Source: Physics 212 Spring 2022, University of Illinois at Urbana-Champaign
Tags: rotating loop, AC generator, angular frequency, infinite wire induction, mutual induction, magnetic flux integral, motional EMF near a wire, force on a coil, PHYS 212, electromagnetism
Difficulty: Intermediate to Advanced Prerequisites: Parts 1 and 2 (Faraday's Law Fundamentals, Motional EMF), integration, Ampère's law (B field of an infinite wire), natural logarithm.
The conceptual ideas from Parts 1 and 2 now get put to quantitative use. Rotating loops in constant fields produce sinusoidal EMFs (this is how AC generators work). Loops near current-carrying wires experience non-uniform fields that require integration to find the flux. These are the harder exam problems, but they follow the same underlying logic: find Φ, differentiate, apply Lenz's law. If you can set up the integral for flux, the rest is calculus.
A rotating loop produces sinusoidal EMF with amplitude NBAω. A loop near an infinite wire requires integrating B = μ₀I/(2πr) across the loop's area to find the flux, then differentiating. Doubling the rotation frequency doubles the EMF amplitude but does not change the maximum flux.
Angular frequency (ω)
The rate of rotation in radians per second. Related to the period T by ω = 2π/T and to the frequency f by ω = 2πf.
Think of it as how fast the loop spins, measured in radians per second rather than revolutions per second.
AC generator (alternator)
A device that converts mechanical rotation into alternating EMF using Faraday's law. A loop (or coil of N turns) rotates in a constant magnetic field, producing ε = NBAω sin(ωt).
Flux from an infinite straight wire
The magnetic field of an infinite straight wire carrying current I is B = μ₀I/(2πr), where r is the perpendicular distance from the wire. This field is non-uniform, so calculating the flux through a nearby loop requires integration.
Mutual induction (conceptual)
When the changing current in one conductor induces an EMF in a nearby conductor. The infinite-wire-plus-loop problems are examples of mutual induction, though the formal treatment (mutual inductance M) comes later in the course.
A conducting loop (or any shape of wire) rotates at constant angular velocity ω in a uniform magnetic field B⃗.
The flux: If the loop has area A and its normal makes angle θ = ωt with B⃗: Φ(t) = BA cos(ωt)
The EMF: ε = −dΦ/dt = BAω sin(ωt)
For N turns: ε = NBAω sin(ωt)
Key relationships:
Maximum flux: Φ_max = BA (occurs when the loop face is perpendicular to B, θ = 0 or π)
Maximum EMF: ε_max = BAω (occurs when the loop face is parallel to B, θ = π/2 or 3π/2)
When the flux is at a maximum, the EMF is zero (and vice versa). They are 90° out of phase.
Maximum current: I_max = ε_max / R = BAω / R
A right triangle with base b = 24 cm, height h = 60 cm, resistance R = 2.5 Ω rotates around the y-axis with period T = 1.2 s in a constant field B = 1.7 T pointing in the +z⃗ direction.
Area of the triangle: A = ½ × b × h = ½ × 0.24 × 0.60 = 0.072 m²
Q10: Angular frequency. ω = 2π / T = 2π / 1.2 ≈ 5.236 rad/s
Q11: Maximum induced current. ε_max = BAω = (1.7)(0.072)(5.236) ≈ 0.641 V I_max = ε_max / R = 0.641 / 2.5 ≈ 0.256 A
Q12: Flux at t₁ = 0.45 s. At t = 0, the triangle is in the x-y plane (as shown), so its area normal points in the +z⃗ direction, aligned with B⃗. Therefore Φ(0) = BA (maximum), and: Φ(t) = BA cos(ωt) = (1.7)(0.072) cos(5.236 × 0.45) = 0.1224 cos(2.356) cos(2.356) ≈ cos(3π/4) ≈ −0.707 Φ ≈ 0.1224 × (−0.707) ≈ −0.0866 T·m² Magnitude: |Φ| ≈ 0.0866 T·m²
Q14: Relationship between Φ and I at t₀ = 0.3 s. t₀ = 0.3 s corresponds to ωt₀ = 5.236 × 0.3 = 1.571 ≈ π/2. At θ = π/2: cos(π/2) = 0, so Φ₀ = 0. At θ = π/2: sin(π/2) = 1, so ε is at its maximum, meaning I₀ = I_max. Answer: (b) Φ₀ = 0 and I₀ = I_max.
