University Physics II · Final Examination

A thin, smooth insulating ring of radius R is held fixed, and gravity can be neglected. Point charges +Q are fixed at the two ends of one diameter of the ring, and a small bead carrying charge +q, with q > 0, is threaded on the ring so that it can slide freely round it. The bead is placed at a point on the ring a quarter of a turn from each fixed charge. Which statement is correct?

1. AThe electric field of the fixed charges is zero at that point, so the bead would stay there even if the ring were removed.
2. BThe field there is not zero but points radially outwards, at right angles to the ring, and a bead nudged a short way along the ring is pushed back towards that point.
3. CThe field there is not zero but points radially outwards, and a bead nudged a short way along the ring is pushed on towards the nearer fixed charge.
4. DThe bead cannot be in equilibrium at that point, because both fixed charges repel it; only a negatively charged bead could rest there.

In a region of space the electric potential is V(x, y) = αx²y − βy, with α = 20 V/m³ and β = 30 V/m. What is the electric field at the point (x, y) = (1.0 m, 2.0 m)?

1. A(−80 î + 10 ĵ) V/m
2. B(+80 î − 10 ĵ) V/m
3. C(−40 î + 10 ĵ) V/m
4. D(+10 î − 80 ĵ) V/m

A cordless drill runs from a battery pack. A high-resistance voltmeter across the pack's terminals reads 18.0 V while the drill is switched off and 16.8 V while the drill draws a steady current of 8.0 A. Model the pack as a constant emf in series with a constant internal resistance. At what rate is energy dissipated as heat inside the pack while the drill runs?

1. A144 W
2. B134 W
3. C9.6 W
4. D2.16 × 10³ W

Electrons leave a small source inside a long solenoid, each moving at 30° to the solenoid's axis. Inside the solenoid the magnetic field is uniform and parallel to the axis, and there is no electric field, so each electron follows a helix whose axis is parallel to the field. What is the ratio of the pitch of the helix, the distance an electron moves parallel to the axis during one revolution, to the radius of the helix?

1. A2π/√3
2. Bπ√3
3. C2π
4. D2π√3

A long straight hollow cylindrical conductor has inner radius 2.0 cm and outer radius 4.0 cm, and carries a steady current of 12 A distributed uniformly over the metal's cross-section. Take μ₀ = 4π × 10⁻⁷ T⋅m/A. What is the magnitude of the magnetic field at a distance of 3.0 cm from the axis?

1. Azero
2. B8.0 × 10⁻⁵ T
3. C4.5 × 10⁻⁵ T
4. D3.3 × 10⁻⁵ T

A strong cylindrical magnet is dropped down the centre of a long vertical copper pipe and soon falls at a steady speed far below that of free fall. Copper is not ferromagnetic. Which statement correctly explains the motion?

1. AThe magnet magnetises the copper, which then attracts it upwards in the way an iron pipe would.
2. BEddy currents flow only in the pipe just below the magnet, where the flux is increasing; above the magnet the flux is decreasing, so no current is induced there.
3. CEddy currents circulate in opposite senses in the pipe just below and just above the magnet; each exerts an upward force on the magnet, and the gravitational energy the magnet loses is dissipated in the pipe.
4. DEddy currents circulate in the same sense above and below the magnet, and the steady speed is reached once the pipe has warmed enough for its resistance to stop them growing.

A series RLC circuit is driven by a sinusoidal source at its resonant frequency f₀. The source frequency is then raised to 2f₀ with the source amplitude unchanged. Which statement describes the circuit at 2f₀? (X_L and X_C are the inductive and capacitive reactances.)

1. AX_L = 2X_C, and the current leads the source emf
2. BX_L = 4X_C, and the current lags the source emf
3. CX_L = 4X_C, and the current leads the source emf
4. DX_L = X_C still, because L and C have not changed, so the impedance is still R

The electric field of a plane electromagnetic wave in vacuum is E = E₀ sin(kz + ωt) ĵ, where E₀, k and ω are positive constants and c is the speed of light. Which expression gives the magnetic field of the wave?

1. AB = (E₀/c) sin(kz + ωt) î
2. BB = −(E₀/c) sin(kz + ωt) î
3. CB = cE₀ sin(kz + ωt) î
4. DB = (E₀/c) sin(kz + ωt) k̂

Light travelling in glass of refractive index 1.52 meets a flat boundary with water of refractive index 1.33. Above what angle of incidence, measured from the normal, is the light totally internally reflected?

