Magnetic Flux Linkage NΦ, Faraday's & Lenz's Laws, rms Voltage/Current & Transformers
Electromagnetic induction is the generation of an electromotive force (EMF, ε) whenever a conductor cuts magnetic field lines or whenever the magnetic flux linkage (NΦ = BAN cos θ, in Webers, Wb) through a coil changes over time. While Faraday's Law gives the magnitude of the induced EMF as directly proportional to the rate of change of magnetic flux linkage (ε = N ΔΦ / Δt), Lenz's Law gives its direction: the induced EMF and any resulting current always act in a direction that opposes the change in flux that produced it, which is a direct consequence of the Principle of Conservation of Energy. Rotating a rectangular coil at angular speed ω inside a uniform magnetic field generates an alternating EMF ε = BANω sin(ωt), whose root-mean-square values (V_rms = V₀ / √2 and I_rms = I₀ / √2) allow step-up and step-down transformers to transmit electrical power across the National Grid with minimal I²R heating losses.
- Magnetic Flux (Φ = BA cos θ) & Flux Linkage (NΦ = BAN cos θ)
- Magnetic flux Φ through a loop of area A is defined as Φ = BA cos θ (units: Webers, Wb = T m²), where θ is the angle between the magnetic field lines and the normal (perpendicular line) to the plane of the coil. Warning: If the magnetic field is perpendicular to the plane of the coil, θ = 0° and cos(0°) = 1, giving maximum flux Φ = BA; if the field is parallel to the plane of the coil, θ = 90° and Φ = 0. For a coil of N turns, Magnetic Flux Linkage = NΦ = BAN cos θ (units: Wb or Wb turns).
- Faraday's Law of Electromagnetic Induction & Moving Rod (ε = BLv)
- Faraday's Law: the magnitude of the induced EMF in a circuit is directly proportional to (equal to) the rate of change of magnetic flux linkage, or the rate at which magnetic flux is cut: ε = N (ΔΦ / Δt). Graphical rule: the gradient of a Flux Linkage (NΦ) vs Time (t) graph equals the induced EMF (-ε), and the area under an EMF (ε) vs Time (t) graph equals the total change in flux linkage Δ(NΦ). For a straight conductor of length L moving at speed v perpendicular to a magnetic field B, in time Δt it sweeps out area ΔA = L v Δt, cutting flux ΔΦ = B L v Δt, so induced EMF ε = ΔΦ / Δt = BLv.
- Lenz's Law & Conservation of Energy (Falling Magnet & Eddy Currents)
- Lenz's Law: the direction of an induced EMF (and any induced current in a complete circuit) is always such as to oppose the change in magnetic flux that produces it (represented by the minus sign in ε = -N ΔΦ / Δt). Why? By Conservation of Energy: if a North pole is pushed into a solenoid, the end of the solenoid must become a North pole to repel the approaching magnet, requiring mechanical work (W = Fd) to be done pushing the magnet in, which is converted into the electrical energy of the induced current. (If it became a South pole, the magnet would accelerate itself in and create electrical energy from nothing). Similarly, a bar magnet dropped through a vertical copper tube induces circular eddy currents that repel the magnet below and attract it above, slowing its fall to a low terminal velocity.
Electric Field Simulator & Interactive NPN Transistor Builder
Adjust charges and distances to visualize Coulomb’s Law field lines, or switch to the Transistor Lab to wire and test an NPN circuit.
1. Coulomb’s Law Point-Charge Calculation
Formula: F = k · |Q₁ · Q₂| / r² where k = 1/(4πε₀) = 8.99 × 10⁹ N·m²·C⁻²
3.60 N
Attractive (towards each other)7.19e+6 N/C
At r/2 = 0.10 m2. Uniform Parallel-Plate Field Calculator
Formula: E = V / d (units: V m⁻¹ or N C⁻¹)
Key Concepts & Exam Definitions (6)
- Magnetic Flux (Φ = BA cos θ) & Flux Linkage (NΦ = BAN cos θ)
- Magnetic flux Φ through a loop of area A is defined as Φ = BA cos θ (units: Webers, Wb = T m²), where θ is the angle between the magnetic field lines and the normal (perpendicular line) to the plane of the coil. Warning: If the magnetic field is perpendicular to the plane of the coil, θ = 0° and cos(0°) = 1, giving maximum flux Φ = BA; if the field is parallel to the plane of the coil, θ = 90° and Φ = 0. For a coil of N turns, Magnetic Flux Linkage = NΦ = BAN cos θ (units: Wb or Wb turns).
