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: $\oint<em>C E \cdot dl = -\frac{\partial}{\partial t} \int</em>S B \cdot ds$

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: $\oint<em>C H \cdot dl = \int</em>S (J + \frac{\partial D}{\partial t}) \cdot ds$

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

• $\sum V_i = 0$

• $V<em>{emf} = -\frac{d}{dt} \int</em>S B \cdot ds = \oint E \cdot dl$

• $\oint E \cdot l = \sum V_i = 0$

Induced EMF

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

Three Types of EMF

• $V<em>{emf} = -\frac{d}{dt} \int</em>S B \cdot ds = \oint E \cdot dl$

Faraday's Law (Revisited)

• $V<em>{emf} = -N \frac{d\Phi}{dt} = -N \frac{d}{dt} \int</em>S B \cdot ds$ (V)

Motional EMF

Definition

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

• $V<em>{emf} = \oint</em>C (u \times B) \cdot dl$ (motional emf) (6.26)

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

Three Types of EMF (Comprehensive)

• $V<em>{emf} = -\frac{d}{dt} \int</em>S B \cdot ds + \oint (u \times B) \cdot dl = \oint E \cdot dl$

Induced EMF and Magnetic Field

• $V<em>{emf} = - \frac{d}{dt} \int</em>S B \cdot ds$

• 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.

• $V<em>{emf} = - \frac{d}{dt} \int</em>S B \cdot ds$

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

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:

  • $V<em>{emf}^{tr} = R</em>i I + RI$

  • $I = \frac{V<em>{emf}^{tr}}{R + R</em>i}$

Induced EMF: Different Scenarios

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

Induced EMF: Potential Difference

• $V = - \int<em>{P</em>1}^{P_2} E \cdot dl$

• $\oint E \cdot dl = \int<em>{P</em>1}^{P<em>2} E \cdot dl + \int</em>{P<em>2}^{P</em>1} E \cdot dl$

• $\oint<em>C E \cdot dl = -\frac{d}{dt} \int</em>S B \cdot ds \neq 0$

• $\int<em>{C1} E \cdot dl \neq \int</em>{C2} E \cdot dl$

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

• $\Phi = \int<em>S B \cdot ds = \int</em>S [B<em>0(\hat{y} 2 + \hat{z}3) \sin(\omega t)] \cdot ds = 3 \pi a^2 B</em>0 \sin(\omega t)$

Solution (b): Transformer EMF

• $V<em>{emf}^{tr} = -N \frac{d\Phi}{dt} = -N \frac{d}{dt} (3 \pi a^2 B</em>0 \sin(\omega t)) = -3 \pi N \omega a^2 B_0 \cos(\omega t)$

• 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.

• $V<em>{emf}^{tr} = V</em>1 - V_2 = -188.5$ (V).

Solution (d): Induced Current

• $I = \frac{V<em>2 - V</em>1}{R} = \frac{188.5 \cos(10^3 t)}{10^3} = 0.19 \cos(10^3 t)$ (A).

Ideal Transformer

Primary Side Voltage

• $V<em>1 = -N</em>1 \frac{d\Phi}{dt}$ (6.20)

Secondary Side Voltage

• $V<em>2 = -N</em>2 \frac{d\Phi}{dt}$

Voltage Ratio

• $\frac{V<em>2}{V</em>1} = \frac{N<em>2}{N</em>1}$

Impedance Transformation

• When the load is an impedance $Z<em>L$ and $V</em>1$ is a sinusoidal source, the phasor-domain equivalent is: $Z<em>{in} = (\frac{N</em>1}{N<em>2})^2 Z</em>L$ (6.21)

Flux Generation

• In a transformer, the directions of $I<em>1$ and $I</em>2$ 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 $V<em>1$ and $V</em>2$ across the 2-\Omega and 4-\Omega resistors.

Solution

• Flux through the loop: $\Phi = \int<em>S B \cdot ds = \int</em>S (-2(0.3t)) \cdot ds = -0.3t \times 4 = -1.2t$ (Wb).

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

• Total voltage is distributed across two resistors in series.

• $I = \frac{V<em>{emf}}{R</em>1 + R_2} = \frac{1.2}{2 + 4} = 0.2$ A.

• $V<em>1 = IR</em>1 = 0.2 \times 2 = 0.4$ V.

• $V<em>2 = IR</em>2 = 0.2 \times 4 = 0.8$ 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: $V<em>{emf} = - \frac{d}{dt} \int</em>S B \cdot ds$

Configuration Analysis (Fig. 4a)

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

• Current: $I = \frac{V<em>{emf}}{R</em>1 + R_2} = \frac{0.1 \pi a^2}{3}$

• 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)

• $V = - IR2$

• $V_{IR} = - 2 / 3 * 0.1 * pi * a^2$

• $V_{IR} = -33.3 mV$