1. A41.1°
2. B48.8°
3. C61.0°
4. DNo angle: total internal reflection requires the second medium to be air or vacuum

An electron is accelerated from rest through a potential difference of 150 V. Take h = 6.63 × 10⁻³⁴ J⋅s, mₑ = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C and c = 3.00 × 10⁸ m/s. What is the electron's de Broglie wavelength?

1. A8.29 nm
2. B0.142 nm
3. C0.00243 nm
4. D0.100 nm

The storage capacitor of a laboratory pulse generator is a hollow metal globe of radius b = 0.200 m with a metal ball of radius a = 0.120 m supported by an insulating rod at its centre. Transformer oil, of dielectric constant κ = 2.2 and dielectric strength 1.2 × 10⁷ V/m, fills the globe exactly up to the horizontal plane through its centre, so that the oil occupies the lower half of the gap between ball and globe; the upper half of the gap contains air, of dielectric constant 1.00 and dielectric strength 3.0 × 10⁶ V/m. The ball is charged positive, to a potential difference of 100 kV relative to the globe, and the capacitor is then disconnected from the supply. Treat the wall of the globe as thin and neglect the supporting rod. Take ε₀ = 8.85 × 10⁻¹² F/m.

1. (a)

   Show that Starting from Gauss's law, show that if air alone filled the whole gap the capacitance would be C₀ = 4πε₀ab/(b − a). Hence, by explaining why the field in the half-filled gap is radial and has the same magnitude in the oil and in the air at the same distance from the centre, show that the half-filled capacitor has C = 2π(1 + κ)ε₀ab/(b − a).

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   1. 1

      Gaussian sphere of radius r, a < r < b, concentric with the ball: by symmetry E is radial and uniform over it, so 4πr²E = Q/ε₀ and E = Q/(4πε₀r²)

   2. 2

      V = ∫ E dr from r = a to r = b = (Q/(4πε₀))(1/a − 1/b), so C₀ = Q/V = 4πε₀ab/(b − a)

   3. 3

      A radial field E = A/r² in each half lies along the flat oil–air boundary with no component across it; because the ball and the globe are equipotentials, ∫ E dr from a to b is the same V along every radius, which forces the same A in the oil and in the air and keeps the tangential field continuous at the boundary, so the two halves share one potential difference and act as capacitors in parallel

   4. 4

      The air half has capacitance C₀/2 and the oil half κC₀/2, so C = (1 + κ)C₀/2 = 2π(1 + κ)ε₀ab/(b − a)

2. (b)

   Calculate Calculate the capacitance, the charge on the ball, and the magnitude of the electric field in the gap next to the surface of the ball, when the potential difference is 100 kV.

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   1. 1

      C = 2π(3.2)(8.85 × 10⁻¹² F/m)(0.120 m)(0.200 m)/(0.080 m) = 5.34 × 10⁻¹¹ F (53.4 pF)

   2. 2

      Q = CV = (5.34 × 10⁻¹¹ F)(1.00 × 10⁵ V) = 5.34 × 10⁻⁶ C

   3. 3

      E(a) = Vb/(a(b − a)) = (1.00 × 10⁵ V)(0.200 m)/((0.120 m)(0.080 m)) = 2.08 × 10⁶ V/m, the same in the oil and in the air, and below the 3.0 × 10⁶ V/m that air withstands

3. (c)

   Determine Determine the free charge per unit area on the half of the ball's surface in contact with the oil and on the half in contact with the air, and check that together they carry the charge found in (b).

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   1. 1

      Just outside a conductor the free surface charge density is σ = κε₀E, with κ = 1.00 for the air, and E(a) is the same on both halves

   2. 2

      σ_air = (8.85 × 10⁻¹² F/m)(2.08 × 10⁶ V/m) = 1.84 × 10⁻⁵ C/m², and σ_oil = κσ_air = 4.06 × 10⁻⁵ C/m²

   3. 3

      Each half has area 2πa² = 9.05 × 10⁻² m², so Q = 2πa²(σ_oil + σ_air) = 5.34 × 10⁻⁶ C, agreeing with (b); the oil-covered half carries κ/(1 + κ) = 69% of it

4. (d)

   Predict With the capacitor still disconnected, the oil is drained out through a valve in the bottom of the globe and air takes its place. Predict the potential difference that the charge on the ball would produce with air alone in the gap, and whether the air could withstand it.

