WAEC SSCE - Physics - 1994 (Essay)
Question 1 Report
(a) State the laws of electromagnetic induction
(b) Explain how one of the laws illustrates the principle of conservation of energy
(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.
(ii) State two reasons why the efficiency of an electric motor is less than 100%.
(a) Laws of electromagnetic induction
Faraday's law: Whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced in the circuit, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, \(E \propto -\dfrac{d\Phi}{dt}\).
Lenz's law: The induced current (or e.m.f.) always flows in such a direction as to oppose the change producing it.
(b) How Lenz's law illustrates conservation of energy
Lenz's law states that the induced current opposes the change that causes it. Therefore, to keep the flux changing (for example, to keep pushing a bar magnet into a coil) an external agent must do work against the opposing force set up by the induced current. The mechanical (kinetic) energy supplied by the external agent is exactly converted into the electrical energy of the induced current (which finally appears as heat in the resistance of the circuit). If the induced current instead aided the motion, the system would accelerate itself and create energy from nothing. Hence Lenz's law is simply a statement of the principle of conservation of energy.
(c)(i) Simple d.c. electric motor
It consists of a rectangular coil of insulated wire (the armature) mounted on an axle between the poles of a permanent magnet, a split-ring commutator fixed to the coil, and two carbon brushes that press against the commutator and feed current from a d.c. supply (battery).
Action: When current flows through the coil, side AB and side CD each lie in the magnetic field and carry current, so by the motor rule (Fleming's left-hand rule) each side experiences a force. On side AB the force acts upward and on side CD it acts downward (the two forces are equal and opposite), producing a couple (turning moment) that rotates the coil. As the coil passes the vertical position the split-ring commutator reverses the direction of current in the coil, so the forces on the two sides continue to turn the coil in the same direction, giving continuous rotation.
(c)(ii) Two reasons why efficiency is less than 100%
- Some energy is wasted as heat in the coil windings because of their electrical resistance (\(I^{2}R\) heating).
- Energy is lost through friction at the bearings and between the brushes and commutator (and through air resistance), so the useful mechanical output is always less than the electrical input.
Answer Details
(a) Laws of electromagnetic induction
Faraday's law: Whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced in the circuit, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, \(E \propto -\dfrac{d\Phi}{dt}\).
Lenz's law: The induced current (or e.m.f.) always flows in such a direction as to oppose the change producing it.
(b) How Lenz's law illustrates conservation of energy
Lenz's law states that the induced current opposes the change that causes it. Therefore, to keep the flux changing (for example, to keep pushing a bar magnet into a coil) an external agent must do work against the opposing force set up by the induced current. The mechanical (kinetic) energy supplied by the external agent is exactly converted into the electrical energy of the induced current (which finally appears as heat in the resistance of the circuit). If the induced current instead aided the motion, the system would accelerate itself and create energy from nothing. Hence Lenz's law is simply a statement of the principle of conservation of energy.
(c)(i) Simple d.c. electric motor
It consists of a rectangular coil of insulated wire (the armature) mounted on an axle between the poles of a permanent magnet, a split-ring commutator fixed to the coil, and two carbon brushes that press against the commutator and feed current from a d.c. supply (battery).
Action: When current flows through the coil, side AB and side CD each lie in the magnetic field and carry current, so by the motor rule (Fleming's left-hand rule) each side experiences a force. On side AB the force acts upward and on side CD it acts downward (the two forces are equal and opposite), producing a couple (turning moment) that rotates the coil. As the coil passes the vertical position the split-ring commutator reverses the direction of current in the coil, so the forces on the two sides continue to turn the coil in the same direction, giving continuous rotation.
(c)(ii) Two reasons why efficiency is less than 100%
- Some energy is wasted as heat in the coil windings because of their electrical resistance (\(I^{2}R\) heating).
- Energy is lost through friction at the bearings and between the brushes and commutator (and through air resistance), so the useful mechanical output is always less than the electrical input.
Question 2 Report
(a) Explain what is meant by a magnetic field
(b)(i) Describe an experiment to show that a magnetic field exists around a straight wire carrying current
(ii) Draw a labelled diagram showing the pattern and direction of the magnetic field rroduced around the wire. (Neglect the earth's magnetic field).
(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field
(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth's magnetic field.
(a) Magnetic field. A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced (for example by a magnetic material, a moving charge, or another magnet). It is a vector quantity, represented by lines of magnetic flux whose direction at any point is the direction in which a free north pole would move.
