4.7 Magnetism and electromagnetism — coverage pack

Answer

• The poles repel.
• The interaction is a non-contact force.

Method: Both approaching ends are north poles, so they are like poles and repel. The magnets exert this force across a gap, making it a non-contact force.

Answer

• The magnetic field induces magnetism in the iron pin.
• The end nearest the south pole becomes a north pole, so there is attraction.
• The pin quickly loses most or all of its magnetism when the field is removed.

Method: Identify the pin as an induced magnet. Its near end becomes the opposite pole to the magnet's south pole, producing attraction. Iron does not retain most of this induced magnetism once the external field is removed.

Answer

• Q must be a permanent magnet because repulsion can occur only between like magnetic poles.
• P could be an induced magnetic material because induced magnetism causes attraction to either pole.
• P could instead present the south pole of a permanent magnet, so attraction alone is not conclusive.

Method: Use repulsion as the decisive test: a like pole is required, so Q already has a permanent pole. Attraction does not distinguish a permanent south pole from an induced magnetic material, because either can be attracted in the stated observation.

Answer

• The field points away from the north pole, towards the south pole.

Method: Use the field-line convention outside a magnet: arrows run from the north pole to the south pole.

Answer

• Place the compass beside the magnet and mark the direction indicated by the north-seeking end.
• Move the compass so its tail is at the previous mark and repeat.
• Join the marks with a smooth line and add an arrow from north to south.
• Repeat from several starting positions to build the pattern.

Method: The north-seeking end shows the field direction at one point. Repeated adjacent readings trace one field line; repeating the process from other starting points maps the whole pattern.

Answer

• Sketch B is consistent with the field being strongest at the poles.
• Crowded field lines represent a stronger field and field lines must not cross.
• The arrows outside the magnet should run from north to south.
• A compass north-seeking end would align tangentially with those arrows at its position.

Method: Test each sketch against three rules: strength is greatest where lines are closest, one point cannot have two directions, and the external direction is north to south. A compass then provides the local direction shown by the tangent to a valid line.

Answer

• Increase the current.
• Add an iron core.

Method: A larger current produces a stronger field from every turn. Easily magnetised iron concentrates and strengthens the solenoid's field.

Answer

• The field lines are concentric circles centred on the wire.
• Point the right thumb upwards, in the current direction.
• Viewed from above, the curled fingers show an anticlockwise field.

Method: Use the right-hand grip rule with the thumb along the upward current. From above, the fingers curl anticlockwise around the wire, matching the circular field lines.

Answer

• Both coils have a strong, nearly uniform field inside and a bar-magnet-shaped external field.
• Y has the stronger field because its iron core becomes magnetised.
• Reversing current reverses the direction of the field made by every turn.
• The north and south poles therefore swap.
• The iron core is still easily magnetised, so it continues to strengthen the reversed field.

Method: First compare like-for-like turn count and current: only the iron core differs, so Y is stronger. Current direction sets field direction by the grip rule, whereas the core controls the extra strength; reversing one does not remove the effect of the other.

Answer

• First finger: magnetic field direction.
• Thumb: force or motion direction.

Method: Recall the ordered labels: first finger is field, second finger is conventional current, and thumb is force.

Answer

• $0.26\,\text{N}$

Method: Use $F=BIl=0.45\times3.2\times0.18=0.2592\,\text{N}$. To two significant figures, the force is $0.26\,\text{N}$.

Answer

• $0.45\,\text{T}$
• The force becomes upward.

Method: Rearrange $F=BIl$ to $B=F/(Il)=0.63/(4.0\times0.35)=0.45\,\text{T}$. Fleming's rule shows that reversing one of current or field reverses the force, so the downward force becomes upward.

Answer

• Electrical energy is transferred to kinetic energy.
• The motor effect produces the force.

Method: A motor uses current to produce motion, so its useful transfer is electrical to kinetic. The force on the current-carrying coil is the motor effect.

Answer

• The two active sides carry currents in opposite directions.
• Each side experiences a motor-effect force in the magnetic field.
• The two forces act in opposite directions.
• Because they act on different sides of the axis, they form a turning effect on the coil.

Method: Apply Fleming's left-hand rule to each side. Reversing current while keeping the field fixed reverses the force, so the separated forces form a couple and rotate the coil.

Answer

• Increasing current increases the force on each active side of the coil.
• A stronger magnet gives a larger magnetic flux density and therefore a larger force.
• The larger forces produce a larger turning effect, tending to increase the rotation rate.
• After half a turn, each side of the coil has exchanged positions.
• The commutator reverses current so the force on each side also reverses.
• The turning effect therefore remains in the same rotational direction rather than reversing every half-turn.

Method: Use $F=BIl$ qualitatively: raising $I$ or $B$ raises the forces and hence the turning effect. Track one side through half a rotation; because its position swaps, its current must also swap to keep its force driving the same sense of rotation.

Answer

• The motor effect.
• The cone transfers energy to the surrounding air as sound waves.

Method: Current in a magnetic field gives a force by the motor effect. The moving cone makes pressure variations in the air, transferring energy as sound.

Answer

• The signal current flows through a coil in a permanent magnetic field.
• The motor effect exerts a force on the coil.
• As the alternating current reverses, the force direction reverses.
• The attached cone therefore moves backwards and forwards, producing pressure variations in the air.

