D.4 Induction HL only

1. Its plane is perpendicular to the field lines, so θ in Φ = BA cos θ is 90°. — A student who measures θ from the plane picks this. With the plane perpendicular to the field, the field passes straight through the loop: the angle to the normal is 0° and the flux is greatest, BA.
2. It is held at rest in the field, so the field through it is not changing. — A student who links flux to motion or change picks this. A loop at rest in a steady field has a constant, non-zero flux, Φ = BA cos θ. What is zero is the RATE of change of flux, and so the induced emf.
3. Its plane is parallel to the field lines, so no field lines pass through the loop. — Φ = BA cos θ, with θ measured from the normal. When the plane of the loop is parallel to the field, the normal is perpendicular to it, θ = 90° and cos θ = 0, so Φ = 0: the field lines run alongside the loop and none pass through it.
4. At no orientation: while a loop is inside a field, flux must pass through it. — A student who treats flux as the same as the field picks this. Flux also depends on orientation: Φ = BA cos θ is zero when the plane of the loop is parallel to the field, even though the field is still present.

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1. The rate at which the magnetic flux through the coil changes — Faraday's law is ε = −N ΔΦ/Δt: for a fixed number of turns the size of the emf is proportional to ΔΦ/Δt, the rate of change of flux. How much flux there is, or how strong the field is, does not matter if the flux is not changing.
2. The magnetic flux passing through the coil at that instant — A student who thinks the emf depends on the amount of flux picks this. A coil can have a large steady flux and no emf at all; the emf depends on how fast the flux is changing.
3. The change in magnetic flux, whatever time it takes — A student who leaves out the Δt picks this. The same change of flux made in half the time induces twice the emf, so the emf is proportional to ΔΦ/Δt, not to ΔΦ alone.
4. The speed of the coil relative to the magnet producing the field — A student who thinks induction needs motion picks this. A stationary coil in a field whose strength is changing, for example inside a solenoid with a changing current, has an emf induced in it with nothing moving; the emf depends on the rate of change of flux, however that change is produced.

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1. The rod moves horizontally in the same direction as the field lines. — A student who thinks any motion in a field induces an emf picks this. Moving along the field lines, the rod sweeps across no field lines, so no emf is induced.
2. The rod slides horizontally along the direction of its own length. — A student who thinks moving at right angles to the field is enough picks this. Sliding along its own length, the rod sweeps out no area and cuts no flux; the magnetic force on its electrons acts across the rod, not along it, so no emf appears between the ends.
3. The rod is held at rest while the magnetic field is kept steady. — A student who thinks a field alone induces an emf picks this. With the rod at rest in a steady field, no flux changes, so no emf is induced however strong the field is.
4. The rod moves vertically upwards at a steady speed while staying horizontal. — Moving vertically, the rod moves perpendicular to its own length and to the horizontal field, so it sweeps across field lines and ε = BvL is induced between its ends. No complete circuit is needed for an emf; without one, there is simply no current.

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1. The magnet's store of magnetism, which is gradually used up by the induced current it produces — A student who thinks the magnet is an energy store that runs down picks this. The magnet can be used indefinitely; the energy comes from the work done moving it against the force that the induced current produces.
2. Work done by the person pushing the magnet against the opposing force from the coil — By Lenz's law the induced current makes the coil repel the approaching magnet. The person must do work against this force, and that work is transferred by the induced current to the internal energy of the resistor. This is why Lenz's law follows from energy conservation.
3. The coil's attraction of the magnet, which pulls it in and does work on it — A student who thinks the induced current helps the motion picks this. If the coil attracted the magnet, both the magnet's kinetic energy and the current would grow with no work done, creating energy. The coil repels the approaching magnet.
4. The changing flux itself, so the magnet is no harder to push in when the coil is connected — A student who thinks induction makes electrical energy without extra work picks this. With the coil connected, the induced current makes the coil repel the magnet, so the person must push harder than for an unconnected coil; that extra work is the energy that warms the resistor.