- Faraday's Law of Electromagnetic Induction & Moving Rod (ε = BLv)
- Faraday's Law: the magnitude of the induced EMF in a circuit is directly proportional to (equal to) the rate of change of magnetic flux linkage, or the rate at which magnetic flux is cut: ε = N (ΔΦ / Δt). Graphical rule: the gradient of a Flux Linkage (NΦ) vs Time (t) graph equals the induced EMF (-ε), and the area under an EMF (ε) vs Time (t) graph equals the total change in flux linkage Δ(NΦ). For a straight conductor of length L moving at speed v perpendicular to a magnetic field B, in time Δt it sweeps out area ΔA = L v Δt, cutting flux ΔΦ = B L v Δt, so induced EMF ε = ΔΦ / Δt = BLv.
- Lenz's Law & Conservation of Energy (Falling Magnet & Eddy Currents)
- Lenz's Law: the direction of an induced EMF (and any induced current in a complete circuit) is always such as to oppose the change in magnetic flux that produces it (represented by the minus sign in ε = -N ΔΦ / Δt). Why? By Conservation of Energy: if a North pole is pushed into a solenoid, the end of the solenoid must become a North pole to repel the approaching magnet, requiring mechanical work (W = Fd) to be done pushing the magnet in, which is converted into the electrical energy of the induced current. (If it became a South pole, the magnet would accelerate itself in and create electrical energy from nothing). Similarly, a bar magnet dropped through a vertical copper tube induces circular eddy currents that repel the magnet below and attract it above, slowing its fall to a low terminal velocity.
- Rotating Coil AC Generator: ε = BANω sin(ωt)
- When a coil of N turns and area A rotates at constant angular velocity ω = 2πf inside a uniform magnetic field B (starting at t = 0 when the coil plane is perpendicular to B, θ = 0): Flux linkage is NΦ = BAN cos(ωt). Differentiating with respect to time gives the induced alternating EMF: ε = -d(NΦ)/dt = BANω sin(ωt), where Peak EMF ε₀ = BANω = 2πfBAN. Notice the 90° (π/2 rad) phase shift: when the coil is perpendicular to the field lines (vertical), NΦ is maximum (±BAN), but its gradient is zero so induced EMF ε is zero. When the coil is parallel to the field lines (horizontal), NΦ is zero, but the sides cut flux lines perpendicularly so induced EMF ε is maximum (±BANω)!
- Alternating Current (AC), rms Values & Oscilloscope (CRO) Traces
- Because sinusoidal AC reverses direction every half-cycle, its mean voltage and current over a full cycle are zero, but power dissipated in a resistor (P = I²R) is always positive and varies between 0 and Peak Power P₀ = I₀V₀ with Mean Power P_mean = ½I₀V₀. The Root-Mean-Square (rms) value of an AC current is the value of the steady direct current (DC) that would dissipate thermal energy at the exact same average rate in a given resistor: I_rms = I₀ / √2 and V_rms = V₀ / √2, so P_mean = I_rms × V_rms = ½I₀V₀ (for UK mains, V_rms = 230 V, f = 50 Hz, so Peak Voltage V₀ = 230√2 = 325 V and Peak-to-Peak Voltage = 650 V). On an oscilloscope: vertical y-gain (V div⁻¹) measures V₀ or V_pk-pk; horizontal time-base (s div⁻¹) measures period T (and f = 1/T).
- Transformers, Efficiency Sources of Loss & High-Voltage Grid Transmission
- An AC current in the primary coil (N_p turns) creates a continuously changing magnetic flux in the soft iron core, which links the secondary coil (N_s turns) and induces an alternating EMF by Faraday's law: V_s / V_p = N_s / N_p. For an ideal 100% efficient transformer (P_out = P_in): V_p I_p = V_s I_s, and Efficiency = (V_s I_s) / (V_p I_p) × 100%. Four causes of energy loss & fixes: 1. Eddy currents induced in the iron core (I²R heating) → fixed by using a LAMINATED core (thin iron sheets insulated from each other at 90° to the induced EMF). 2. Joule heating (I²R) in coil windings → fixed by using thick, low-resistivity copper wire. 3. Hysteresis loss (energy needed to repeatedly magnetise/demagnetise the core) → fixed by using magnetically soft iron. 4. Flux leakage → fixed by winding coils on top of one another on a closed loop core. In the National Grid, step-up transformers boost V to 400 kV so that for a given power P = VI, the transmission current I = P / V is very small, massively reducing power wasted in cables (P_loss = I²R ∝ 1/V²).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State Faraday's Law and Lenz's Law of electromagnetic induction.