Method: Follow the causal chain from alternating current to alternating motor-effect force, then to coil and cone vibration, and finally to air-pressure variations.

Answer

• Signal A produces the higher-pitched sound because the cone vibrates at a higher frequency.
• Signal A produces the quieter sound because its smaller current gives a smaller motor-effect force and cone amplitude.

Method: The cone follows the frequency of the alternating force, so doubling signal frequency raises pitch. Force size depends on current, so the smaller current amplitude produces smaller pressure variations and hence lower loudness.

Answer

• The reading is zero.
• There is no relative motion or change in magnetic field through the coil.

Method: The generator effect needs a changing magnetic field around the conductor. Once the magnet is stationary, that change stops, so no potential difference is induced.

Answer

• Insertion and withdrawal produce pulses in opposite directions.
• At equal speeds their magnitudes are equal, for the same motion range.
• Moving the magnet faster increases the magnitude.
• Using a stronger magnet or more coil turns also increases the magnitude.

Method: Reversing the relative motion reverses the induced potential difference. Its size depends on how quickly the magnetic field changes and on field strength and turn count.

Answer

• The changing magnetic field induces a current in the complete coil circuit.
• That current produces its own magnetic field.
• The induced field opposes the magnet's motion or the change producing it.
• Pulling the magnet out reverses the change, so the induced current reverses.
• Faster motion produces a larger induced potential difference and current.
• The stronger opposing magnetic effect means more mechanical energy is transferred each second.

Method: Apply the opposition principle to the change, not merely to the magnet's field. Reversing motion reverses the change and current. Increasing speed raises the induction rate, so the opposing force and mechanical power input rise.

Answer

• An alternator generates ac.
• A dynamo generates dc.

Method: Match the devices to their outputs: continuous slip rings retain the alternator's reversal, while a split-ring commutator makes the dynamo output unidirectional.

Answer

• Frequency $=5\,\text{Hz}$.
• Period $=0.20\,\text{s}$.
• Doubling rotation rate doubles the frequency and halves the period.

Method: One complete rotation produces one complete ac cycle, so $f=5\,\text{Hz}$. Then $T=1/f=1/5=0.20\,\text{s}$. At twice the rotation rate, $f=10\,\text{Hz}$ and $T=0.10\,\text{s}$.

Answer

• P has an alternating graph with equal positive and negative half-cycles.
• Q has pulses on one side of zero, so its polarity does not reverse.
• Both outputs vary from zero to a maximum magnitude as the coil rotates.
• Q's split ring swaps the external connections every half-turn.
• That swap reverses the connection exactly when the coil's induced potential reverses, keeping the external polarity unchanged.

Method: Begin with the same alternating induced potential in each rotating coil. Slip rings pass that reversal to P. Q's split ring swaps connections at each half-turn, rectifying the external output into varying dc pulses.

Answer

• Input: pressure variations in a sound wave.
• Output: variations in electrical potential difference or current.

Method: A microphone is a transducer from sound to electrical signal, so identify air-pressure variation as the input and a varying electrical signal as the output.

Answer

• Sound-pressure variations make the diaphragm vibrate.
• The attached coil moves relative to a magnetic field.
• The changing magnetic field through the coil induces a potential difference by the generator effect.
• A complete circuit carries a current that varies with the sound.

Method: Trace energy and cause in order: sound moves the diaphragm, the diaphragm moves the coil through the field, and the generator effect produces the varying electrical signal.

Answer

• P produces the larger signal amplitude because the louder sound drives a larger diaphragm and coil vibration than Q.
• R produces the higher signal frequency because its higher pitch makes the diaphragm vibrate more frequently than S.
• Moving the coil relative to the field induces the potential difference.
• Faster change of magnetic field gives a larger instantaneous induced potential difference.

Method: Use the controlled pairs separately: loudness changes vibration and signal amplitude when pitch is fixed, while pitch changes vibration and signal frequency when loudness is fixed. The generator effect converts the coil's motion through the field into voltage.

Answer

• It is step-up.
• The secondary potential difference is three times the primary potential difference.

Method: The turn ratio is $n_{\mathrm{s}}/n_{\mathrm{p}}=900/300=3$. The secondary has more turns, so $V_{\mathrm{s}}=3V_{\mathrm{p}}$ and the transformer is step-up.

Answer

• $15.3\,\text{V}$

Method: Use $V_{\mathrm{p}}/V_{\mathrm{s}}=n_{\mathrm{p}}/n_{\mathrm{s}}$. Thus $V_{\mathrm{s}}=V_{\mathrm{p}}n_{\mathrm{s}}/n_{\mathrm{p}}=230\times80/1200=15.3\,\text{V}$.

Answer

• Secondary current $=1.50\,\text{A}$.
• Primary current $=75.0\,\text{A}$.
• $n_{\mathrm{s}}:n_{\mathrm{p}}=50:1$.
• The smaller transmission current reduces heating losses in cables.

Method: Use $P=VI$. The secondary current is $I_{\mathrm{s}}=18000/12000=1.50\,\text{A}$ and, for equal ideal power, $I_{\mathrm{p}}=18000/240=75.0\,\text{A}$. The turn ratio equals the potential-difference ratio: $n_{\mathrm{s}}/n_{\mathrm{p}}=12000/240=50$. For the same transmitted power, a larger potential difference means a smaller current, reducing cable heating.