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1. It is greatest when the coil's plane is perpendicular to the field, where the flux through it is greatest. — A student who thinks the emf follows the amount of flux picks this. When the plane is perpendicular to the field the flux is greatest but momentarily not changing, so the emf is zero at that position.
2. It is the same at every position, because the coil turns at a steady rate in a uniform field. — A student who expects steady motion to give a steady emf picks this. The sides move at constant speed, but the component of their velocity across the field varies through each turn, so the emf varies sinusoidally.
3. It is greatest when the plane of the coil is parallel to the field, where the flux is zero. — The flux Φ = BA cos θ varies sinusoidally, and the emf depends on how fast Φ changes. The cosine changes fastest where it passes through zero, so the emf is greatest when the plane is parallel to the field and the flux is zero; the long sides then move straight across the field lines.
4. It is zero at every position, because the field through the coil is uniform and does not change. — A student who treats flux as the same as field picks this. B is constant, but the angle between the field and the coil changes as it rotates, so Φ = BA cos θ changes and an emf is induced.

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1. 3.3 × 10⁻² s — 1800 rev min⁻¹ ÷ 60 = 30 revolutions per second. The emf completes one cycle per revolution, so its period is T = 1/f = 1/30 s = 3.3 × 10⁻² s: a lower rotation frequency gives a longer period.
2. 1.2 × 10⁻² s — A student who scales the period in the same way as the frequency gets 20 ms × 30/50 = 12 ms. T = 1/f, so reducing the frequency from 50 Hz to 30 Hz INCREASES the period.
3. 2.0 × 10⁻² s — A student who thinks changing the rotation frequency changes only the size of the emf keeps 20 ms. The emf repeats once per revolution, so its period changes whenever the rotation rate changes.
4. 5.6 × 10⁻⁴ s — A student who uses 1800 as the frequency in hertz gets T = 1/1800 s = 5.6 × 10⁻⁴ s. Revolutions per minute must be divided by 60: f = 30 Hz.

Working f = 1800 rev min⁻¹ / 60 s min⁻¹ = 30 Hz. The emf completes one cycle per revolution, so T = 1/f = 1/(30 Hz) = 0.0333 s = 3.3 × 10⁻² s.

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1. 5.2 × 10⁻⁴ Wb — A student who measures θ from the plane of the loop uses cos 30°: 0.30 × 2.0 × 10⁻³ × 0.866 = 5.2 × 10⁻⁴ Wb. θ in Φ = BA cos θ is measured from the normal, so here θ = 90° − 30° = 60°.
2. 6.0 × 10⁻⁴ Wb — A student who uses Φ = BA whatever the orientation gets 0.30 × 2.0 × 10⁻³ = 6.0 × 10⁻⁴ Wb. That is the flux only when the field passes straight through the loop; tilted at 30° to the field, the loop has Φ = BA cos 60°.
3. 3.0 × 10⁻² Wb — A student who converts 20 cm² to m² by dividing by 100 uses A = 0.20 m² and gets 0.30 × 0.20 × 0.5 = 3.0 × 10⁻² Wb. 1 m² = 10⁴ cm², so A = 2.0 × 10⁻³ m².
4. 3.0 × 10⁻⁴ Wb — A = 0.040 m × 0.050 m = 2.0 × 10⁻³ m². The plane is at 30° to the field, so the normal is at θ = 60° to the field. Φ = BA cos θ = 0.30 × 2.0 × 10⁻³ × cos 60° = 3.0 × 10⁻⁴ Wb.

Working A = 0.040 m × 0.050 m = 2.0 × 10⁻³ m². The angle between the field and the plane is 30°, so the angle between the field and the normal is θ = 60°. Φ = BA cos θ = (0.30 T)(2.0 × 10⁻³ m²)(cos 60°) = (0.30)(2.0 × 10⁻³)(0.50) = 3.0 × 10⁻⁴ Wb.

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1. 1.4 V — A student who measures θ from the plane takes the starting flux as BA cos 90° = 0 and the final flux as BA cos 30°, so ΔΦ = 0.866BA and ε = 1.4 V. The plane starts perpendicular to the field, so the flux starts at its maximum, BA.
2. 0.80 V — At the start θ = 0° (field along the normal), so Φ₁ = BA. After turning 60°, θ = 60°, so Φ₂ = BA cos 60° = 0.5BA. ΔΦ = 0.5 × 0.40 × 2.0 × 10⁻³ = 4.0 × 10⁻⁴ Wb. ε = NΔΦ/Δt = 100 × 4.0 × 10⁻⁴ / 0.050 = 0.80 V.
3. 0.040 V — A student who stops at NΔΦ gets 100 × 4.0 × 10⁻⁴ = 0.040, which is in webers, not volts. Faraday's law needs the RATE of change: divide by the 0.050 s taken.
4. 0.0080 V — A student who leaves out the number of turns finds the emf of one turn: 4.0 × 10⁻⁴ / 0.050 = 0.0080 V. Each of the 100 turns has this emf and they are in series, so ε = 0.80 V.