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   1. 1

      The charge stays at 5.34 × 10⁻⁶ C while the capacitance falls to C₀ = 2C/(1 + κ) = 33.4 pF, so the potential difference would rise by the factor (1 + κ)/2 = 1.6, to 1.60 × 10⁵ V

   2. 2

      The field at the surface of the ball would rise by the same factor, to 3.33 × 10⁶ V/m, above the 3.0 × 10⁶ V/m that air withstands, so the air cannot withstand it: breakdown starts next to the ball while the last of the oil is still draining, before the potential difference reaches 1.60 × 10⁵ V

An ideal battery of emf ε = 9.0 V, a switch and a resistor R₁ = 30 kΩ are connected in series from the battery's positive terminal to a node P. From node P back to the battery's negative terminal run two branches in parallel: one contains a resistor R₂ = 60 kΩ, the other an initially uncharged capacitor C = 220 μF. The switch is closed at t = 0.

1. (a)

   Determine Determine the current in R₁ immediately after the switch is closed, and a long time afterwards.

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   1. 1

      At t = 0 the uncharged capacitor has no potential difference across it, so P is at the potential of the negative terminal and I₁ = 9.0 V/30 kΩ = 3.0 × 10⁻⁴ A (0.30 mA)

   2. 2

      A long time later the capacitor carries no current, so R₁ and R₂ are in series and I₁ = 9.0 V/90 kΩ = 1.0 × 10⁻⁴ A (0.10 mA)

2. (b)

   Derive Using Kirchhoff's loop and junction rules, derive the differential equation obeyed by the capacitor voltage V(t), and hence obtain V(t) = V∞(1 − e^(−t/τ)), giving V∞ and τ in terms of ε, R₁, R₂ and C.

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   1. 1

      Loop rule round the battery, R₁ and the capacitor: ε − I₁R₁ − V = 0, so I₁ = (ε − V)/R₁

   2. 2

      Junction rule at P: I₁ = V/R₂ + C dV/dt

   3. 3

      Combining: C dV/dt = ε/R₁ − V(1/R₁ + 1/R₂), which rearranges to τ dV/dt = V∞ − V with V∞ = εR₂/(R₁ + R₂) and τ = CR₁R₂/(R₁ + R₂)

   4. 4

      Separates the variables and integrates with V(0) = 0: ln((V∞ − V)/V∞) = −t/τ, so V(t) = V∞(1 − e^(−t/τ))

3. (c)

   Calculate Calculate τ, and the time at which the capacitor voltage reaches 4.5 V.

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   1. 1

      τ = (220 × 10⁻⁶ F)(20 kΩ) = 4.4 s, with V∞ = 6.0 V

   2. 2

      4.5 V = (6.0 V)(1 − e^(−t/τ)) gives e^(−t/τ) = 0.25, so t = τ ln 4 = 6.1 s

4. (d)

   Evaluate A student argues that because the capacitor charges through R₁, the time constant must be R₁C = 6.6 s. Evaluate this argument.

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   1. 1

      The argument fails because R₂ diverts part of the current arriving at P: seen from the capacitor, the network is equivalent to a source εR₂/(R₁ + R₂) = 6.0 V behind R₁ and R₂ in parallel, 20 kΩ

   2. 2

      So τ = 4.4 s rather than 6.6 s, and the student's value would place the 4.5 V crossing at 9.1 s instead of 6.1 s, an error of 50% in a timing design

5. (e)

   Calculate Calculate the energy stored in the capacitor a long time after the switch is closed.

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   1. 1

      U = ½CV∞² = ½(220 × 10⁻⁶ F)(6.0 V)² = 3.96 × 10⁻³ J (4.0 mJ)

A long straight insulated wire lies in the plane of a flat rectangular coil of N = 200 turns. The coil's two long sides, each of length ℓ = 0.200 m, are parallel to the wire at distances a = 0.020 m and b = 0.080 m from it. The coil's ends are connected to an oscilloscope of very high input resistance. Take μ₀ = 4π × 10⁻⁷ T⋅m/A.

1. (a)

   Derive Starting from Ampère's law and the definition of magnetic flux, derive an expression for the magnetic flux through one turn of the coil when the wire carries a steady current I.

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   1. 1

      Ampère's law round a circle of radius r centred on the wire gives B(2πr) = μ₀I, so B = μ₀I/(2πr), directed perpendicular to the plane of the coil

   2. 2

      Because B varies across the coil, takes a strip of width dr and length ℓ parallel to the wire, so dΦ = Bℓ dr

   3. 3

      Φ = (μ₀Iℓ/2π)∫ dr/r from r = a to r = b = (μ₀Iℓ/2π) ln(b/a)

2. (b)

   Calculate Calculate the mutual inductance of the wire and the coil.