Explain how Lenz's Law is an example of the Principle of Conservation of Energy.
Why is the induced EMF in a rotating generator coil maximum when the magnetic flux linkage through the coil is zero?
If the rotational speed (frequency f) of an AC generator coil is doubled, state and explain two changes to the sinusoidal EMF output trace on an oscilloscope.
Define the root-mean-square (rms) value of an alternating current, and calculate the peak voltage of the 230 V UK mains supply.
State three causes of energy loss in a real transformer and how each is minimised in transformer design.
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
An aircraft with a wingspan of 60.0 m flies horizontally at a constant speed of 250 m s⁻¹ in a region where the Earth's magnetic field has a flux density of 5.20 × 10⁻⁵ T directed at an angle of 68.0° below the horizontal (angle of dip). Calculate the EMF induced between the wingtips of the aircraft, and explain why this EMF cannot be measured simply by connecting a voltmeter inside the cabin between the two wingtips with copper wires. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Only the vertical component of the Earth's magnetic field is cut perpendicularly by the horizontally moving wings: B_vertical = B sin(68.0°) = (5.20 × 10⁻⁵ T) × 0.9272 = 4.821 × 10⁻⁵ T [1 mark]. 2. Use ε = B_vertical × L × v to calculate the induced EMF across the wings: ε = (4.821 × 10⁻⁵ T) × 60.0 m × 250 m s⁻¹ [1 mark]. 3. ε = 0.723 V [1 mark]. 4. The copper wires connecting the wingtips to the voltmeter inside the cabin move along with the aircraft at the exact same speed (250 m s⁻¹) and cut the same vertical magnetic flux at the same rate, inducing an equal and opposite EMF (0.723 V) in the leads that cancels out the wingtip EMF around the closed loop (voltmeter reads 0 V) [1 mark].
- Q2
A circular search coil of 500 turns and diameter 24.0 mm is placed with its plane perpendicular to a uniform magnetic field of flux density 0.350 T. The coil has a total resistance of 12.0 Ω and is connected to a data logger. In a time of 40.0 ms, the coil is flipped through 180° so its plane is again perpendicular to the field. Calculate: (a) the initial magnetic flux linkage, (b) the magnitude of the mean EMF induced during the 180° flip, and (c) the total electric charge that flows around the coil during the flip. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Calculate cross-sectional area A = πr² = π × (0.0120 m)² = 4.524 × 10⁻⁴ m², and initial flux linkage (θ = 0°, cos 0° = +1): (NΦ)_initial = BAN = 0.350 × (4.524 × 10⁻⁴) × 500 = +0.0792 Wb turns (7.92 × 10⁻² Wb) [1 mark]. 2. (b) When flipped through 180° (cos 180° = -1), final flux linkage is (NΦ)_final = -0.0792 Wb turns, so the magnitude of the change in flux linkage is Δ(NΦ) = 2 × BAN = 2 × 0.07917 = 0.1583 Wb turns [1 mark]. 3. By Faraday's Law, mean induced EMF ε = Δ(NΦ) ÷ Δt = 0.1583 Wb ÷ 0.0400 s = 3.96 V [1 mark]. 4. (c) Mean induced current I = ε ÷ R = 3.958 V ÷ 12.0 Ω = 0.330 A [1 mark]. 5. Total charge ΔQ = I Δt = Δ(NΦ) ÷ R = 0.1583 ÷ 12.0 = 0.0132 C (13.2 mC, independent of how fast the coil was flipped) [1 mark].