Working A = 0.040 m × 0.050 m = 2.0 × 10⁻³ m². Initially the plane is perpendicular to the field, so θ = 0° and Φ₁ = BA = (0.40 T)(2.0 × 10⁻³ m²) = 8.0 × 10⁻⁴ Wb. After rotating 60°, θ = 60° and Φ₂ = BA cos 60° = 4.0 × 10⁻⁴ Wb. ΔΦ = 4.0 × 10⁻⁴ Wb. Average ε = N ΔΦ/Δt = 100 × (4.0 × 10⁻⁴ Wb)/(0.050 s) = 0.80 V.

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1. 0.0090 V — A student who thinks the emf equals the flux through the circuit uses BA = 0.150 × 0.40 × 0.15 = 0.0090, which is a flux in webers. The emf depends on how fast the flux changes, BvL, not on the flux at that instant.
2. 0.060 V — A student who uses the change of flux without dividing by the time finds the flux swept in 0.50 s: 0.150 × 0.40 × 1.0 = 0.060 Wb. Dividing by 0.50 s gives the emf, 0.12 V.
3. 120 V — A student who substitutes 150 for the field, without converting mT to T, gets 150 × 2.0 × 0.40 = 120 V. B = 150 mT = 0.150 T, and the emf is 0.12 V.
4. 0.12 V — v = 1.0 m / 0.50 s = 2.0 m s⁻¹. The rod, the field and the velocity are mutually perpendicular, so ε = BvL = 0.150 × 2.0 × 0.40 = 0.12 V. The rod's distance from the resistor does not affect the emf.

Working v = 1.0 m / 0.50 s = 2.0 m s⁻¹. B = 150 mT = 0.150 T. The rod (L = 0.40 m), its velocity and the field are mutually perpendicular, so ε = BvL = (0.150 T)(2.0 m s⁻¹)(0.40 m) = 0.12 V.

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1. It falls more slowly, because currents induced in the tube exert forces that oppose its motion. — The moving magnet changes the flux through each ring of the tube, inducing circulating currents. By Lenz's law these currents oppose the relative motion, so the magnet falls more slowly and soon reaches a low terminal speed; its lost gravitational potential energy becomes internal energy of the tube.
2. It falls in the same way, because copper is not magnetic and so exerts no force on it. — A student who thinks non-magnetic metals are unaffected picks this. Copper is not attracted by a stationary magnet, but a moving magnet induces currents in it, and these currents exert forces that oppose the motion.
3. It falls faster, because the currents induced in the tube pull it downwards. — A student who thinks induced currents help the change picks this. A downward pull would increase both the magnet's kinetic energy and the induced currents with no extra work done, violating energy conservation. The force opposes the motion.
4. It is pulled sideways onto the wall and sticks there, as a magnet attracts metals. — A student who thinks magnets attract all metals picks this. Copper is not ferromagnetic, so the magnet is not attracted to the wall; its interaction with the tube arises only from induced currents while it moves.

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1. Constant from 0 to 4 s, zero from 4 to 8 s, then reversed and larger in size from 8 to 10 s — The flux through the small coil follows the current. From 0 to 4 s it rises at a steady rate, giving a constant emf; from 4 to 8 s it is steady, so the emf is zero; from 8 to 10 s it falls by the same amount in half the time, so the emf is reversed and larger in size (twice as large).
2. Constant from 0 to 4 s, zero from 4 to 8 s, then reversed but the same size from 8 to 10 s — A student who thinks the emf depends only on the change of flux reasons that the rise and the fall are the same change, so the emfs match. The fall happens in 2 s rather than 4 s, so the flux changes twice as fast and the emf is larger.
3. Rising from 0 to 4 s, steady at its largest from 4 to 8 s, then falling to zero from 8 to 10 s — A student who thinks the emf follows the amount of flux describes the flux, not the emf. While the current is steady at 2.0 A the flux is largest but unchanging, so the emf is zero.
4. Zero throughout, because the small coil is at rest and nothing moves relative to it — A student who thinks an emf needs motion picks this. The field of the solenoid changes with its current, so the flux through the stationary coil changes and an emf is induced: a time-varying field needs no motion.

