Unit 4: Electromagnetism

14–20% of AP Exam · 6 Topics · ~15 Practice Questions

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4.1 Magnetic Flux

Magnetic flux measures the total B-field piercing a surface:

$$\Phi_B = \int \vec{B} \cdot d\vec{A}$$

For uniform field and flat surface: \(\Phi_B = BA\cos\theta\), where \(\theta\) is the angle between \(\vec{B}\) and the surface normal \(\hat{n}\). Units: webers (Wb = T·m²).

  • \(\theta = 0°\) (B ∥ normal): max flux \(\Phi_B = BA\)
  • \(\theta = 90°\) (B ∥ surface): zero flux
  • \(\theta = 180°\) (B anti-parallel): \(\Phi_B = -BA\)
Exam Tip The induced EMF depends on the rate of change of flux, not the flux value. A large but constant flux produces zero EMF.

4.2 Faraday's Law

Changing magnetic flux through a loop induces an EMF:

$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$

For a coil with N turns: \(\mathcal{E} = -N\frac{d\Phi_B}{dt}\)

Flux can change through three mechanisms:

  1. Changing B: moving a magnet, turning an electromagnet on/off
  2. Changing area: deforming loop, sliding rod on rails
  3. Changing orientation: rotating the loop (electric generators!)
Common Mistake Faraday's law magnitude = \(|d\Phi_B/dt|\). The minus sign (Lenz's law) determines direction. Keep magnitude and direction analysis separate.
Practice

Q4.1: Solenoid + Outer Loop

A long solenoid (500 turns/m, radius 2.0 cm) carries \(I(t) = 5.0t\) A. A 10-turn loop of radius 3.0 cm surrounds it concentrically. Find the induced EMF at t = 2.0 s.

Show Solution
B inside solenoid: \(B = \mu_0 n I = \mu_0(500)(5t) = \mu_0\cdot2500t\). Only B inside solenoid contributes to flux through loop.
\(\Phi_B = B\cdot\pi r_{\text{sol}}^2 = \mu_0\cdot2500t\cdot\pi(0.02)^2\). \(d\Phi_B/dt = \mu_0\cdot2500\cdot\pi(0.02)^2\).
\(\mathcal{E} = -N\frac{d\Phi_B}{dt} = -10\cdot(4\pi\times10^{-7})(2500)\pi(4\times10^{-4}) = -3.95\times10^{-5}\ \text{V}\).

4.3 Lenz's Law

Lenz's Law: The induced current's magnetic flux opposes the change in the original flux. This is the minus sign in Faraday's law — nature resists changes in magnetic flux.

3-step strategy:

  1. Identify direction of original \(\Phi_B\) and whether it's increasing or decreasing
  2. The induced B-field must oppose the change:
    • \(\Phi_B\) increasing → induced B opposite to original B
    • \(\Phi_B\) decreasing → induced B same direction as original B
  3. Use RHR to find current direction that creates the induced B

alen: N pole approaching loop → increasing flux into loop → induced B points out → current = CCW as viewed from magnet.

Critical "Opposes the change" ≠ "opposes the flux." If flux is decreasing, induced B helps the original B (to slow the decrease).
Practice

Q4.2: Lenz's Law

A bar magnet's south pole is pulled away from a circular loop. Determine the induced current direction as viewed from the magnet side.

Show Solution
S pole faces loop → B lines go into the loop. Pulling away means flux into loop is decreasing.
To oppose the decrease, induced B must also point into loop (reinforce original).
RHR: thumb into page, fingers curl → clockwise as viewed from magnet side.

4.4 Motional EMF

When a conductor moves through B, charges experience magnetic force → separation → induced EMF:

$$\mathcal{E} = B\ell v \quad \text{(rod ⟂ B, motion ⟂ both)}$$

Sliding rod on rails: The rod's motion changes loop area → changing flux:

$$\mathcal{E} = B\ell v, \quad I = \frac{B\ell v}{R}, \quad P = \frac{B^2\ell^2 v^2}{R}$$

Magnetic force on induced current opposes motion: \(F_B = I\ell B = B^2\ell^2 v/R\). To maintain constant v, apply external force \(F_{\text{ext}} = F_B\).