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   1. 1

      M = NΦ/I = Nμ₀ℓ ln(b/a)/(2π)

   2. 2

      M = 200 × (2.0 × 10⁻⁷ T⋅m/A)(0.200 m) ln 4 = 1.11 × 10⁻⁵ H (11.1 μH)

3. (c)

   Calculate The wire carries I(t) = I₀ sin(2πft) with I₀ = 15 A and f = 50 Hz. Calculate the amplitude of the emf induced in the coil.

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   1. 1

      ε = −M dI/dt = −2πfMI₀ cos(2πft), so the amplitude is 2πfMI₀

   2. 2

      ε₀ = 2π(50 Hz)(1.11 × 10⁻⁵ H)(15 A) = 5.2 × 10⁻² V (52 mV)

4. (d)

   Explain The coil is now closed through a resistor. At an instant when the wire current is in its positive direction and increasing, explain the direction of the induced current in the long side nearer the wire, and explain whether the wire attracts or repels the coil.

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   1. 1

      The flux of the wire's field through the coil is increasing, so by Lenz's law the induced current makes a field opposing it inside the coil, which requires the near side to carry current antiparallel to the wire's current

   2. 2

      Antiparallel currents repel and parallel currents attract; the near side, antiparallel and closer, feels the stronger force, so the net force pushes the coil away from the wire

5. (e)

   Predict The coil is slid within its plane until the wire lies along the coil's centre line, parallel to and midway between its long sides. Predict the emf induced by the same alternating current, giving the reason.

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   1. 1

      The wire's field points into the plane on one side of the wire and out of it on the other, with equal magnitudes at equal distances

   2. 2

      The two halves of the coil are mirror images about the wire, so the net flux is zero for any current: M = 0 and no emf is induced

A series circuit contains a resistor R = 25 Ω, an inductor L = 80 mH of negligible resistance and a capacitor C = 50 μF. It is driven by a sinusoidal source of rms emf 120 V whose frequency can be varied.

1. (a)

   Calculate Calculate the resonant frequency, and the rms current and average power delivered at resonance.

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   1. 1

      f₀ = 1/(2π√(LC)) = 1/(2π√((0.080 H)(50 × 10⁻⁶ F))) = 79.6 Hz

   2. 2

      At resonance Z = R, so I_rms = 120 V/25 Ω = 4.80 A and P = I_rms²R = 576 W

2. (b)

   Determine The source frequency is set to 50 Hz. Determine the impedance, the rms current, the phase angle between the current and the source emf, and the average power delivered, stating whether the current leads or lags the emf.

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   1. 1

      X_L = 2π(50 Hz)(0.080 H) = 25.1 Ω and X_C = 1/(2π(50 Hz)(50 × 10⁻⁶ F)) = 63.7 Ω

   2. 2

      Z = √((25 Ω)² + (25.1 Ω − 63.7 Ω)²) = 45.9 Ω, so I_rms = 120 V/45.9 Ω = 2.61 A

   3. 3

      tan φ = (X_L − X_C)/R = −1.54, so the phase angle is 57.0°, with the current leading the emf because below resonance the circuit is capacitive

   4. 4

      P = I_rms²R = (2.61 A)²(25 Ω) = 171 W, equivalently V_rms I_rms cos φ with power factor 0.544

3. (c)

   Sketch Sketch the phasor diagram for the circuit at 50 Hz, with the current phasor drawn horizontally, showing the rms voltage phasors across R, L and C and the source emf, each labelled with its magnitude.

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   1. 1

      V_R = 65.3 V drawn along the current phasor

   2. 2

      V_L = 65.7 V drawn 90° ahead of the current and V_C = 166 V drawn 90° behind it, visibly longer than V_L, for phasors rotating anticlockwise

   3. 3

      The source emf, 120 V, drawn as the vector sum of the three, lying 57° behind the current phasor so that the current leads

4. (d)

   Explain At resonance the rms voltage across the capacitor is 192 V, well above the 120 V rms of the source. Explain how this is consistent with Kirchhoff's loop rule and with conservation of energy.