Working The flux through the small coil is proportional to the solenoid current, so the emf is proportional to the rate of change of current. 0 to 4.0 s: ΔI/Δt = (2.0 A)/(4.0 s) = +0.50 A s⁻¹, constant, so the emf is constant. 4.0 s to 8.0 s: ΔI/Δt = 0, so the emf is zero. 8.0 s to 10.0 s: ΔI/Δt = (−2.0 A)/(2.0 s) = −1.0 A s⁻¹, so the emf is reversed and twice the size it was from 0 to 4.0 s.

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1. Constant while entering, zero while wholly inside, then in the same direction while leaving — A student who thinks the induced current always opposes the field predicts the same direction both times. Lenz's law opposes the CHANGE: entering, the flux increases; leaving, it decreases, so the induced emf reverses.
2. Rising steadily while entering, largest while wholly inside, then falling to zero while leaving — A student who thinks the emf follows the amount of flux has described the flux through the coil. While the coil is wholly inside, the flux is greatest but not changing, so the emf is zero.
3. Constant while entering, zero while wholly inside, then reversed while leaving — While entering, the flux rises at a steady rate, so the emf is constant. While wholly inside a uniform field the flux is constant, so the emf is zero. While leaving, the flux falls at the same rate, so the emf has the same size but the opposite direction.
4. Constant and in one direction for the whole time that any part of the coil is in the field — A student who thinks any motion in a field induces an emf picks this. While the coil is wholly inside a uniform field its flux does not change, so no emf is induced even though it is moving.

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1. 1.2 V — A student who counts both sides perpendicular to the motion gets 2NBvL = 1.2 V. While the coil is entering, the trailing side is still outside the field, so only the leading side cuts field lines and has an emf.
2. 0.60 V — While entering, only the leading side cuts field lines: the trailing side is still outside the field, and the sides parallel to the motion move along their own length. The leading side moves perpendicular to its length and to the field. Each turn has ε = BvL = 0.20 × 0.60 × 0.050 = 0.0060 V; for 100 turns ε = 0.60 V. Check with Faraday: the coil takes 0.050/0.60 s to enter while the flux per turn rises by BL², giving the same result.
3. 0.050 V — A student who uses the total change of flux linked with the coil, NBL² = 100 × 0.20 × 0.050² = 0.050 Wb, without dividing by the time taken to enter gets this. The coil takes 0.083 s to enter, so ε = 0.050/0.083 = 0.60 V.
4. 0.0060 V — A student who leaves out the number of turns gets BvL = 0.20 × 0.60 × 0.050 = 0.0060 V, the emf of a single turn. The 100 turns are in series, so the emf is 100 times larger.

Working While entering, only the leading side (L = 0.050 m) cuts field lines (the trailing side is still outside the field; the sides parallel to the motion move along their own length). ε = NBvL = 100 × (0.20 T)(0.60 m s⁻¹)(0.050 m) = 0.60 V. Check with Faraday's law: time to enter Δt = 0.050 m / 0.60 m s⁻¹ = 0.0833 s; ΔΦ per turn = BL² = 0.20 × 0.0025 = 5.0 × 10⁻⁴ Wb; ε = NΔΦ/Δt = 100 × 5.0 × 10⁻⁴ / 0.0833 = 0.60 V.