Exam Tip Motional EMF can be derived from Faraday (\(\Phi_B = B\ell x\), so \(d\Phi_B/dt = B\ell v\)) or from \(F_B = qvB\) in the moving rod. Both are valid.

Electric Generators: A coil rotating in a uniform B-field produces sinusoidal EMF:

$$\mathcal{E} = NBA\omega\sin\omega t = \mathcal{E}_{\text{max}}\sin\omega t$$

where N = turns, A = coil area, ω = angular frequency. Peak EMF: \(\mathcal{E}_{\text{max}} = NBA\omega\). This is the principle behind AC generators (alternators).

Practice

Q4.3: Sliding Rod

A 0.30 m rod at 4.0 m/s on frictionless rails across 0.50 T B ⟂ to plane. R = 2.0 Ω. Find EMF, current, and force to maintain speed.

Show Solution
\(\mathcal{E} = B\ell v = (0.5)(0.3)(4) = 0.60\ \text{V}\). \(I = \mathcal{E}/R = 0.30\ \text{A}\).
\(F = I\ell B = (0.3)(0.3)(0.5) = 0.045\ \text{N}\) opposing motion. External force = 0.045 N in direction of motion.

4.5 Inductance

Self-inductance L describes opposition to current change:

$$L = \frac{N\Phi_B}{I}, \quad \mathcal{E}_L = -L\frac{dI}{dt}$$

Units: henries (H = V·s/A). For a solenoid: \(L = \mu_0 n^2 A\ell = \mu_0 N^2 A/\ell\).

Inductor behavior at extremes:

  • t = 0 (switch closed): inductor = open circuit (I cannot jump)
  • t → ∞ (steady DC): inductor = short circuit (V_L = 0)

Energy stored: \(U = \frac{1}{2}LI^2\) — dual to capacitor \(U = \frac{1}{2}CV^2\).

Common Mistake At t=0 inductor is open, capacitor is short. At t→∞ inductor is short, capacitor is open. They are opposites — test both extremes to verify your understanding.

Energy density of the magnetic field:

$$u_B = \frac{B^2}{2\mu_0}$$

This is the magnetic analog of \(u_E = \frac{1}{2}\varepsilon_0 E^2\). For a solenoid: \(U = u_B \cdot A\ell = \frac{B^2}{2\mu_0}A\ell = \frac{1}{2}LI^2\).

Mutual Inductance (M): When changing current in coil 1 induces EMF in coil 2:

$$\mathcal{E}_2 = -M\frac{dI_1}{dt}$$

M depends on geometry of both coils and their relative position. Units: henries (H). The basis for transformers.

Practice

Q4.4: Solenoid Inductance

Solenoid: 800 turns, length 25 cm, radius 4.0 cm. Find L. If current changes at 3.0 A/s, find induced EMF.

Show Solution
\(A = \pi(0.04)^2 = 5.027\times10^{-3}\ \text{m}^2\). \(L = \mu_0 N^2 A/\ell = (4\pi\times10^{-7})(640000)(5.027\times10^{-3})/0.25 = 16.2\ \text{mH}\).
\(|\mathcal{E}| = L|dI/dt| = (0.0162)(3) = 0.0486\ \text{V}\).

4.6 RL & LC Circuits

RL Circuits — time constant \(\tau_L = L/R\):

Current growth (connected to battery):

$$I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-Rt/L}\right) = I_f\left(1 - e^{-t/\tau_L}\right)$$

Current decay (battery removed):

$$I(t) = I_0 e^{-Rt/L} = I_0 e^{-t/\tau_L}$$

RL and RC are mathematically identical — I in RL ↔ V in RC. Both first-order exponentials.