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   1. 1

      The loop rule holds for instantaneous voltages, or for phasors added with their phases: at resonance V_L and V_C are equal in size and 180° apart, so they cancel and the whole source voltage appears across R

   2. 2

      No energy is created: the inductor and capacitor exchange stored energy back and forth each cycle with zero average power, and the only average power drawn from the source is I²R dissipated in the resistor

A compact telephoto camera lens is modelled as two thin lenses on a common axis: a converging lens of focal length f₁ = +10.0 cm, followed 6.0 cm behind it by a diverging lens of focal length f₂ = −12.0 cm. The image sensor sits behind the second lens. Use the real-is-positive convention, 1/u + 1/v = 1/f with magnification m = −v/u, in which a virtual object has a negative object distance u.

1. (a)

   Determine Light from a very distant object enters the first lens. Determine where the final image forms, and state whether it is real or virtual.

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   1. 1

      The first lens alone would form its image at its focal point, 10.0 cm behind it, which is 4.0 cm behind the second lens

   2. 2

      The second lens intercepts rays that are still converging, so that point is a virtual object for it, with u₂ = −4.0 cm

   3. 3

      1/v₂ = 1/f₂ − 1/u₂ = −1/12.0 + 1/4.0, so v₂ = +6.0 cm: a real image 6.0 cm behind the second lens, 12.0 cm behind the first

2. (b)

   Show that The distant object subtends an angle of 0.50° at the lens. Show that the final image is 0.131 cm tall, and hence that the combination behaves as a single lens of focal length 15.0 cm.

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   1. 1

      The first image has height h₁ = f₁θ = (10.0 cm)(8.73 × 10⁻³ rad) = 0.0873 cm

   2. 2

      The second lens magnifies it by m₂ = −v₂/u₂ = −(6.0 cm)/(−4.0 cm) = +1.50, so the final image is 1.50 × 0.0873 cm = 0.131 cm tall

   3. 3

      A single lens images a distant object at height fθ, so f_eff = (0.131 cm)/(8.73 × 10⁻³ rad) = 15.0 cm

3. (c)

   Compare Compare this combination with a single converging lens that forms an image of the same size of the same distant object, in terms of the distance from the front of the lens to the sensor.

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   1. 1

      A single lens giving the same image size must have f = 15.0 cm, so its sensor sits 15.0 cm behind the lens

   2. 2

      The combination places the sensor only 12.0 cm behind its front lens, 3.0 cm (20%) shorter for the same image size, because the diverging lens shifts the effective lens position 3.0 cm in front of the first lens

4. (d)

   Predict The camera is refocused on an object 2.00 m in front of the first lens, with the lens separation unchanged. Predict whether the sensor must move towards or away from the second lens, and by how much.

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   1. 1

      First lens: 1/v₁ = 1/10.0 − 1/200, so v₁ = 10.53 cm and the virtual object for the second lens is at u₂ = −(10.53 − 6.0) cm = −4.53 cm

   2. 2

      1/v₂ = −1/12.0 + 1/4.53 gives v₂ = 7.27 cm, so the sensor must move 1.27 cm away from the second lens

A student uses a transmission grating ruled with 450 lines per millimetre, illuminated at normal incidence, to look for the deuterium line in the light of a discharge lamp that contains hydrogen and deuterium in comparable amounts. The red Balmer line of ordinary hydrogen has wavelength 656.28 nm; the corresponding line of deuterium has wavelength 656.10 nm. The lamp also emits the blue-green Balmer line of hydrogen, of wavelength 486.13 nm.

1. (a)

   Calculate Calculate the grating spacing, and the highest order in which light of wavelength 656 nm can be observed.

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   1. 1

      d = (1.00 × 10⁻³ m)/450 = 2.22 × 10⁻⁶ m

   2. 2

      d sin θ = mλ with sin θ < 1 requires m < d/λ = 3.39, so the highest order is m = 3

2. (b)

   Calculate Calculate the angle at which the hydrogen line appears in second order.

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   1. 1

      sin θ = 2(656.28 × 10⁻⁹ m)/(2.222 × 10⁻⁶ m) = 0.591, so θ = 36.2°

3. (c)

   Show that By differentiating the grating equation, show that two lines separated in wavelength by a small amount Δλ appear in order m with an angular separation Δθ = mΔλ/(d cos θ).

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   1. 1

      Differentiates d sin θ = mλ with d and m fixed: d cos θ dθ = m dλ

   2. 2

      For small separations replaces the differentials by differences, Δθ = mΔλ/(d cos θ), with Δθ in radians

4. (d)

   Calculate Calculate the angular separation of the hydrogen and deuterium lines in second order.

