Electromagnetic Induction

LessonAbout 1.5 h

Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

1. Faraday’s Law

Faraday’s first law: changing flux linking a coil induces an emf
Faraday’s second law: induced emf magnitude equals flux linkage change rate
Flux linkage: product of magnetic flux and coil turns

Faraday’s Law

\[ e = N \dfrac{d\Phi}{dt} \]

Where: \(e\) = induced voltage (\(\text{V}\)), \(N\) = number of turns, \(d\Phi/dt\) = rate of change of flux (\(\text{Wb/s}\))

EE Board Exam October 1998

Example. A magnetic coil produces \(100{,}000\ \text{maxwells}\) with \(2{,}000\ \text{turns}\) and with a current of \(2\ \text{A}\). The current is cut-off and the flux collapses in \(0.01\ \text{s}\). What is the average voltage that will appear across the coil?

Solution.

\[ e = N \dfrac{\Delta\Phi}{\Delta t} = 2000 \left[\dfrac{100{,}000 \times 10^{-8}\ \text{Wb}}{0.01\ \text{s}}\right] = 200\ \text{V} \]

Answer: \(200\ \text{V}\)

2. Induced EMF in Short Length of Wire

Dynamically induced emf: generated by conductor motion across magnetic flux
Flux cutting: conductor cutting magnetic lines at relative velocity \(v\)

Induced EMF in Moving Conductor

\[ e = \beta L v \sin\alpha \]

\[ e = \dfrac{\Phi}{t} \]

Where: \(e\) = induced voltage (\(\text{V}\)), \(\beta\) = magnetic flux density (\(\text{T}\) or \(\text{Wb/m}^2\)), \(L\) = conductor length (\(\text{m}\)), \(v\) = velocity (\(\text{m/s}\)), \(\alpha\) = angle of motion to magnetic field

EE Board Exam April 1997

Example. A conductor, \(50\ \text{mm}\) long, moves at a velocity of \(2.5\ \text{m/s}\) across a magnetic field of \(0.90\ \text{Wb/m}^2\). What is the voltage generated?

Solution.

\[ e = \beta L v = (0.90)(0.050)(2.5) = 0.1125\ \text{V} \]

Answer: \(0.1125\ \text{V}\)

3. Self-Inductance

Self-inductance is a measure of counter emf generated in a circuit per unit change in current.

Inductor: two-terminal component coiling wire around core or free air
Henry: SI unit of self-inductance (\(\text{H}\)), named after Joseph Henry

Self-Induced Voltage

\[ e = L \dfrac{di}{dt} \]

Where: \(e\) = self-induced voltage (\(\text{V}\)), \(L\) = self-inductance (\(\text{H}\)), \(di/dt\) = rate of change of current (\(\text{A/s}\))

3.1 General Formula

Self-Inductance Formulas

\[ L = \dfrac{N\Phi}{I} \]

\[ L = \dfrac{N^2}{\mathfrak{R}} \]

\[ L = \dfrac{\mu_0 \mu_r A N^2}{\ell} \]

Where: \(L\) = self-inductance (\(\text{H}\)), \(N\) = turn count, \(\Phi\) = magnetic flux (\(\text{Wb}\)), \(I\) = current (\(\text{A}\)), \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability, \(A\) = core cross-sectional area (\(\text{m}^2\)), \(\ell\) = mean magnetic path length (\(\text{m}\))

Example. A solenoid \(20\ \text{cm}\) long and with a cross-sectional area of \(4\ \text{cm}^2\) is uniformly wound with \(1000\ \text{turns}\) of wire in air. Calculate its self-inductance.