Q15: Effect of doubling the frequency. If ω doubles:
Φ_max = BA (unchanged, because B and A do not change)
ε_max = BAω (doubles, because ω doubles)
I_max = ε_max / R (doubles) Answer: (c) Φ_max remains the same and I_max doubles.
An infinite wire carries current I(t) along the y-axis. A rectangular loop of width W and length L is fixed in the x-y plane at a distance d from the wire.
Setting up the flux integral: The field from the wire at distance r is B = μ₀I/(2πr), pointing in the +z⃗ direction (out of the page, by the right-hand rule, for current in the +y⃗ direction and positions at +x).
The loop spans from x = d to x = d + L. A thin vertical strip at position x with width dx has area dA = W dx. The flux through this strip is:
dΦ = B(x) · W dx = (μ₀I / 2πx) · W dx
Integrating:
Φ = (μ₀IW / 2π) ∫ from d to (d+L) dx/x = (μ₀IW / 2π) ln[(d + L) / d]
The EMF: If I varies with time: ε = −dΦ/dt = −(μ₀W / 2π) ln[(d + L) / d] · dI/dt
The factor (μ₀W / 2π) ln[(d + L) / d] is a geometric constant that depends on the loop dimensions and its position. The only time-dependent part is dI/dt.
Given: I₁ = 3.1 A at t₁ = 17 s, I₄ = −3.1 A at t₄ = 34 s, with linear changes. W = 25 cm, L = 73 cm, d = 29 cm.
Q5: Flux at t = t₁ = 17 s. Φ = (μ₀ × 3.1 × 0.25 / 2π) ln[(0.29 + 0.73) / 0.29] = (4π × 10⁻⁷ × 3.1 × 0.25 / 2π) ln[1.02/0.29] = (2 × 10⁻⁷ × 3.1 × 0.25) ln(3.517) = 1.55 × 10⁻⁷ × 1.257 ≈ 1.95 × 10⁻⁷ T·m²
Q6: EMF at t = 8.5 s (during the linear increase from 0 to I₁). The current increases from 0 to 3.1 A over 17 s, so dI/dt = 3.1/17 ≈ 0.1824 A/s. ε = −(μ₀ × 0.25 / 2π) ln(3.517) × 0.1824 The geometric factor: (2 × 10⁻⁷ × 0.25) × 1.257 = 6.285 × 10⁻⁸ ε = −6.285 × 10⁻⁸ × 0.1824 ≈ −1.15 × 10⁻⁸ V
The sign indicates the current direction. With positive EMF defined as clockwise, a negative EMF means the induced current is counterclockwise.
Q7 and Q9 follow the same pattern, using the appropriate dI/dt from the graph for each time interval.
Q8: Direction at t = t₃ = 28 s. At t₃ = 28 s, the current is decreasing (passing through zero, going from positive to negative). dI/dt is negative. The flux is decreasing (since I is decreasing). The induced EMF is positive (opposing the decrease), meaning the induced current is clockwise.
An infinite wire carries a steady current I₁ in the +y⃗ direction. A short conducting wire of length W, aligned with the y-axis, is located at distance d from the infinite wire and moves in the −x⃗ direction at velocity v.
The motional EMF on the moving wire: The field at distance d from the infinite wire is B = μ₀I₁/(2πd), pointing in the +z⃗ direction (out of the page).
The motional EMF across the wire is: ε = vBW = v × [μ₀I₁/(2πd)] × W
As the wire moves closer to the infinite wire (d decreases), B increases, and so does the EMF.