(b)(i) Experiment to show a field around a straight wire (Oersted's experiment). A thick straight copper wire is passed vertically through a small hole in the middle of a horizontal piece of stiff cardboard. Several small plotting compasses are placed on the card around the wire, and the wire is connected through a switch to a battery. When a large current is switched on and the card is tapped gently, the compass needles swing round and settle so that they all point along circles centred on the wire. Iron filings sprinkled on the card set into the same pattern of concentric circles. This shows that a magnetic field exists around the wire. When the current is reversed, every compass needle reverses its direction, showing that the field direction depends on the current direction.
(b)(ii) Pattern and direction of the field. The lines of force are concentric circles centred on the wire and lying in the plane at right angles to it. Their direction is given by the right-hand grip rule: gripping the wire with the right hand so that the thumb points along the conventional current, the curled fingers give the field direction. With the current coming out of the page (shown by the central dot), the field lines run anticlockwise, as drawn below.
(c) Two parallel wires carrying current in the same direction. Each wire produces its own set of concentric circular field lines. At the midpoint of the line joining the two wires the two fields are equal in magnitude but opposite in direction, so they cancel exactly, giving a neutral point N. (Outside the pair the fields reinforce, which is why two such wires attract each other.)
(d) Protecting a delicate magnetic material (magnetic shielding). The instrument is enclosed in a thick ring or box of soft iron, which has a very high magnetic permeability. Because the flux prefers the easy path through the iron, the earth's field lines are drawn into the shield and channelled round its walls instead of crossing the enclosed space. The cavity inside is therefore left almost field-free, screening the delicate material from the earth's magnetic field.
Answer Details
(a) Magnetic field. A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced (for example by a magnetic material, a moving charge, or another magnet). It is a vector quantity, represented by lines of magnetic flux whose direction at any point is the direction in which a free north pole would move.
(b)(i) Experiment to show a field around a straight wire (Oersted's experiment). A thick straight copper wire is passed vertically through a small hole in the middle of a horizontal piece of stiff cardboard. Several small plotting compasses are placed on the card around the wire, and the wire is connected through a switch to a battery. When a large current is switched on and the card is tapped gently, the compass needles swing round and settle so that they all point along circles centred on the wire. Iron filings sprinkled on the card set into the same pattern of concentric circles. This shows that a magnetic field exists around the wire. When the current is reversed, every compass needle reverses its direction, showing that the field direction depends on the current direction.
(b)(ii) Pattern and direction of the field. The lines of force are concentric circles centred on the wire and lying in the plane at right angles to it. Their direction is given by the right-hand grip rule: gripping the wire with the right hand so that the thumb points along the conventional current, the curled fingers give the field direction. With the current coming out of the page (shown by the central dot), the field lines run anticlockwise, as drawn below.
(c) Two parallel wires carrying current in the same direction. Each wire produces its own set of concentric circular field lines. At the midpoint of the line joining the two wires the two fields are equal in magnitude but opposite in direction, so they cancel exactly, giving a neutral point N. (Outside the pair the fields reinforce, which is why two such wires attract each other.)
(d) Protecting a delicate magnetic material (magnetic shielding). The instrument is enclosed in a thick ring or box of soft iron, which has a very high magnetic permeability. Because the flux prefers the easy path through the iron, the earth's field lines are drawn into the shield and channelled round its walls instead of crossing the enclosed space. The cavity inside is therefore left almost field-free, screening the delicate material from the earth's magnetic field.
Question 3 Report
(a) What is meant by the statement: The specific heat capacity of copper is \(400 J kg^{-1}K^{-1}\)?
(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results
(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \(4.2 J g^{-1}K^{-1}\)) (Specific heat capacity of copper = \(0.4 Jg^{-1}K^{-1}\)).
(a) "The specific heat capacity of copper is \(400\ \text{J kg}^{-1}\text{K}^{-1}\)" means that 400 joules of heat energy are required to raise the temperature of 1 kilogram of copper by 1 kelvin (1 °C).
(b)(i) Experiment (method of mixtures).
- Weigh the copper ball to get its mass \(m_c\). Heat it in boiling water for some minutes so it reaches a known steady temperature \(\theta_1\) (about 100 °C, read from a thermometer in the water).
- Weigh a calorimeter (mass \(m_{cal}\)), pour in a known mass of water \(m_w\), and record the initial temperature \(\theta_2\) of the calorimeter and water.
