Electromagnetic Induction and Faraday's Law

Maxwell's Equations

Reference

• Maxwell's equations relate electric and magnetic fields and their sources.

• Table 6-1 lists Maxwell's equations in both differential and integral forms.

Maxwell's Equations

Gauss's Law

• Differential Form: \nabla \cdot D = \rho_v (6.1)

• Integral Form: \oint_S D \cdot ds = Q

Faraday's Law

• Differential Form: \nabla \times E = -\frac{\partial B}{\partial t} (6.2)

• Integral Form: \ointC E \cdot dl = -\frac{\partial}{\partial t} \intS B \cdot ds

Gauss's Law for Magnetism

• Differential Form: \nabla \cdot B = 0 (6.3)

• Integral Form: \oint_S B \cdot ds = 0

Ampère's Law

• Differential Form: \nabla \times H = J + \frac{\partial D}{\partial t} (6.4)

• Integral Form: \ointC H \cdot dl = \intS (J + \frac{\partial D}{\partial t}) \cdot ds

Note: The integral form of Faraday's law is for a stationary surface S.

Faraday's Law

Basic Principle

• Magnetic fields can produce an electric current in a closed loop.

• This only occurs if the magnetic flux linking the surface area of the loop changes with time.

• The key to this induction process is change; a static field won't induce current (I = 0).

  • Time varying field induces current (I \neq 0).

Circuit Theory

• \sum V_i = 0

• V{emf} = -\frac{d}{dt} \intS B \cdot ds = \oint E \cdot dl

• \oint E \cdot l = \sum V_i = 0

Induced EMF

• V_{ab} will appear as a separation of charge in a loop when I \neq 0

Three Types of EMF

• V{emf} = -\frac{d}{dt} \intS B \cdot ds = \oint E \cdot dl

Faraday's Law (Revisited)

• V{emf} = -N \frac{d\Phi}{dt} = -N \frac{d}{dt} \intS B \cdot ds (V)

Motional EMF

Definition

• Motional EMF (V_m) is generated when a conductor moves through a static magnetic field.

• V{emf} = \ointC (u \times B) \cdot dl (motional emf) (6.26)

• Only the segments of the circuit that cross magnetic field lines contribute to V_{emf}.

Three Types of EMF (Comprehensive)

• V{emf} = -\frac{d}{dt} \intS B \cdot ds + \oint (u \times B) \cdot dl = \oint E \cdot dl

Induced EMF and Magnetic Field

• V{emf} = - \frac{d}{dt} \intS B \cdot ds

• Increasing B induces EMF and current (I).

• Decreasing B induces EMF and current (I).

• B_{ind} opposes the change in B.

Contour and Polarity of Induced EMF

• The contour C determines the direction and polarity of the induced EMF.

• V{emf} = - \frac{d}{dt} \intS B \cdot ds

• \oint E \cdot dl = - \frac{d}{dt} \int_S B \cdot ds

Stationary Loop in Time-Varying B

• Transformer EMF (V_{emf}^{tr}) is induced in a stationary loop in a time-varying magnetic field.

• V_{emf}^{tr} = -N \frac{\partial B}{\partial t} \cdot ds (transformer emf), (6.9)

• Equivalent Circuit:

  • V{emf}^{tr} = Ri I + RI

  • I = \frac{V{emf}^{tr}}{R + Ri}

Induced EMF: Different Scenarios

• Relationship between electric field (E) and magnetic field (B) when EMF is induced.

Induced EMF: Potential Difference

• V = - \int{P1}^{P_2} E \cdot dl

• \oint E \cdot dl = \int{P1}^{P2} E \cdot dl + \int{P2}^{P1} E \cdot dl

• \ointC E \cdot dl = -\frac{d}{dt} \intS B \cdot ds \neq 0

• \int{C1} E \cdot dl \neq \int{C2} E \cdot dl

Induced EMF: Generalized Form

• \oint E \cdot l = -\frac{d}{dt} \int_S B \cdot ds + \oint (u \times B) \cdot dl

Induced EMF: Changing B(t)

• V_{emf}^{tr} is due to a changing B field.

Example 6-1: Inductor in a Changing Magnetic Field

Problem Setup

• An inductor with N turns of wire in a circular loop of radius a.

• Loop is in the x-y plane, centered at the origin.

• Connected to a resistor R.

• Magnetic field: B = B_0(\hat{y}2 + \hat{z}3) \sin(\omega t), where \omega is the angular frequency.