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1. An emf is induced in the coil only once its ends are joined, so only then can the coil affect the magnet at all. — A student who treats emf and current as the same thing picks this. The emf is induced whether or not the ends are joined; what the complete circuit adds is a current, and it is the current that exerts a force on the magnet.
2. With the ends joined, the coil repels the magnet at every point of its motion, reducing its amplitude. — A student who thinks the coil always repels the magnet picks this. The force opposes the motion: it repels the magnet while it moves down towards the coil and attracts it while it moves up and away. A force that always repelled, depending only on the magnet's position, would remove no energy from the oscillation.
3. With the ends joined a current flows, and its force opposes the motion, transferring energy to the resistor. — An emf is induced in both cases, but only with a complete circuit does a current flow. By Lenz's law the current's magnetic force opposes the magnet's motion whichever way it moves, so work is done against it and the oscillation energy is transferred to internal energy of the resistor and coil.
4. The current takes its energy from the magnet's magnetism, so the weaker magnet oscillates less. — A student who thinks induction uses up the magnet picks this. The magnet does not lose its magnetism; the energy of the current comes from the oscillation's kinetic and elastic potential energy, through work done against the opposing magnetic force.

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1. An emf is induced opposite in direction to the current, making the current fall faster. — A student who thinks a self-induced emf always opposes the current picks this. It opposes the CHANGE: while the current is falling, the emf acts to keep it flowing, so it is in the same direction as the current.
2. An emf is induced in the same direction as the current, tending to keep the current flowing. — The falling current reduces the flux through the coil's own turns, so an emf is induced in the coil (self-induction). By Lenz's law it opposes the decrease, so it acts in the same direction as the current, tending to maintain it.
3. No emf is induced, because a coil's own changing field cannot induce an emf in it. — A student who thinks induction needs a separate magnet or coil picks this. The coil's own changing current changes the flux through its own turns, and Faraday's law applies to that change: an emf is induced in the coil itself.
4. An emf is induced whose size is proportional to the current, so it falls to zero as the current falls. — A student who thinks the emf depends on the amount of flux picks this. The flux is proportional to the current, but the emf depends on the RATE of change of flux. A steadily falling current makes the flux fall at a constant rate, so the emf is constant, in the same direction as the current, for as long as the current keeps falling steadily.

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1. 0.24 V — A student who uses v = fr gets v = 25 × 0.020 = 0.50 m s⁻¹ and ε = 2 × 40 × 0.10 × 0.50 × 0.060 = 0.24 V. In one revolution a side travels 2πr, so v = 2πfr.
2. 0.0096 V — A student who thinks the emf is set by the amount of flux gives the greatest flux linkage, NBA = 40 × 0.10 × 2.4 × 10⁻³ = 0.0096, which is in webers. The peak emf depends on how fast the flux changes, and occurs when the flux is zero.
3. 0.038 V — A student who leaves out the number of turns finds the peak emf of one turn, 2BvL = 0.038 V. The 40 turns are in series, so the peak emf is 40 times larger.
4. 1.5 V — Each 6.0 cm side is 0.020 m from the axis, so it moves at v = 2πfr = 2π × 25 × 0.020 = 3.14 m s⁻¹. When the plane is parallel to the field both sides move perpendicular to it, each with emf BvL, and these add around the loop: ε = 2NBvL = 2 × 40 × 0.10 × 3.14 × 0.060 = 1.5 V.

Working r = 4.0 cm / 2 = 0.020 m; v = 2πfr = 2π × 25 s⁻¹ × 0.020 m = 3.14 m s⁻¹. At the peak the two 6.0 cm sides move perpendicular to the field; each has ε = BvL = (0.10 T)(3.14 m s⁻¹)(0.060 m) = 0.0188 V, and the two add around each turn. Peak ε = 2NBvL = 2 × 40 × 0.0188 V = 1.5 V. (The 4.0 cm sides have no emf along them.)

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1. 1.6 V — A student who reads the period as the 10 ms between successive zeros gets a frequency of 100 Hz and 4.0 × 40/100 = 1.6 V. The emf passes through zero twice per cycle; the period is 20 ms.
2. 4.0 V — A student who thinks the peak emf depends only on N, B and A keeps 4.0 V. At a lower rotation frequency the flux changes more slowly, so the peak emf falls in proportion.
3. 5.0 V — A student who thinks the peak emf is proportional to the time period finds that the period rises from 20 ms (50 Hz) to 25 ms (40 Hz) and gets 4.0 × 25/20 = 5.0 V. The peak emf is proportional to the frequency, so slower rotation, with its smaller rate of change of flux, gives a smaller peak emf.
4. 3.2 V — The period is the time from one maximum to the next, 25 ms − 5 ms = 20 ms, so the original frequency is 50 Hz. The peak emf is proportional to the rotation frequency: 4.0 V × 40/50 = 3.2 V.