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   1. 1

      Δθ = 2(0.18 × 10⁻⁹ m)/((2.222 × 10⁻⁶ m) cos 36.2°) = 2.0 × 10⁻⁴ rad (0.0115°)

5. (e)

   Determine Determine the minimum number of grating lines that must be illuminated to resolve the two lines in second order, and the corresponding minimum width of the illuminated region of the grating.

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   1. 1

      Resolving power λ/Δλ = mN, so N ≥ (656.19 nm)/(2 × 0.18 nm) = 1.82 × 10³ lines (1823)

   2. 2

      Width ≥ 1823/(450 lines per mm) = 4.1 mm

6. (f)

   Compare Compare working in third order with working in second order for this measurement.

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   1. 1

      Third order spreads the lines further: m/cos θ rises from 2.48 at 36.2° to 6.47 at 62.4°, so Δθ = 5.2 × 10⁻⁴ rad, about 2.6 times larger, and only 1216 lines (2.7 mm of grating) are needed to resolve the pair

   2. 2

      Against that, gives a supported cost of third order: the pair lies at 62.4°, near the limit of usable angles, and higher orders are usually fainter; or the third-order spectrum overlaps the fourth, so the 486.13 nm line appears in fourth order at 61.0°, only 1.3° from the red pair, and each line must be identified with care

In a photoelectric experiment, monochromatic light falls on a clean metal cathode in an evacuated tube. For each wavelength the reverse potential difference between anode and cathode is increased until the photocurrent just falls to zero, and this stopping potential V_s is recorded. Each stopping potential has an uncertainty of ±0.03 V, and each wavelength is known to ±1 nm. Take e = 1.60 × 10⁻¹⁹ C and c = 3.00 × 10⁸ m/s.

1. (a)

   State State the photoelectric equation relating the stopping potential V_s to the frequency f of the light.

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   1. 1

      eV_s = hf − φ, where φ is the work function of the cathode, so V_s against f is a straight line of gradient h/e and intercept −φ/e

2. (b)

   Evaluate Calculate the frequency of the light for each reading, and evaluate whether all six readings are consistent with a single straight line.

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   1. 1

      f = c/λ gives 10.00, 8.82, 7.89, 7.14, 6.52 and 6.00 × 10¹⁴ Hz for readings 1 to 6

   2. 2

      Readings 1, 2, 3, 5 and 6 lie within ±0.03 V of a straight line, but reading 4 lies about 0.10 V below it, more than three times its uncertainty, so it is anomalous and is excluded from the fit

3. (c)

   Determine Determine the Planck constant from the gradient of a graph of V_s against f.

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   1. 1

      Plots V_s against f, or fits a line, using readings 1, 2, 3, 5 and 6

   2. 2

      Gradient = (1.95 V − 0.28 V)/((10.00 − 6.00) × 10¹⁴ Hz) = 4.2 × 10⁻¹⁵ V⋅s, or 4.13 × 10⁻¹⁵ V⋅s by least squares (accept 4.0 to 4.3 × 10⁻¹⁵ V⋅s)

   3. 3

      h = e × gradient = 6.6 × 10⁻³⁴ J⋅s from the least-squares gradient, or 6.7 × 10⁻³⁴ J⋅s from the two end points (accept 6.4 to 6.9 × 10⁻³⁴ J⋅s)

4. (d)

   Determine Determine the work function of the cathode, in electronvolts, and the threshold wavelength beyond which no photoelectrons are emitted.

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   1. 1

      Extrapolating to f = 0 gives an intercept of about −2.19 V, so φ = 2.2 eV (accept 2.1 to 2.3 eV)

   2. 2

      f₀ = φ/h = 5.3 × 10¹⁴ Hz, so λ₀ = c/f₀ = 5.7 × 10² nm (accept 540 to 590 nm, or a value consistent with the candidate's own gradient and intercept)

5. (e)

   Estimate Estimate the uncertainty in your value of the Planck constant, stating the assumptions you make.

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   1. 1

      Takes the ±0.03 V uncertainty as the dominant one and draws the steepest and shallowest lines through the error bars of the extreme readings at 300 nm and 500 nm, with gradients of about 4.33 and 4.03 × 10⁻¹⁵ V⋅s

   2. 2

      Half the spread gives h = (6.6 ± 0.2) × 10⁻³⁴ J⋅s, an uncertainty of about 4%; the ±1 nm wavelength uncertainty shifts each frequency by at most 0.33% and the gradient by about 1%, small beside the 4% from the voltages

6. (f)

   Justify Justify the claim that these data cannot be explained by a classical wave model, in which the energy an electron absorbs grows with the intensity of the light and the time it is exposed.

