to \(1.5 \times 10^{3} \mathrm{~m}^{-1}\) and \(\mu_{0}\) is the permeability of free space. Calculate, for a current of 3.5 A in the solenoid,
structured2 marks
Question 5(b)(ii)
5(b)(ii)
the flux linkage in the coil C . flux linkage = Wb
Mediumstructured2 marks
Answer
flux linkage \(=6.6 \times 10^{-3} \times 28 \times 10^{-4} \times 160\)
Question 5(c)
5(c)
4 marks
Question 5(c)(i)
5(c)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage)
Question 5(c)(ii)
5(c)(ii)
The current in the solenoid in (b) is reversed in direction in a time of 0.80 s . Calculate the average e.m.f. induced in coil C. e.m.f. = V
Mediumstructured2 marks
Answer
e.m.f. \(=\left(2 \times 3.0 \times 10^{-3}\right) / 0.80\)
Question 5
5
A uniform magnetic field of flux density B makes an angle \(\theta\) with a flat plane PQRS , as shown in Fig. 5.1. The plane PQRS has area A.
structured7 marks
Question 5(a)
5(a)
State
structured1 marks
Question 5(a)(ii)
5(a)(ii)
an expression, in terms of A, B and \(\theta\), for the magnetic flux \(\Phi\) through the plane PQRS.
Mediumstructured1 marks
Answer
\((\Phi=) B A \sin \theta\)
Question 5(b)
5(b)
A vertical aluminium window frame DEFG has width 52 cm and length 95 cm , as shown in Fig. 5.2. The frame is hinged along the vertical edge DG. The horizontal component \(B_{\mathrm{H}}\) of the Earth's magnetic field is \(1.8 \times 10^{-5} \mathrm{~T}\). For the closed window, the frame is normal to the horizontal component \(B_{\mathrm{H}}\). The window is opened so that the plane of the window rotates through \(90^{\circ}\).
structured3 marks
Question 5(b)(i)
5(b)(i)
Explain why, when the window is opened, the change in magnetic flux linkage due to the vertical component of the Earth's magnetic field is zero.
Mediumstructured1 marks
Answer
plane of frame is always parallel to \(B_{\vee}\) / flux linkage always zero
Question 5(b)(ii)
5(b)(ii)
Calculate, for the window opening through an angle of \(90^{\circ}\), the change in magnetic flux linkage. change in flux linkage = Wb
Mediumstructured2 marks
Answer
\(\Delta \Phi=1.8 \times 10^{-5} \times 52 \times 10^{-2} \times 95 \times 10^{-2}\)
Question 5(c)
5(c)
3 marks
Question 5(c)(i)
5(c)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage) (allow rate of cutting of flux)
Question 5(c)(ii)
5(c)(ii)
The window in (b) is opened in a time of 0.30 s . Use your answer in (b)(ii) to calculate the average e.m.f. induced in the window frame. e.m.f. =
Mediumstructured0 marks
Answer
e.m.f. \(=\left(8.9 \times 10^{-6}\right) / 0.30\)
Question 5(c)(iii)
5(c)(iii)
State the sides of the window frame between which the e.m.f. is induced. between side and side
Mediumstructured1 marks
Answer
This question part was removed from the assessment. All candidates were awarded 1 mark.
Question 7
7
A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1. The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off.
structured5 marks
Question 7(b)
7(b)
The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid.
structured5 marks
Question 7(b)(i)
7(b)(i)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1
Question 7(b)(ii)
7(b)(ii)
On the axes of Fig. 7.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil when the current in the solenoid is switched on and then switched off.
Mediumstructured3 marks
Answer
pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1
Question 5
5
6 marks
Question 5(b)
5(b)
The Hall probe in (a) is now replaced with a small coil of wire connected to a sensitive voltmeter. The coil is arranged so that its plane is normal to the magnetic field of the wire.
structured6 marks
Question 5(b)(i)
5(b)(i)
State Faraday's law of electromagnetic induction and hence explain why the voltmeter indicates a zero reading.
Mediumstructured3 marks
Answer
e.m.f. induced is proportional / equal to rate of change of (magnetic) flux (linkage) constant field in coil / flux (linkage) of coil does not change
Question 5(b)(ii)
5(b)(ii)
State three different ways in which an e.m.f. may be induced in the coil. 1. 2.
Mediumstructured3 marks
Answer
e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil move coil towards / away from wire (1 mark each, max 3)
Question 6
6
2 marks
Question 6(a)
6(a)
State Faraday's law of electromagnetic induction.
Easystructured2 marks
Answer
(induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1