Chapter 6: Electromagnetic Induction

Rule 1: Magnetic flux — Φ = B A cos θ (B = field, A = area, θ = angle between B and the normal to the surface). For a coil of N turns, the total flux linked is N Φ.

• Easy : B = 0.20 T and A = 0.05 m² with the field along the normal ⇒ Φ = 0.010 Wb.
• Medium : If the normal makes 60° with B, Φ = 0.010 × 0.5 = 5.0 × 10⁻³ Wb.
• Exam : A circular coil of radius 0.10 m in a normal field of 0.50 T has Φ = 0.50 × π × 0.01 = 1.57 × 10⁻² Wb.

Rule 2: Ways of changing the flux — The flux changes if B changes, if the area A changes, or if the angle θ changes. Any one of these can induce an emf; a steady flux induces none.

• Easy : Moving a magnet in or out of a coil changes B.
• Medium : Pulling a loop out of a field changes A in the field.
• Exam : Rotating a coil in a field changes θ, which is the principle of the generator.

Spot the difference

Rule 1: Induced emf — ε = − N dΦ ÷ dt (N = number of turns). For a constant rate, ε = −N ΔΦ ÷ Δt. The induced current is I = ε ÷ R and the charge that flows is q = N ΔΦ ÷ R, which does not depend on the time taken.

• Easy : N = 200, B rises from 0.10 T to 0.50 T in 0.20 s through A = 0.010 m² ⇒ ΔΦ = 0.004 Wb and ε = 200 × 0.004 ÷ 0.20 = 4.0 V.
• Medium : For a coil of resistance 8 Ω, I = 0.50 A and the charge that flows is q = 200 × 0.004 ÷ 8 = 0.10 C.
• Exam : If Φ = 5t² − 2 (in Wb, t in s) for a single turn, then ε = −10t and its magnitude at t = 2 s is 20 V.

Rule 2: What affects the emf — A faster change of flux gives a larger emf, and more turns give a proportionally larger emf.

• Easy : Doubling the speed of the magnet doubles the emf.
• Medium : Doubling both the number of turns and the rate of change makes the emf four times as large.
• Exam : When the flux is steady, dΦ ÷ dt = 0 and there is no emf even if Φ is large.

Spot the difference

Rule 1: Finding the direction — If the flux through the coil is increasing, the induced current creates a field opposing the existing one; if the flux is decreasing, it creates a field in the same direction as the existing one.

• Easy : A north pole moving towards a coil makes the near face of the coil a north pole, which repels the magnet.
• Medium : When the magnet is pulled away, the near face becomes a south pole, which attracts the magnet and resists its removal.
• Exam : If a field into the page is increasing through a loop, the induced current creates a field out of the page, so it flows anticlockwise.

Rule 2: Why energy is conserved — Work must be done against the opposing force to move the magnet, and this work becomes electrical energy and heat in the coil. If the induced current helped the change instead, the effect would grow by itself and energy would be created from nothing.

• Easy : This is why a magnet falling in a copper tube reaches a steady slow speed.
• Medium : The mechanical work done in moving the magnet equals the electrical energy produced (apart from losses).
• Exam : Lenz's law gives the direction only; Faraday's law gives the size.

Spot the difference

Rule 1: Motional emf — ε = B l v (B, l and v mutually perpendicular). In a closed circuit of resistance R, I = B l v ÷ R, and the magnetic force opposing the motion is F = B I l = B² l² v ÷ R.

• Easy : l = 0.5 m, v = 4 m s⁻¹, B = 0.30 T ⇒ ε = 0.60 V.
• Medium : With R = 3 Ω, I = 0.20 A and F = 0.30 × 0.20 × 0.5 = 0.030 N.
• Exam : An aeroplane with wingspan 40 m at 250 m s⁻¹ in the vertical component of the Earth's field, 5 × 10⁻⁵ T, develops ε = 5 × 10⁻⁵ × 40 × 250 = 0.5 V.

Rule 2: Energy view — The power that must be supplied to keep the rod moving at a steady speed is P = F v, and it equals the electrical power I² R.

• Easy : P = 0.030 × 4 = 0.12 W.
• Medium : I²R = 0.04 × 3 = 0.12 W, so the two agree.
• Exam : If the circuit is open, no current flows, no force acts and no power is needed, although the emf still exists.

In the figure : A rod sliding on rails in a field B directed into the page; the induced current flows through the closed circuit.

Spot the difference

Rule 1: Induced emf — ε₂ = − M dI₁ ÷ dt (mutual) and ε = − L dI ÷ dt (self). For a long solenoid of n turns per metre, area A and length l, L = μ₀ n² A l.

• Easy : M = 0.05 H and dI₁/dt = 20 A s⁻¹ ⇒ ε₂ = 1.0 V in magnitude.
• Medium : L = 0.40 H and dI/dt = 10 A s⁻¹ ⇒ ε = 4.0 V.
• Exam : For n = 1000 m⁻¹, A = 2 cm² and l = 0.5 m, L = 4π × 10⁻⁷ × 10⁶ × 2 × 10⁻⁴ × 0.5 = 1.26 × 10⁻⁴ H.

Rule 2: Energy in an inductor — U = ½ L I². The self-induced emf opposes the rise or the fall of the current, so L acts like inertia in a circuit.

• Easy : L = 0.40 H and I = 5 A ⇒ U = 5.0 J.
• Medium : If this current falls to zero in 0.10 s, the average emf is 0.40 × 5 ÷ 0.10 = 20 V.
• Exam : An iron core increases L, because it increases the flux for the same current.

Spot the difference

Rule 1: The emf — The flux is Φ = NBA cos ωt, so ε = N B A ω sin ωt = ε₀ sin ωt, with peak value ε₀ = N B A ω and ω = 2πf. The effective (rms) value is ε₀ ÷ √2.

• Easy : N = 100, B = 0.20 T, A = 0.02 m² and f = 50 Hz (ω = 314 rad s⁻¹) ⇒ ε₀ = 100 × 0.2 × 0.02 × 314 = 126 V.
• Medium : The rms value is 126 ÷ 1.414 = 88.9 V.
• Exam : At ωt = 30° the emf is 126 × 0.5 = 63 V.

Rule 2: Position of the coil — The emf is largest when the plane of the coil is parallel to the field (the coil cuts the field lines fastest), and zero when the plane is perpendicular to the field (the flux is largest but is not changing at that instant).

• Easy : Doubling the speed of rotation doubles both ε₀ and the frequency.
• Medium : In India the mains frequency is 50 Hz, so the polarity changes 100 times per second.
• Exam : The emf is zero when the flux is greatest.

In the figure : The sinusoidal emf of a generator over one period T; the peak is ε₀.

Spot the difference