Objectives

• (a) Find the magnetic flux linking a single turn of the inductor.

• (b) Find the transformer EMF, given N = 10, B_0 = 0.2 T, a = 10 cm, and \omega = 10^3 rad/s.

• (c) Determine the polarity of V_{emf}^{tr} at t = 0.

• (d) Find the induced current in the circuit for R = 1 k\Omega (wire resistance is negligible).

Solution (a): Magnetic Flux

• \Phi = \intS B \cdot ds = \intS [B0(\hat{y} 2 + \hat{z}3) \sin(\omega t)] \cdot ds = 3 \pi a^2 B0 \sin(\omega t)

Solution (b): Transformer EMF

• V{emf}^{tr} = -N \frac{d\Phi}{dt} = -N \frac{d}{dt} (3 \pi a^2 B0 \sin(\omega t)) = -3 \pi N \omega a^2 B_0 \cos(\omega t)

• Given values: N = 10, a = 0.1 m, \omega = 10^3 rad/s, B_0 = 0.2 T

• V_{emf}^{tr} = -188.5 \cos(10^3 t) (V)

Solution (c): Polarity at t = 0

• At t = 0, \frac{d\Phi}{dt} > 0 and V_{emf}^{tr} = -188.5 V.

• The current I is in the direction shown to satisfy Lenz's law.

• Terminal 2 is at a higher potential than terminal 1.

• V{emf}^{tr} = V1 - V_2 = -188.5 (V).

Solution (d): Induced Current

• I = \frac{V2 - V1}{R} = \frac{188.5 \cos(10^3 t)}{10^3} = 0.19 \cos(10^3 t) (A).

Ideal Transformer

Primary Side Voltage

• V1 = -N1 \frac{d\Phi}{dt} (6.20)

Secondary Side Voltage

• V2 = -N2 \frac{d\Phi}{dt}

Voltage Ratio

• \frac{V2}{V1} = \frac{N2}{N1}

Impedance Transformation

• When the load is an impedance ZL and V1 is a sinusoidal source, the phasor-domain equivalent is: Z{in} = (\frac{N1}{N2})^2 ZL (6.21)

Flux Generation

• In a transformer, the directions of I1 and I2 are such that the flux generated by one opposes the flux generated by the other.

Example 6-2: Lenz's Law

Problem Setup

• A loop in the x-y plane with an area of 4 m².

• Magnetic flux density: B = -2(0.3t) (T).

• Internal resistance of the wire is negligible.

• Determine voltages V1 and V2 across the 2-\Omega and 4-\Omega resistors.

Solution

• Flux through the loop: \Phi = \intS B \cdot ds = \intS (-2(0.3t)) \cdot ds = -0.3t \times 4 = -1.2t (Wb).

• Transformer EMF: V_{emf} = - \frac{d\Phi}{dt} = 1.2 (V).

• Total voltage is distributed across two resistors in series.

• I = \frac{V{emf}}{R1 + R_2} = \frac{1.2}{2 + 4} = 0.2 A.

• V1 = IR1 = 0.2 \times 2 = 0.4 V.

• V2 = IR2 = 0.2 \times 4 = 0.8 V.

Faraday’s Law Question 4

Problem Setup

• A toroidal frame with a circular cross-section of radius a, containing a uniform time-varying magnetic field B = 0.1t Wb/m².

• A loop of wire with two series resistors of 1 \Omega and 2 \Omega is placed across the toroid.

Objective

• Determine the voltmeter reading for an ideal voltmeter connected across the 2 \Omega resistor in two configurations (Figs. 4a and 4b).

Solution Approach

• Neglect fringe effects.

• Use Faraday's Law to determine the induced EMF.

Faraday’s Law Application

• Induced EMF: V{emf} = - \frac{d}{dt} \intS B \cdot ds

Configuration Analysis (Fig. 4a)

• V_{emf} = - \frac{d}{dt} (0.1t \pi a^2) = -0.1 \pi a^2

• Current: I = \frac{V{emf}}{R1 + R_2} = \frac{0.1 \pi a^2}{3}

• Voltage across the 2 \Omega resistor: V = IR_2 = \frac{2}{3} (0.1 \pi a^2)

• V = -66.6 mV

Configuration Analysis (Fig. 4b)

• V + IR_1 = 0

Faraday’s Law Computation (Fig. 4a)

• V = - IR2

• V_{IR} = - 2 / 3 * 0.1 * pi * a^2

• V_{IR} = -33.3 mV