- Quickly transfer the hot copper ball into the calorimeter, stir gently, and record the highest (final steady) temperature \(\theta_3\).
- Apply the principle of mixtures: heat lost by copper ball = heat gained by water + heat gained by calorimeter.
Every quantity except the specific heat capacity of copper \(c_c\) is known or measured, so \(c_c\) is calculated.
(b)(ii) Two precautions.
- Transfer the hot ball quickly and lag (insulate) the calorimeter to reduce heat loss to the surroundings.
- Stir the water gently before taking the final temperature so the mixture is uniform.
(b)(iii) Calculation. Let the final steady temperature be \(\theta\).
Heat lost by copper ball \(= 20\times0.4\times(200-\theta)=8(200-\theta)\)
Heat gained by water \(= 50\times4.2\times(\theta-30)=210(\theta-30)\)
Heat gained by calorimeter \(= 60\times0.4\times(\theta-30)=24(\theta-30)\)
\[ 8(200-\theta) = (210+24)(\theta-30) \] \[ 1600 - 8\theta = 234\theta - 7020 \] \[ 8620 = 242\theta \] \[ \theta = 35.6\ ^{\circ}\text{C} \]The final steady temperature of the mixture is about 35.6 °C.
Answer Details
(a) "The specific heat capacity of copper is \(400\ \text{J kg}^{-1}\text{K}^{-1}\)" means that 400 joules of heat energy are required to raise the temperature of 1 kilogram of copper by 1 kelvin (1 °C).
(b)(i) Experiment (method of mixtures).
- Weigh the copper ball to get its mass \(m_c\). Heat it in boiling water for some minutes so it reaches a known steady temperature \(\theta_1\) (about 100 °C, read from a thermometer in the water).
- Weigh a calorimeter (mass \(m_{cal}\)), pour in a known mass of water \(m_w\), and record the initial temperature \(\theta_2\) of the calorimeter and water.
- Quickly transfer the hot copper ball into the calorimeter, stir gently, and record the highest (final steady) temperature \(\theta_3\).
- Apply the principle of mixtures: heat lost by copper ball = heat gained by water + heat gained by calorimeter.
Every quantity except the specific heat capacity of copper \(c_c\) is known or measured, so \(c_c\) is calculated.
(b)(ii) Two precautions.
- Transfer the hot ball quickly and lag (insulate) the calorimeter to reduce heat loss to the surroundings.
- Stir the water gently before taking the final temperature so the mixture is uniform.
(b)(iii) Calculation. Let the final steady temperature be \(\theta\).
Heat lost by copper ball \(= 20\times0.4\times(200-\theta)=8(200-\theta)\)
Heat gained by water \(= 50\times4.2\times(\theta-30)=210(\theta-30)\)
Heat gained by calorimeter \(= 60\times0.4\times(\theta-30)=24(\theta-30)\)
\[ 8(200-\theta) = (210+24)(\theta-30) \] \[ 1600 - 8\theta = 234\theta - 7020 \] \[ 8620 = 242\theta \] \[ \theta = 35.6\ ^{\circ}\text{C} \]The final steady temperature of the mixture is about 35.6 °C.
Question 4 Report
(a) Describe an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string
(b) Draw diagrams showing a vibrating string fixed at both ends emitting (i) fundamental frequency (ii) second overtone indicate the nodes and antinodes on the diagrams
(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concave mirror (ii) converging lens
(a) Experiment: how the frequency of the note depends on the tension
Apparatus: a sonometer (a hollow wooden box carrying a thin, uniform stretched wire), two movable bridges, a fixed peg, a frictionless pulley at one end, a scale-pan carrying known masses, a set of tuning forks of known frequency, and a small paper rider.
Procedure: The wire is stretched over the two bridges so that the vibrating (effective) length \(l\) between them is kept constant throughout, and the same uniform wire is used so that the mass per unit length \(\mu\) is constant. One end is fixed to the peg; the other passes over the pulley and carries the scale-pan. A known mass \(m\) is placed on the pan, giving a tension \(T = mg\) in the wire. A tuning fork of known frequency \(f\) is struck and its stem pressed on the box. The mass (tension) is adjusted until the wire vibrates in unison with the fork in its fundamental mode; at resonance a small paper rider placed at the middle of the wire is violently thrown off. The tension \(T\) and the fork frequency \(f\) are recorded. The experiment is repeated with four more forks of different frequency, each time altering the mass to obtain resonance, while \(l\) and \(\mu\) are held fixed.