Q27: EMF at t = 0 (d = 50 cm). ε = vBW = 0.12 × [μ₀ × 3.1/(2π × 0.50)] × 0.24 = 0.12 × [4π × 10⁻⁷ × 3.1/(2π × 0.50)] × 0.24 = 0.12 × [6.2 × 10⁻⁷ / 0.50] × 0.24 (Wait, let me redo this cleanly.) = 0.12 × [(2 × 10⁻⁷ × 3.1) / 0.50] × 0.24 = 0.12 × 1.24 × 10⁻⁶ × 0.24 = 0.12 × 2.976 × 10⁻⁷ ≈ 3.57 × 10⁻⁸ V
The sign depends on which end is at higher potential. Using F⃗ = qv⃗ × B⃗, with v⃗ = −v x⃗ and B⃗ = +B ẑ, the force on positive charges is in the +y⃗ direction (toward point b, the top). So point b is at higher potential than point a, and ε is negative by the problem's convention (positive means a higher than b). Actually, the problem says positive if the potential at point a is higher than at point b. The force pushes positive charges from a (bottom) to b (top), so b is higher. Therefore ε is negative.
Q28: EMF at t₁ = 2.8 s. At t₁ = 2.8 s, the wire has moved closer: d = 0.50 − 0.12 × 2.8 = 0.50 − 0.336 = 0.164 m. The calculation is the same with d = 0.164 m. Since d is smaller, B is larger, and the EMF magnitude is larger.
When a full rectangular loop moves toward or away from an infinite wire, the two vertical sides of the loop are at different distances from the wire, so they experience different values of B. The net EMF is the difference:
ε = vW × [μ₀I/(2π)] × [1/d₁ − 1/d₂]
where d₁ and d₂ are the distances of the near and far sides of the loop from the wire, and W is the side length parallel to the wire.
Alternatively, compute the flux as in the integral above and differentiate with respect to time, using the chain rule since d changes with time.
If the loop moves in the +y⃗ direction (parallel to the infinite wire), the distances d₁ and d₂ of the loop's sides from the wire do not change. The flux is constant. There is no induced EMF or current.
This is analogous to the "loop fully inside a uniform field" scenario from Part 2. The key is that the flux through the loop is not changing.
Quantity | Formula | Notes |
|---|---|---|
Rotating loop flux | Φ(t) = BA cos(ωt) | θ = ωt measured from alignment with B |
Rotating loop EMF | ε(t) = BAω sin(ωt) | Maximum when loop face is parallel to B |
Angular frequency | ω = 2π/T = 2πf | T = period, f = frequency |
Maximum EMF | ε_max = NBAω | N = number of turns |
Maximum current | I_max = ε_max / R | |
B field of infinite wire | B = μ₀I / (2πr) | r = perpendicular distance |
Flux through rect. loop near wire | Φ = (μ₀IW / 2π) ln[(d+L)/d] | W = side parallel to wire, L = side perpendicular, d = distance of near side |
EMF from time-varying current | ε = −(μ₀W / 2π) ln[(d+L)/d] × dI/dt | Geometric factor × rate of current change |
Motional EMF near infinite wire | ε = vW μ₀I / (2πd) | Single moving wire at distance d |
The rotating-loop equations describe exactly how an AC generator works. Power stations spin coils in magnetic fields to produce the 50 Hz (or 60 Hz) AC electricity that comes out of your wall socket. The amplitude of the voltage depends on NBAω, which is why generators use many turns of wire, strong magnets, and high rotation speeds. The infinite-wire-plus-loop scenario is a simplified model for mutual induction between parallel conductors, which matters for understanding crosstalk in electrical wiring and the design of transformers.
Students often think that doubling the rotation frequency doubles both the maximum flux and the maximum EMF. The maximum flux (Φ_max = BA) depends only on the field strength and loop area, not on how fast the loop spins. Only the EMF amplitude (which includes the ω factor) doubles.
