This document contains June 2022 OCR A Level Physics B (Advancing Physics) H557/02 Scientific literacy in physics Question Paper. Perfect for Revision.

SECTION A

Answer all the questions.

1 This question is about the experiment to measure the charge on an electron performed by

American physicists Robert Millikan and Harvey Fletcher in the early years of the twentieth

century.

Consider the forces on an oil drop falling at terminal velocity through air, as shown in Fig. 1.1.

D

W

 Fig. 1.1

 (a) Ignoring any upthrust forces, state why the drag force D must be equal and opposite to the

weight W of the drop when it is falling at terminal velocity.

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 (b) The drag force is given by the equation D = 6πηairrv where r is the radius of the oil drop and

v is the terminal velocity. The symbol ηair represents the viscosity of air; this is a measure of

the resistance that air has to the motion of objects in it.

 Show that the terminal velocity of a drop of mass 1.8 × 10–15 kg is about 7 × 10–5 m s–1. The

buoyancy of air may be ignored.

 density of oil = 8.7 × 102 kg m–3

 viscosity of air = 1.8 × 10–5 Pa s

 [2]

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 (c) The oil drop is given a negative charge and falls into a region of uniform electric field, as

shown in the diagram.

5.4 mm

+ V

0V

Fig. 1.2 (not to scale)

 (i) Draw six lines on Fig. 1.2 to represent the uniform field between the plates. Assume

that the oil drop does not distort the field. [2]

 (ii) The drop of mass 1.8 × 10–15 kg is held stationary between the plates when the p.d.V

between the plates is 200 V. The plate separation is 5.4 mm.

 Calculate the charge on the oil drop.

charge = ...................................................... C [2]

 (iii) When Millikan and Fletcher performed this experiment with many oil drops, they

found that the charge on the drop was always a multiple of about 1.6 × 10–19 C (to

two significant figures). Why does this suggest that there is a fundamental quantity of

electric charge and how does the modern model of the structure of hadrons suggest

that 1.6 × 10–19 C may not be the smallest quantity of charge that a particle can

possess?

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2 This question is about a simple generator as shown in Fig. 2.1.

S

N

X Y X Y rotor

laminated

iron core

220-turn

coil

laminated core with

cross-sectional area A

Key iron

insulator

copper coil

Face view Section through X-Y

Fig. 2.1a Fig. 2.1b

 (a) (i) The core is made of sheets of iron which have been laminated (separated by thin

sheets of electrically insulating material). The layers of iron and insulator are all parallel

to the plane of Fig. 2.1a and stacked as shown in Fig. 2.1b.

 Explain why iron is used in the core and why laminating the core improves the

performance of the generator.

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 (ii) The permeance of a magnetic circuit can be compared to the conductivity of an

electrical circuit. Suggest and explain a change to the generator in Fig. 2.1 which would

increase the permeance of the magnetic circuit.

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 Fig. 2.2 shows how the flux density in the iron core within the coil changes as the magnet

rotates.

0.20

0.15

0.10

0.05

0.00

– 0.05

– 0.10

– 0.15

– 0.20

flux density /T

0 5 10 15 20 25 30 35 40

time /ms

Fig. 2.2

 (b) (i) The cross-sectional area A of the iron core inside the coil is 1.5 × 10–4 m2.

 Explain, without calculation, how the maximum emf across the 220-turn coil can be

estimated from the cross-sectional area of the coil and the data in Fig. 2.2.

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 (ii) The emf ε across each turn of the coil is given by the equation

 ε = 2πfφmaxsin 2πft

 Use the equation and data from Fig. 2.2 and part (b) (i) to calculate the maximum emf

induced across the coil