Sample results:
| Tension T / N | √T / N1/2 | Frequency f / Hz |
|---|---|---|
| 10 | 3.16 | 100 |
| 40 | 6.32 | 200 |
| 90 | 9.49 | 300 |
| 160 | 12.65 | 400 |
| 250 | 15.81 | 500 |
A graph of \(f\) against \(\sqrt{T}\) is plotted:
The graph is a straight line passing through the origin of slope \(\approx 31.6\ \text{Hz N}^{-1/2}\). This shows that the frequency is directly proportional to the square root of the tension, \( f \propto \sqrt{T}\), in agreement with \( f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}\).
(b) Vibrating string fixed at both ends
(i) Fundamental frequency (first harmonic): the string vibrates in a single loop, with a node (N) at each fixed end and one antinode (A) at the middle.
(ii) Second overtone (third harmonic): the string vibrates in three loops, giving four nodes (N) and three antinodes (A).
(c) Formation of a virtual image (ray diagrams)
(i) Concave mirror: with the object O placed between the pole P and the principal focus F, the two reflected rays diverge. One incident ray parallel to the axis reflects through F; a second ray directed to the pole P reflects symmetrically about the axis. Producing the reflected rays backwards behind the mirror (broken lines) locates a virtual, erect and magnified image I.
(ii) Converging lens: with the object O placed between the lens and its principal focus F (the magnifying-glass position), the refracted rays diverge. One incident ray parallel to the axis refracts through the far focus F'; a second ray through the optical centre passes straight on. Producing the refracted rays backwards on the same side as the object (broken lines) locates a virtual, erect and magnified image I.
Answer Details
(a) Experiment: how the frequency of the note depends on the tension
Apparatus: a sonometer (a hollow wooden box carrying a thin, uniform stretched wire), two movable bridges, a fixed peg, a frictionless pulley at one end, a scale-pan carrying known masses, a set of tuning forks of known frequency, and a small paper rider.
Procedure: The wire is stretched over the two bridges so that the vibrating (effective) length \(l\) between them is kept constant throughout, and the same uniform wire is used so that the mass per unit length \(\mu\) is constant. One end is fixed to the peg; the other passes over the pulley and carries the scale-pan. A known mass \(m\) is placed on the pan, giving a tension \(T = mg\) in the wire. A tuning fork of known frequency \(f\) is struck and its stem pressed on the box. The mass (tension) is adjusted until the wire vibrates in unison with the fork in its fundamental mode; at resonance a small paper rider placed at the middle of the wire is violently thrown off. The tension \(T\) and the fork frequency \(f\) are recorded. The experiment is repeated with four more forks of different frequency, each time altering the mass to obtain resonance, while \(l\) and \(\mu\) are held fixed.
Sample results:
| Tension T / N | √T / N1/2 | Frequency f / Hz |
|---|---|---|
| 10 | 3.16 | 100 |
| 40 | 6.32 | 200 |
| 90 | 9.49 | 300 |
| 160 | 12.65 | 400 |
| 250 | 15.81 | 500 |
A graph of \(f\) against \(\sqrt{T}\) is plotted:
The graph is a straight line passing through the origin of slope \(\approx 31.6\ \text{Hz N}^{-1/2}\). This shows that the frequency is directly proportional to the square root of the tension, \( f \propto \sqrt{T}\), in agreement with \( f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}\).
(b) Vibrating string fixed at both ends
(i) Fundamental frequency (first harmonic): the string vibrates in a single loop, with a node (N) at each fixed end and one antinode (A) at the middle.
(ii) Second overtone (third harmonic): the string vibrates in three loops, giving four nodes (N) and three antinodes (A).
(c) Formation of a virtual image (ray diagrams)
(i) Concave mirror: with the object O placed between the pole P and the principal focus F, the two reflected rays diverge. One incident ray parallel to the axis reflects through F; a second ray directed to the pole P reflects symmetrically about the axis. Producing the reflected rays backwards behind the mirror (broken lines) locates a virtual, erect and magnified image I.
(ii) Converging lens: with the object O placed between the lens and its principal focus F (the magnifying-glass position), the refracted rays diverge. One incident ray parallel to the axis refracts through the far focus F'; a second ray through the optical centre passes straight on. Producing the refracted rays backwards on the same side as the object (broken lines) locates a virtual, erect and magnified image I.
