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Question 1008: [Electromagnetism > Hall probe]

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.

(a) The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved.

On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage V_H.

(b) The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid.

(i) State Faraday’s law of electromagnetic induction.

(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.

Reference: Past Exam Paper – June 2014 Paper 41 & 43 Q7

Solution 1008:

(a) For the graph, the Hall voltage V_H increases from zero when the current is switched on. V_H then has a non-zero constant value. V_H then returns to zero when the current is switched off.

{The Hall voltage depends on the magnetic field and the magnetic field depends on the current flowing in the solenoid. When the switch is off, no current flows, thus no magnetic field is present and V_H is zero. When the switch is on, current flows and V_H is not zero. V_H is constant because the current is not changing (it is d.c., not a.c.).}

(b)

(i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of (magnetic) flux (linkage).

(ii) The graph consists of a pulse as the current is being switched on. There is zero e.m.f when there is current in the coil. Then, there is a pulse in the opposite direction when switching off.

{Recall Faraday’s law above. An e.m.f. is only induced when there is a ‘change’ in magnetic flux. This only occurs when the current is being switched on or off. When the current has a constant value of current (this may be zero or non-zero – as long as it is constant), no e.m.f. is induced.

The pulses are opposite because when the current is switched on, the change is from ‘off’ to ‘on’, and when the current is switched off, the change is from ‘on’ to ‘off’. This changes are opposite to each other.}

Question 1009: [Measurements > Uncertainties]

(a) State the most appropriate instrument, or instruments, for the measurement of the following.

(i) the diameter of a wire of diameter about 1 mm

(ii) the resistance of a filament lamp

(iii) the peak value of an alternating voltage

(b) The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm.

Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures.

Reference: Past Exam Paper – June 2009 Paper 21 Q1

Solution 1009:

(a)

(i) Micrometer (screw gauge) / travelling microscope

(ii) EITHER ohm-meter OR voltmeter and ammeter OR multimeter / avo on ohm setting

(iii) EITHER (calibrated) c.r.o. OR a.c. voltmeter and ×√2

(b)

Density ρ = mass m / volume V = 580 / 6³ = 2.685gcm^-3

% uncertainty in the mass, Δm = (10 / 580) × 100 = 1.7%

% uncertainty in the volume, ΔV = 3 × (0.1 / 6) × 100 = 5.0%

{Density ρ = m / V.

Percentage uncertainties: Δρ/ρ × 100% = [Δm/m + ΔV/V] × 100%

% uncertainty in density = 1.7 + 5.0 = 6.7%

% uncertainty in density = Δρ/ρ × 100% = 6.7% = 6.7 / 100

Density ρ = 2.685gcm^-3

So, Δρ = 6.7% × ρ = (6.7 / 100) × 2.685 = 0.18}

Uncertainty in density Δρ = 0.18gcm^-3

{Uncertainty is given to only 1s.f.}

So, density = 2.7 ± 0.2gcm^-3

Question 1010: [Dynamics > Collisions]

Two balls X and Y are moving towards each other with speeds of 5 m s^–1 and 15 m s^–1 respectively.

They make a perfectly elastic head-on collision and ball Y moves to the right with a speed of 7 m s^–1.

What is the speed and direction of ball X after the collision?

A 3 m s^–1 to the left

B 13 m s^–1 to the left

C 3 m s^–1 to the right

D 13 m s^–1 to the right

Reference: Past Exam Paper – June 2015 Paper 13 Q12

Solution 1010:

Answer: B.

Let the positive direction be towards the right.

For a perfectly elastic collision, both momentum and kinetic energy are conserved.

For a perfectly elastic collision,

Relative speed of approach = Relative speed of separation

‘Approach’ means getting closer to each other.

Relative speed of approach = 5 + 15 = 20 ms^-1

‘Separation’ means getting away from each other.

Assume that the velocity of X is +v (assume that it continues to move to the right after the collision. If that’s not the case, the value of v would be negative, indicating that X moves to the left after the collision). Y also moves to the right with a speed of (+)7ms^-1. From this assumption, for the 2 balls to be separating, Y must be faster than X, else they would be approaching. Since both are moving in the same direction, we need to take the difference between the 2 speeds to obtain the speed of separation.

Relative speed of separation = 7 – v

7 – v = 20

Speed v = 7 – 20 = – 13 ms^-1

Thus, X moves at 13 m s^–1 to the left.

Question 1011: [Waves > Stationary waves]

An organ pipe of length l is open at both ends. Notes are produced by the pipe when stationary waves are set up.

The speed of sound in the air column is v.

What is the lowest (fundamental) frequency of the note produced by the pipe?

A 2v / l                        B v / l                         C v / 2l                        D v / 4l

Reference: Past Exam Paper – November 2014 Paper 13 Q29

Solution 1011:

Answer: C.

Since the pipe is open at both ends, antinodes (points of maximum displacement) will be present at each end. For a note of lowest frequency, the stationary wave will consist of the 2 antinodes (mentioned above) and a single node in the middle of the pipe (at length = l / 2).

Consider a wavelength λ. This can be obtained by measuring the length between 3 consecutives antinodes (or nodes).

The distance between the 1^st and the 3^rd antinodes gives the wavelength.

The distance between the 1^st and 2^nd antinodes give half the wavelength.

Thus, in the organ pipe, l = λ / 2 for the fundamental frequency

Speed v = f λ

Fundamental frequency f₀ = v / λ = v / 2l                   [since l = λ / 2]