Working Time between successive maxima (5 ms and 25 ms) = 20 ms, so T = 0.020 s and f₁ = 1/T = 50 Hz. Peak emf ∝ f, so ε₂ = 4.0 V × (40 Hz / 50 Hz) = 3.2 V.

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1. 0.40 V — The emf comes from the gradient of the flux–time graph. From 0.50 s to 0.70 s the flux per turn falls by 0.40 mWb = 0.40 × 10⁻³ Wb in 0.20 s, a rate of 2.0 × 10⁻³ Wb s⁻¹. With 200 turns, |ε| = N ΔΦ/Δt = 200 × 2.0 × 10⁻³ = 0.40 V.
2. 0.080 V — 200 × 0.40 × 10⁻³ = 0.080 is N times the change in flux, with no division by the 0.20 s over which it happens. Faraday's law involves the rate of change of flux — the gradient of the graph — not the change itself. Divide by 0.20 s to get 0.40 V.
3. 0.0020 V — 0.40 × 10⁻³ Wb / 0.20 s = 2.0 × 10⁻³ V is the emf induced in one turn. The graph gives the flux through each turn, and the emfs of the 200 turns in series add: multiply by N to get 0.40 V.
4. 400 V — 200 × 0.40/0.20 = 400 treats the axis reading 0.40 as a flux in weber. The axis is labelled Φ / mWb, so the change is 0.40 × 10⁻³ Wb and the emf is 0.40 V. Read the unit and prefix off the axis label before substituting.

Working Between t = 0.50 s and t = 0.70 s the flux per turn falls in a straight line from 0.40 mWb to 0, so ΔΦ/Δt = (0 − 0.40 × 10⁻³ Wb)/(0.20 s) = −2.0 × 10⁻³ Wb s⁻¹ (the gradient of that segment of the graph). Faraday's law: ε = −N ΔΦ/Δt = −200 × (−2.0 × 10⁻³) = +0.40 V. Magnitude 0.40 V (2 s.f.).

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1. The field region is three times as wide as the coil. — The emf is +ε₀ while the coil enters (flux rising), zero while it is wholly inside (flux constant) and −ε₀ while it leaves (flux falling). Entering takes 0.20 s, so the coil's side is L = 0.20v. The emf is zero for 0.40 s, during which the leading edge moves from a distance L inside the region to the far edge, so W − L = 0.40v = 2L. Hence W = 3L.
2. The field region is exactly twice as wide as the coil. — This takes the 0.40 s of zero emf as the time for the whole field region to pass, giving W = 0.40v = 2L. But the flux is constant only once the trailing edge is inside, when the leading edge is already a distance L into the region. The zero-emf time corresponds to W − L, so W = 3L.
3. The flux through the coil is zero from t = 0.20 s to t = 0.60 s. — Zero emf means zero rate of change of flux, not zero flux. Between 0.20 s and 0.60 s the coil is wholly inside the field and the flux through it is constant at its greatest value, BA; that is exactly why no emf is induced.
4. The coil is outside the field from t = 0.20 s to t = 0.60 s. — Moving through a uniform field does not by itself induce an emf; the flux through the coil must change. From 0.20 s to 0.60 s the coil is wholly inside the field with constant flux, so the emf is zero even though the coil is moving in the field.

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1. It flows in the rod from P to Q, driving the rod along the rails faster. — A current from P to Q would give a magnetic force on the rod to the right, speeding it up and creating energy from nothing. Lenz's law and energy conservation require the opposite: the current flows from Q to P, the force on the rod is to the left, and whoever pulls the rod does the work that heats R.
2. It is zero, because the uniform field through the circuit is steady. — A steady, uniform field can still give a changing flux: the flux is BA, and the area of the circuit enclosed by the rod, the rails and R grows as the rod moves. The changing flux (equivalently, the rod cutting field lines at speed v) induces an emf BvL and a current flows.
3. It flows in the rod from Q to P, so that end P is at the higher potential. — As the rod moves right, the area of the circuit and the flux into the page both increase. By Lenz's law the induced current produces a field out of the page inside the circuit, so it circulates anticlockwise: up the rod from Q to P, left along the top rail, down through R and back along the bottom rail. The rod acts as the source, and current leaves it at P, so P is its positive end.
4. It flows in the rod from Q to P and exerts no force on the rod. — The direction is right, but a current-carrying rod in a magnetic field always experiences a force. Here it acts to the left, opposing the motion, which is why a pull is needed to keep the rod moving at constant speed; the work done by that pull is the energy dissipated in R.