When integrating flux from an infinite wire, students sometimes forget to integrate. They use Φ = BA with B evaluated at one distance, but the field is non-uniform across the loop's width. You must integrate.
In problems with piecewise-linear current, students sometimes try to use the value of I rather than dI/dt to find the EMF. The EMF depends on how fast the current is changing, not on what the current is at that moment. A large, steady current produces zero EMF.
A loop moving parallel to an infinite wire (same distance maintained) has no induced EMF. Students sometimes think that being near a current-carrying wire is enough, but the flux is not changing in this scenario.
⚠️ The rotating-loop problem almost always appears on the exam. Know the phase relationship: flux is maximum when EMF is zero, and vice versa.
⚠️ The effect of doubling frequency on Φ_max and ε_max is a common multiple-choice question.
⚠️ The infinite-wire flux integral (with the natural log) is a standard calculation. Practise setting it up and evaluating it.
⚠️ For piecewise-linear I(t) graphs, the EMF in each segment is constant (since dI/dt is constant within each linear segment). Read the slope directly from the graph.
⚠️ Be careful about which direction the loop moves relative to the wire. Moving toward/away from the wire changes the flux; moving parallel to it does not.
True or false: In a rotating loop, the EMF is at its maximum at the same instant the flux is at its maximum.
Fill in the blank: The maximum EMF produced by a rotating coil of N turns is ε_max = ______.
True or false: If the current in an infinite wire is constant, a nearby stationary loop has an induced EMF.
Fill in the blank: The magnetic flux through a rectangular loop near an infinite wire involves a ______ function (name the function) because the field is non-uniform.
True or false: Doubling the frequency of rotation of a generator doubles the maximum flux through the coil.
Answers: 1. False (they are 90° out of phase). 2. NBAω. 3. False (dI/dt = 0 means no EMF). 4. Natural logarithm (ln). 5. False (Φ_max = BA is independent of ω).
Q: A square loop of side 10 cm and resistance 5 Ω rotates at 120 rev/s in a 0.5 T field. What is the maximum induced current?
A: ω = 2π × 120 = 754 rad/s. A = (0.1)² = 0.01 m². ε_max = BAω = (0.5)(0.01)(754) = 3.77 V. I_max = 3.77/5 = 0.754 A.
Q: An infinite wire carries 5 A. A rectangular loop (W = 20 cm, L = 30 cm) has its near edge 10 cm from the wire. What is the magnetic flux through the loop?
A: Φ = (μ₀IW/2π) ln[(d+L)/d] = (4π×10⁻⁷ × 5 × 0.20)/(2π) × ln[(0.10+0.30)/0.10] = (2×10⁻⁷ × 5 × 0.20) × ln(4) = 2×10⁻⁷ × 1.386 = 2.77×10⁻⁷ T·m².
Q: For the same loop, if the current drops from 5 A to 0 A in 0.2 s (linearly), what is the average induced EMF?
A: dI/dt = −5/0.2 = −25 A/s. ε = −(μ₀W/2π) ln[(d+L)/d] × dI/dt = −(2×10⁻⁷ × 0.20) × ln(4) × (−25) = (4×10⁻⁸)(1.386)(25) = 1.39×10⁻⁶ V ≈ 1.39 μV.
Q: A rotating loop produces a maximum EMF of 12 V at 60 Hz. If the rotation speed is increased to 90 Hz, what is the new maximum EMF? Does the maximum flux change?
A: ε_max is proportional to ω (and therefore to f). New ε_max = 12 × (90/60) = 18 V. The maximum flux does not change (it depends on B and A, not on frequency).
The rotating-loop analysis leads directly into AC circuits, where sinusoidal voltages drive resistors, capacitors, and inductors. The infinite-wire flux integral reappears when you study mutual inductance (M), defined by Φ₂ = MI₁. The geometric factor (μ₀W/2π) ln[(d+L)/d] from this chapter is, in fact, the mutual inductance M between the wire and the loop. Self-inductance (L = NΦ/I) is the same idea applied to a coil's own flux.
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