Wednesday, May 20, 2015

Physics 9702 Doubts | Help Page 147

  • Physics 9702 Doubts | Help Page 147

Question 725: [Electromagnetism]
(a) Define the tesla.

(b) A charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B as shown in Fig.1.

Magnetic field is normal to the direction of motion of the particle. The path of the particle in the field is the arc of a circle of radius r.
(i) Explain why path of the particle in the field is the arc of a circle.
(ii) Show that radius r is given by the expression

(c) A uniform magnetic field is produced in the region PQRS, as shown in Fig.2.

Magnetic field is normal to the page.
At point X, a gamma-ray photon interaction causes two particles to be formed. The paths of these particles are shown in Fig.2.
(i) Suggest, with a reason, why each of the paths is a spiral, rather than the arc of a circle.
(ii) State and explain what can be deduced from paths about
1. charges on the two particles,
2. initial speeds of the two particles.

Reference: Past Exam Paper – November 2011 Paper 43 Q6

Solution 725:
(a) The tesla is the unit of magnetic flux density for a field normal to (straight) conductor carrying current of 1 A when the force per unit length (on the conductor) is 1 N m–1.

(i) The force on the particle is always normal to the direction of motion (and the speed of the particle is constant). The magnetic force provides the centripetal force.

mv2 / r = Bqv
r = mv / Bq

(i) The momentum/speed is becoming less, so the radius is becoming smaller

1. The spirals are in opposite directions, so the 2 particles are oppositely charged

2. They have equal initial radii, so they have equal (initial) speeds

Question 726: [Waves > Stationary waves]
(a) Explain how stationary waves are formed.

(b) The arrangement of apparatus used to determine the wavelength of a sound wave is shown in Fig.1.

Loudspeaker emits sound of one frequency. The microphone is connected to cathode-ray oscilloscope (c.r.o.).
The waveform obtained on the c.r.o. for one position of the microphone is shown in Fig.2.

Time-base setting of the c.r.o. is 0.20 ms cm−1.
(i) Use Fig.2 to show that the frequency of the sound is approximately 1300 Hz.
(ii) Explain how apparatus is used to determine the wavelength of the sound.
(iii) Wavelength of the sound wave is 0.26 m. Calculate speed of sound in this experiment.

Reference: Past Exam Paper – November 2014 Paper 21 Q8

Solution 726:
(a) Stationary waves are formed when two waves (of the same kind) travelling in opposite directions overlap. The waves should have the same frequency / wavelength and speed.

{5 small squares in the graph corresponds to 1cm
Distance between adjacent peaks = 4cm (20 small squares)
Time-base setting is 0.20 ms cm−1.
1cm represents 0.20ms. 4cm represents 4 × 0.20 = 0.80ms}
Period T = 0.8 (ms)
Frequency f = 1 / (0.8 × 10–3) = 1250 (Hz)

(ii) The microphone is moved from the plate to loudspeaker or vice versa. The wavelength is the twice the distance between adjacent maxima or minima (seen on c.r.o.).

(iii) Speed v = fλ = 1250 × 0.26 = 330 (325) m s–1

Question 727: [Quantum Physics > Quantisation]
(a) Describe main principles of the determination of the charge on an oil drop by Millikan’s experiment. You may draw a diagram if you wish.

(b) In an experiment to determine fundamental charge, values of charge on oil drops were found by a student to be as shown below.
3.2 × 10–19 C;              6.4 × 10–19 C;             16 × 10–19 C;               9.7 × 10–19 C;
12.8 × 10–19 C;            3.1 × 10–19 C;              6.3 × 10–19 C.
State value, to two significant figures, of the fundamental charge that is suggested by these values of charge on oil drops.

Reference: Past Exam Paper – June 2012 Paper 42 Q6

Solution 727:
(a) An oil drop is charged by friction/beta source and placed between 2 parallel metal plates which are horizontal.
An adjustable potential difference is applied across the plates / field between the plates until the oil drop is stationary.
The oil drop is viewed through a microscope.

For the stationary drop, mg = q x (V/d)
[where m is the mass of the oil drop, g is the acceleration due to gravity, q is the charge on the oil drop, E is the electric field between the plates and d is the separation of the plates]

m is determined from the terminal speed of the drop (when pd is zero)

(b) 3.2 × 10–19 C
{Quantisation of Charge – conclusion from Milikan’s Oil Drop Experiment:
Note that we should use the data available here (from the experiment performed by the student) to try to determine the fundamental charge. We cannot just state the known charge of an electron here.
Since we are looking for the FUNDAMENTAL charge (smallest possible value of charge), the value cannot be enough the largest values – it should be a small value.
Thus, it is obvious that we should neglect the large values and consider only the smallest ones. These are 3.2 × 10–19 C and 3.1 × 10–19 C.

We need to interpret the data values given.
Conclusion 1:
Most of the values are given to 1 decimal place. Taking into account that any experiment may contain some errors (random error here), we can assume that the uncertainty in the values is ± 0.1 × 10–19 C.

Conclusion 2:
The values given are seen to be integral multiple of 3.2 × 10–19 C – that is, if we divide them by 3.2 × 10–19 C, integers are obtained. These are true for 3.2 × 10–19 C, 6.4 × 10–19 C, 16 × 10–19 C and 12.8 × 10–19 C.
As for 9.7 × 10–19 C, 3.1 × 10–19 C and 6.3 × 10–19 C, if we include the uncertainty in the values, they become 9.6 × 10–19 C, 3.2 × 10–19 C and 6.4 × 10–19 C which are then integral multiples of 3.2 × 10–19 C.

Conclusion 3:
From conclusion 2, it can be said that the values are all multiples of 3.2 × 10–19 C. We can either from this point itself conclude that the fundamental charge is 3.2 × 10–19 C [this is the most appropriate conclusion]. As for the value of 3.1 × 10–19 C, we assume that with the uncertainty included, it also becomes 3.2 × 10–19 C.
Or if we consider the average of the 2 smallest values (3.2 × 10–19 C and 3.1 × 10–19 C) present, we obtain 3.15 × 10–19 C which is 3.2 × 10–19 C to 2 significant figures.}

Question 728: [Measurement > Estimates]
In making reasonable estimates of physical quantities, which statement is not correct?
A The frequency of sound can be of the order of GHz.
B The wavelength of light can be of the order of 600 nm.
C The Young modulus can be of the order of 1011 Pa.
D Beta radiation is associated with one unit of negative charge.

Reference: Past Exam Paper – June 2011 Paper 12 Q3

Solution 728:
Answer: A.
Sound that is perceptible by humans has frequencies from about 20 Hz to 20,000 Hz (20 kHz).

1 giga = 1G = 109

The others are reasonable estimates.


  1. Replies
    1. For 11/M/J/11 Q.10, see solution 729 at

  2. Can you please explain questions 22,25,27 and 35 of oct nov 11 variant 13, thank you.

    1. For Q27, see solution 625 at


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