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   1. 1

      In the wave model the maximum energy of an ejected electron should depend on intensity and exposure time, not on frequency, and light of any frequency should eventually eject electrons

   2. 2

      The data show the maximum kinetic energy eV_s rising linearly with frequency, and the line extrapolates to a threshold near 5.3 × 10¹⁴ Hz (about 570 nm) below which no electrons would be emitted: a frequency dependence and a threshold the wave model does not predict, but which follow if each electron absorbs a single quantum of energy hf

A parallel-plate capacitor has circular plates of radius R = 5.0 cm separated by d = 2.0 mm in vacuum. It is being charged by a steady current I = 0.40 A carried by long straight leads that meet the centres of the plates along their common axis. Neglect fringing, so that the electric field is uniform between the plates and zero outside the gap. Take ε₀ = 8.85 × 10⁻¹² F/m and μ₀ = 4π × 10⁻⁷ T⋅m/A.

1. (a)

   Explain Ampère's law without Maxwell's correction gives contradictory results for the magnetic field at a point beside the gap, depending on which surface is chosen to span a circular path around the axis. Explain the contradiction, and how the displacement current resolves it.

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   1. 1

      A flat disc spanning a circle round the axis in the midplane of the gap is crossed by no conduction current, but a surface with the same boundary that bulges out to cut one of the leads is pierced by the current I, so ∮B⋅dl would have to be both zero and μ₀I

   2. 2

      Maxwell added the displacement current ε₀ dΦ_E/dt; the electric flux through the gap is q/ε₀, so ε₀ dΦ_E/dt = dq/dt = I, and ∮B⋅dl = μ₀(I_c + ε₀ dΦ_E/dt) = μ₀I for both surfaces

2. (b)

   Derive Using the Ampère–Maxwell law, derive the magnitude of the magnetic field between the plates at a distance r < R from the axis.

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   1. 1

      By symmetry B is azimuthal and uniform in magnitude on a circle of radius r centred on the axis in a plane between the plates, so ∮B⋅dl = 2πrB

   2. 2

      No conduction current crosses the disc bounded by that circle; the field E = q/(ε₀πR²) is uniform, so Φ_E = πr²E = qr²/(ε₀R²)

   3. 3

      The displacement current through the disc is ε₀ dΦ_E/dt = Ir²/R²

   4. 4

      2πrB = μ₀Ir²/R², so B = μ₀Ir/(2πR²)

3. (c)

   Determine Determine the magnitude of the magnetic field in the midplane of the gap at a distance r > R from the axis, and state how it compares with the field at the same distance from one of the leads.

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   1. 1

      A disc of radius r > R encloses all the changing electric flux, so the displacement current through it is I and 2πrB = μ₀I, giving B = μ₀I/(2πr)

   2. 2

      This is identical to the field at distance r from the long lead: outside the gap the magnetic field cannot distinguish conduction current from displacement current

4. (d)

   Sketch Sketch a graph of the magnitude of B against r in the midplane of the gap, from r = 0 to r = 3R, marking the value at r = R.

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   1. 1

      B rises linearly from zero on the axis to a maximum at r = R, marked as μ₀I/(2πR) = 1.6 × 10⁻⁶ T

   2. 2

      Beyond r = R, B falls as 1/r, continuous at r = R, reaching one-third of the maximum (5.3 × 10⁻⁷ T) at r = 3R

5. (e)

   Show that At the edge of the gap, r = R, show that the Poynting vector S = E × B/μ₀ points radially inwards, and that the total power flowing in through the cylindrical surface of radius R and height d equals the rate of increase of the energy stored in the capacitor, d/dt(q²/2C).

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   1. 1

      E is directed along the axis from the positive plate, the same sense as the current in the leads, and B at r = R circulates in the sense given by the right-hand rule about that current; with the axis as ẑ, E × B lies along ẑ × φ̂ = −r̂, radially inwards

   2. 2

      |S| = EB/μ₀ = (q/(ε₀πR²))(I/(2πR)) = qI/(2π²ε₀R³)

   3. 3

      Power entering through the side of the gap = |S| × 2πRd = qId/(πε₀R²)

   4. 4

      With C = ε₀πR²/d, d/dt(q²/2C) = (q/C)(dq/dt) = qId/(πε₀R²), equal to the power entering: the energy reaches the capacitor through the sides of the gap, not along the wires

6. (f)

   Predict The capacitor is later discharged through the same leads. Predict the direction of the Poynting vector at the edge of the gap during the discharge.