Solution.

\[ L = \dfrac{\mu_0 \mu_r N^2 A}{\ell} = \dfrac{(4\pi \times 10^{-7})(1)(1000)^2 (4 \times 10^{-4})}{0.20} = 0.8\ \text{mH} \]

Answer: \(0.8\ \text{mH}\)

3.2 Inductance with Varying Reluctance

Inductance with Varying Reluctance

\[ L = N \dfrac{d\Phi/dt}{di/dt} \]

Where: \(L\) = inductance (\(\text{H}\)), \(N\) = number of turns, \(d\Phi/dt\) = flux change rate (\(\text{Wb/s}\)), \(di/dt\) = current change rate (\(\text{A/s}\))

3.3 Inductance of Coaxial Cable

Coaxial Cable Inductance

\[ L = (2 \times 10^{-7}) \ln\left(\dfrac{D}{d}\right) \]

Where: \(L\) = inductance per unit length (\(\text{H/m}\)), \(D\) = outer conductor diameter (\(\text{m}\)), \(d\) = inner conductor diameter (\(\text{m}\))

3.4 Inductance of Two Long Cylindrical Conductors, Parallel

Parallel Conductors Inductance

\[ L = (4 \times 10^{-7}) \ln\left(\dfrac{D}{\sqrt{r_1 r_2}}\right) \]

Where: \(L\) = loop inductance per unit length (\(\text{H/m}\)), \(D\) = conductor spacing (\(\text{m}\)), \(r_1, r_2\) = wire radii (\(\text{m}\))

4. Series-Connected Inductors

Total inductance equals sum of individual series inductances
Same current flows through all uncoupled series inductors

Series Inductance

\[ L_t = L_1 + L_2 + L_3 + \dots \]

Where: \(L_t\) = total series inductance (\(\text{H}\)), \(L_1, L_2, L_3\) = individual inductances (\(\text{H}\))

5. Parallel-Connected Inductors

Reciprocal total inductance equals sum of reciprocal inductances
Total current splits among parallel inductor branches

Parallel Inductance

\[ \dfrac{1}{L_t} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \dfrac{1}{L_3} + \dots \]

\[ L_t = \dfrac{L_1 L_2}{L_1 + L_2} \]

Where: \(L_t\) = total parallel inductance (\(\text{H}\)), \(L_1, L_2, L_3\) = branch inductances (\(\text{H}\))

6. Mutual Induction

Mutual induction is the condition of inducing an emf in a coil by flux generated in an adjacent coil.

Mutual inductance: degree of magnetic coupling between two coils
Coefficient of coupling: fraction of total flux linking secondary coil

Mutual Inductance Formulas

\[ M = k \sqrt{L_1 L_2} \]

\[ M = \dfrac{N_1 N_2}{\mathfrak{R}} \]

\[ M = \dfrac{\mu_0 \mu_r A N_1 N_2}{\ell} \]

\[ k = \dfrac{\text{flux linking } L_1 \text{ and } L_2}{\text{flux produced by } L_1} \]

Where: \(M\) = mutual inductance (\(\text{H}\)), \(k\) = coupling coefficient (\(0 \le k \le 1\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(N_1, N_2\) = turn counts, \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability, \(A\) = core area (\(\text{m}^2\)), \(\ell\) = mean path length (\(\text{m}\))

Example. A coil of inductance \(8\ \text{mH}\) produces \(80\ \mu\text{Wb}\) of magnetic flux. Of this total flux, \(60\ \mu\text{Wb}\) are linked to a second coil of inductance \(2\ \text{mH}\). How much is the mutual inductance between the coils?

Solution.

\[ k = \dfrac{\Phi_{\text{link}}}{\Phi_{\text{total}}} = \dfrac{60\ \mu\text{Wb}}{80\ \mu\text{Wb}} = 0.75 \]

\[ M = k \sqrt{L_1 L_2} = 0.75 \sqrt{8(2)} = 3\ \text{mH} \]

Answer: \(3\ \text{mH}\)

7. Series Coils with Mutual Inductance

Series aiding: magnetic fields produced by common current reinforce each other
Series opposing: magnetic fields produced by common current oppose each other

Series Coils with Mutual Inductance

\[ L_{ta} = L_1 + L_2 + 2M \]

\[ L_{to} = L_1 + L_2 - 2M \]

\[ M = \dfrac{L_{ta} - L_{to}}{4} \]