- Q3
A rectangular generator coil of 250 turns and dimensions 0.120 m by 0.080 m rotates at 3000 revolutions per minute in a uniform magnetic field of flux density 0.450 T. The coil is connected to an external load resistor of resistance 40.0 Ω (coil resistance is negligible). Calculate the frequency of rotation, the peak EMF induced, the rms current in the resistor, and the mean power dissipated in the resistor. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Frequency of rotation f = 3000 rev min⁻¹ ÷ 60 s min⁻¹ = 50.0 Hz, so angular speed ω = 2πf = 100π = 314.2 rad s⁻¹ [1 mark]. 2. Coil area A = 0.120 m × 0.080 m = 9.60 × 10⁻³ m²; Peak EMF ε₀ = BANω = 0.450 × (9.60 × 10⁻³) × 250 × 314.2 = 339 V [1 mark]. 3. Calculate rms voltage V_rms = ε₀ ÷ √2 = 339.3 ÷ √2 = 240 V [1 mark]. 4. Calculate rms current I_rms = V_rms ÷ R = 240 V ÷ 40.0 Ω = 6.00 A [1 mark]. 5. Mean power dissipated in the 40.0 Ω resistor: P_mean = I_rms × V_rms = (I_rms)² R = (6.00)² × 40.0 = 1440 W (1.44 kW) [1 mark].
- Q4
A power station generates 120 MW of electrical power, which is transmitted to a town through overhead cables that have a total resistance of 2.50 Ω. Calculate the power wasted as heat in the cables if the power is transmitted at 25.0 kV versus if a step-up transformer is used to transmit the power at 400 kV, and explain why alternating current (AC) rather than direct current (DC) is used in the National Grid. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. At V₁ = 25.0 kV (2.50 × 10⁴ V), transmission current I₁ = P ÷ V₁ = (120 × 10⁶ W) ÷ (2.50 × 10⁴ V) = 4800 A [1 mark]. 2. Power wasted in the 2.50 Ω cables at 25.0 kV: P_loss,1 = I₁² R = (4800)² × 2.50 = 5.76 × 10⁷ W = 57.6 MW (48% of the generated power is lost) [1 mark]. 3. At V₂ = 400 kV (4.00 × 10⁵ V, 16 times higher voltage), transmission current I₂ = (120 × 10⁶ W) ÷ (4.00 × 10⁵ V) = 300 A (16 times lower current) [1 mark]. 4. Power wasted in the cables at 400 kV: P_loss,2 = I₂² R = (300)² × 2.50 = 2.25 × 10⁵ W = 0.225 MW (256 times less power wasted) [1 mark]. 5. Alternating current (AC) is used because transformers only work with a continuously changing magnetic flux (ε = N ΔΦ / Δt); a steady DC current produces a constant magnetic flux (ΔΦ / Δt = 0) so no EMF would be induced in the secondary coil to step the voltage up and down [1 mark].
Common Exam Questions (4)
When a coil is flipped through 180° in a magnetic field, why is the change in flux linkage 2BAN instead of zero?
Magnetic flux is directional relative to the coil's faces. If the magnetic field enters Face A initially, the flux linkage is +BAN; after flipping the coil upside down through 180°, the field enters the opposite face so the flux linkage is -BAN. Therefore, the change in flux linkage is Δ(NΦ) = (+BAN) - (-BAN) = 2BAN.
In transmission line calculations, why must I never use P_loss = V² / R with the 400 kV supply voltage V?
This is the #1 exam trap in Electricity. The 400 kV is the voltage between the live cable and neutral/earth (across the town load + cable), not the potential drop ΔV = IR along the cable itself. Always calculate the current first using I = P_transmitted / V_supply, and then calculate cable power loss using P_loss = I² R_cable.
What happens if you drop a strong neodymium bar magnet down a vertical copper pipe vs a vertical plastic pipe of the same size?
Plastic is an electrical insulator, so although an EMF is induced as the magnet falls, no induced current can flow and the magnet falls with acceleration g = 9.81 m s⁻². Copper is a good non-magnetic conductor: the changing flux induces strong circular eddy currents in the copper walls above and below the magnet; by Lenz's Law, these currents create magnetic fields that oppose the magnet's motion (repelling below, attracting above), so the magnet quickly reaches a very slow terminal velocity.
How do I read Peak Voltage (V₀), rms Voltage (V_rms), and Frequency (f) from an oscilloscope screen?
1. Count the number of vertical grid divisions from the central horizontal equilibrium line to the wave crest (or half the peak-to-peak divisions) and multiply by the y-gain setting (V div⁻¹) to get V₀, then divide by √2 to get V_rms. 2. Count the number of horizontal grid divisions for one complete wave cycle and multiply by the time-base setting (s div⁻¹) to get time period T, then calculate f = 1 / T.
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