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   1. 1

      E keeps its direction while the current, and so B, reverses, so S points radially outwards: energy leaves the gap through its sides

A long solenoid of radius a = 4.0 cm has n = 8000 turns per metre and carries a current I₀ = 20 A. Coaxial with it, outside the winding, a thin insulating ring of radius b = 6.0 cm and mass m = 0.50 g carries a charge Q = 0.50 μC spread uniformly around it. The ring can turn freely about the common axis, and the apparatus is in vacuum. The solenoid current is reduced steadily to zero over T = 0.50 s. Treat the solenoid as ideal, with no magnetic field outside its winding, and neglect the magnetic field of the ring itself. Take μ₀ = 4π × 10⁻⁷ T⋅m/A.

1. (a)

   Describe Describe the electric field induced while the current is falling: its direction, the shape of its field lines, and why it cannot be described by an electric potential.

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   1. 1

      By axial symmetry the induced field is azimuthal, with closed circular field lines centred on the axis and a magnitude that depends only on r; its sense is the one that would drive a current opposing the fall in flux, the same sense as the solenoid current

   2. 2

      Its circulation round a closed field line, ∮E⋅dl = −dΦ_B/dt, is not zero, whereas any field derived from a potential has zero circulation round every closed path, so no potential function exists

2. (b)

   Derive Using Faraday's law in the form ∮E⋅dl = −dΦ_B/dt, derive the magnitude of the induced electric field at distance r from the axis, for r < a and for r > a, in terms of the rate of change dB/dt of the field inside the solenoid.

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   1. 1

      Chooses a circular path of radius r centred on the axis, on which E is tangential and uniform, so ∮E⋅dl = 2πrE

   2. 2

      For r < a the flux enclosed is πr²B, so 2πrE = πr²|dB/dt| and E = (r/2)|dB/dt|

   3. 3

      For r > a only the solenoid's cross-section carries flux, πa²B, so 2πrE = πa²|dB/dt|

   4. 4

      E = (a²/(2r))|dB/dt|, which agrees with the inside result at r = a

3. (c)

   Calculate Calculate the magnetic field inside the solenoid before the current starts to fall, and the magnitude of the induced electric field at the ring while the current is falling.

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   1. 1

      B₀ = μ₀nI₀ = (4π × 10⁻⁷ T⋅m/A)(8000 m⁻¹)(20 A) = 0.201 T, so |dB/dt| = 0.201 T/0.50 s = 0.402 T/s

   2. 2

      At r = b: E = (a²/(2b))|dB/dt| = ((0.040 m)²/(2 × 0.060 m))(0.402 T/s) = 5.4 × 10⁻³ V/m

4. (d)

   Derive Using the rotational form of Newton's second law, τ = dL/dt, derive the angular momentum the ring has acquired once the current has fallen to zero, and show that it depends neither on T nor on b, provided b > a.

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   1. 1

      Every element of the ring's charge feels a tangential force at radius b, so the torque about the axis is τ = QEb = (Qa²/2)|dB/dt|

   2. 2

      Integrating over the whole fall in field, L = ∫τ dt = (Qa²/2)∫|dB/dt| dt = Qa²B₀/2

   3. 3

      The ring radius b has cancelled and only the total change in B appears, so neither b nor the duration T of the decrease affects L

5. (e)

   Calculate Calculate the angular speed of the ring once the current has fallen to zero.

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   1. 1

      L = (0.50 × 10⁻⁶ C)(0.040 m)²(0.201 T)/2 = 8.0 × 10⁻¹¹ kg⋅m²/s

   2. 2

      Moment of inertia mb² = (0.50 × 10⁻³ kg)(0.060 m)² = 1.8 × 10⁻⁶ kg⋅m², so ω = L/(mb²) = 4.5 × 10⁻⁵ rad/s

6. (f)

   Evaluate A student claims that the ring starts to turn because the solenoid's magnetic field pushes on its charges. Evaluate this claim.

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   1. 1

      The claim fails: a magnetic force qv × B needs moving charges and the ring starts from rest, and in any case the ideal solenoid's field is zero at r = b, so no magnetic force acts on the ring at any time

   2. 2

      The torque comes from the induced, non-conservative electric field, which exists at the ring even though B is zero there: what matters is the changing flux enclosed by the ring's path, not the field at the ring's own position