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1. Clockwise, so that the loop pulls the magnet in and makes its approach faster. — If the loop attracted the magnet, the magnet would gain kinetic energy while the loop also gained electrical energy — energy from nothing. Lenz's law says the induced current opposes the change causing it: the loop repels the approaching N pole, which requires an anticlockwise current seen from above.
2. Anticlockwise, and the person moving the magnet does work against the loop. — The N pole is nearest the loop, so the magnet's field at the loop points downward and the downward flux is increasing. By Lenz's law the induced current produces an upward field inside the loop, which from above is an anticlockwise current. Its upper face acts as a north pole and repels the approaching magnet, so the person must do work to push it closer; that work becomes the electrical energy in the loop.
3. There is no current at all, because the loop is not made of a magnetic material. — Induction needs a conductor and a changing flux, not a magnetic material. As the magnet approaches, the flux through the copper loop increases, an emf is induced and, because the loop is closed, a current flows — anticlockwise as seen from above.
4. Anticlockwise, with the energy for the current coming from the magnet's own magnetism. — The direction is right, but the magnet is not weakened by inducing a current. The loop's induced field repels the approaching magnet, so the person pushing it does work against that force; this work, not the magnet's magnetism, is the source of the electrical energy.

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1. At Q the plane of the coil is perpendicular to the field. — At Q the emf is zero, so the flux through the coil is momentarily not changing: the flux is at a maximum in magnitude. Φ = BA cos θ is greatest when θ = 0, that is when the plane of the coil is perpendicular to the field lines, and at that instant the sides of the coil move parallel to the field and cut no field lines.
2. At Q the plane of the coil is parallel to the field. — Zero emf does not mean zero flux; it means the flux is momentarily not changing. With the plane parallel to the field the flux is zero but is changing at its fastest, which is where the emf is greatest (P), not zero. At Q the flux is at its maximum and the plane is perpendicular to the field.
3. The flux through the coil at P is the same as the flux at Q. — The field is uniform, but the flux BA cos θ through the coil changes continuously as θ changes. At P the coil's plane is parallel to the field and the flux is zero; at Q, a quarter of a turn later, the flux is at its maximum. It is this changing flux that produces the sinusoidal emf.
4. The time from P to Q is half of one time period. — One time period is one complete cycle: from a zero, through the peak, the next zero and the trough, back to a zero. From a peak (P) to the next zero (Q) is a quarter of a cycle, a quarter of the time period, corresponding to a quarter turn of the coil.

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1. 80 ms — Doubling the frequency does not double the time period. Frequency is the number of cycles per second and the period is the time for one cycle, so T = 1/f: doubling f halves T, from 40 ms to 20 ms.
2. 20 ms — One full cycle on the graph, from a zero through the peak, the next zero and the trough back to a zero, lasts 40 ms. The emf completes one cycle per revolution, so doubling the rotation frequency halves the time period to 20 ms.
3. 40 ms — Changing the frequency of rotation changes both the peak emf and the time period. The coil now takes half as long to turn once, and the emf completes one cycle per turn, so the period falls from 40 ms to 20 ms.
4. 10 ms — This halves 20 ms, but 20 ms is only the time from one zero to the next — half a cycle. One time period is a full cycle, including both the positive and the negative half: 40 ms on the graph. Halving it gives 20 ms.

Working From the graph, one complete cycle (zero → peak → zero → trough → zero) takes 40 ms: the emf is zero at 0, 20 and 40 ms, but 0 to 20 ms is only half a cycle. So T = 40 ms and f = 1/T = 25 Hz. The emf goes through one cycle per revolution, so doubling the frequency of rotation to 50 Hz halves the time period: T_new = 1/50 = 0.020 s = 20 ms. (The peak emf also doubles, to 6.0 V, but that is not asked.)

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