LC Circuits — no resistance, perpetual energy oscillation:

$$ \omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}} $$

\(q(t) = Q_0\cos(\omega_0 t)\), \(I(t) = -\omega_0 Q_0\sin(\omega_0 t)\). Energy: \(U_{\text{total}} = \frac{Q_0^2}{2C} = \frac{1}{2}LI_{\text{max}}^2\) (constant).

Mechanical analogy: L ↔ mass (inertia), C ↔ 1/k (spring), \(\omega_0 = \sqrt{k/m} \leftrightarrow \omega_0 = 1/\sqrt{LC}\)

Practice

Q4.5: RL Transient

RL circuit: R = 100 Ω, L = 200 mH, ε = 12 V. Find τ_L, I_final, I at t=2τ_L, and U stored at steady state.

Show Solution
\(\tau_L = L/R = 0.2/100 = 2.0\ \text{ms}\). \(I_f = 12/100 = 0.12\ \text{A}\).
\(I(2\tau) = 0.12(1-e^{-2}) = 0.104\ \text{A}\). \(U = \frac{1}{2}LI_f^2 = 0.5(0.2)(0.0144) = 1.44\ \text{mJ}\).
Practice

Q4.6: LC Oscillation

LC: L = 5.0 mH, C = 20 μF, Q_0 = 200 μC. Find f, T, I_max, and total energy.

Show Solution
\(\omega_0 = 1/\sqrt{LC} = 3162\ \text{rad/s}\). \(f = \omega_0/(2\pi) = 503\ \text{Hz}\). \(T = 1.99\ \text{ms}\).
\(I_{\text{max}} = \omega_0 Q_0 = 0.632\ \text{A}\). \(U = Q_0^2/(2C) = 1.0\ \text{mJ}\).

4.7 Applications: Transformers & Eddy Currents

Transformers: Two coils wound on a shared iron core. AC in the primary creates changing flux, inducing EMF in the secondary:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

For an ideal transformer: \(P_p = P_s \Rightarrow V_p I_p = V_s I_s\). Step-up: \(N_s > N_p\) (higher voltage, lower current). Step-down: \(N_s < N_p\). Used in power distribution.

Eddy Currents: Changing flux through bulk conductors induces circulating currents that dissipate energy as heat (\(I^2R\)). Applications: induction braking (trains), metal detectors, induction cooktops. Lamination of iron cores reduces eddy current losses.

4.8 RLC Circuits (Qualitative)

Adding resistance to an LC circuit produces damped oscillation. Current oscillates at:

$$\omega' = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$$

(derived from LRC differential equation; AP expects qualitative understanding only).

  • Underdamped (small R): oscillations decay exponentially
  • Critically damped (R² = 4L/C): fastest return to equilibrium without oscillation
  • Overdamped (large R): slow, non-oscillatory decay

Resonance: At the natural frequency \(\omega_0 = 1/\sqrt{LC}\), the impedance is minimized and current amplitude is maximized — the basis for radio tuning.

Exam Tiplam RLC is primarily tested qualitatively: "does the oscillation frequency increase or decrease when L is increased?" Know the trends.
Practice

Q4.7: Generator EMF

A 200-turn coil of area 0.015 m² rotates at 60 rev/s in a 0.40 T B-field. Find (a) the peak EMF and (b) the frequency of the output.

Show Solution
ω = 2π(60) = 377 rad/s. \(\mathcal{E}_{\text{max}} = NBA\omega = (200)(0.4)(0.015)(377) = 452\ \text{V}\).
Output frequency = rotation frequency = 60 Hz.
Practice

Q4.8: Transformer

A step-down transformer has 500 primary turns and 50 secondary turns. Primary voltage is 120 V (rms). Find the secondary voltage and the secondary current if the primary draws 0.20 A.

Show Solution
\(V_s = V_p \cdot N_s/N_p = 120 \times (50/500) = 12\ \text{V}\).
\(I_s = I_p \cdot N_p/N_s = 0.2 \times (500/50) = 2.0\ \text{A}\) (ideal: power conserved, \(V_p I_p = V_s I_s\)).