Where: \(L_{ta}\) = series aiding total inductance (\(\text{H}\)), \(L_{to}\) = series opposing total inductance (\(\text{H}\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(M\) = mutual inductance (\(\text{H}\))

EE Board Exam April 1992

Example. The total inductance of two coils, A and B, when connected in series aiding is \(0.5\ \text{H}\) and \(0.2\ \text{H}\) when connected in series opposing. Coil A when isolated from coil B has a self-inductance of \(0.2\ \text{H}\). Calculate the self-inductance of coil B.

Solution.

\[ M = \dfrac{L_{ta} - L_{to}}{4} = \dfrac{0.5 - 0.2}{4} = 0.075\ \text{H} \]

\[ L_2 = L_{ta} - 2M - L_1 = 0.5 - 2(0.075) - 0.2 = 0.15\ \text{H} \]

Answer: \(0.15\ \text{H}\)

8. Parallel Coils with Mutual Inductance

Parallel aiding: coupled coils connected in parallel with aiding polarities
Parallel opposing: coupled coils connected in parallel with opposing polarities

Parallel Coils with Mutual Inductance

\[ L_{ta} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 - 2M} \]

\[ L_{to} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 + 2M} \]

Where: \(L_{ta}\) = parallel aiding total inductance (\(\text{H}\)), \(L_{to}\) = parallel opposing total inductance (\(\text{H}\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(M\) = mutual inductance (\(\text{H}\))

EE Board Exam April 1995

Example. Two coils of inductances \(4\ \text{H}\) and \(6\ \text{H}\) are connected in parallel. If their mutual inductance is \(3\ \text{H}\), calculate the equivalent inductance of the combination if mutual inductance assists and if mutual inductance opposes the self-inductance respectively.

Solution.

\[ L_{ta} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 - 2M} = \dfrac{4(6) - 3^2}{4 + 6 - 2(3)} = 3.75\ \text{H} \]

\[ L_{to} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 + 2M} = \dfrac{4(6) - 3^2}{4 + 6 + 2(3)} = 0.94\ \text{H} \]

Answer: \(3.75\ \text{H}\ \text{and}\ 0.94\ \text{H}\)

9. Energy Stored in an Inductor Carrying a Current

Stored magnetic energy is energy potential held within an inductor’s magnetic field.

Energy Stored in an Inductor

\[ W = \dfrac{1}{2} L I^2 \]

Where: \(W\) = stored energy (\(\text{J}\)), \(L\) = inductance (\(\text{H}\)), \(I\) = current (\(\text{A}\))

EE Board Exam October 1990

Example. A \(6.0\ \text{H}\) coil whose resistance is \(12\ \Omega\) is connected in series with a \(24\ \Omega\) resistor and to a \(144\ \text{V}\) battery and a switch. The switch is closed at \(t = 0\). Determine the energy stored in the magnetic field at steady state.

Solution.

\[ I = \dfrac{E}{R + R_{\text{coil}}} = \dfrac{144}{24 + 12} = 4\ \text{A} \]

\[ W = \dfrac{1}{2} L I^2 = \dfrac{1}{2}(6)(4)^2 = 48\ \text{J} \]

Answer: \(48\ \text{J}\)

Formula Summary

Formula Name	Formula	Variables
Faraday’s Law	\(e = N \dfrac{d\Phi}{dt}\)	\(e\) = induced voltage (\(\text{V}\)), \(N\) = number of turns, \(d\Phi/dt\) = rate of change of flux (\(\text{Wb/s}\))
Induced EMF in Moving Conductor	\(e = \beta L v \sin\alpha\) \(e = \dfrac{\Phi}{t}\)	\(e\) = induced voltage (\(\text{V}\)), \(\beta\) = magnetic flux density (\(\text{T}\) or \(\text{Wb/m}^2\)), \(L\) = conductor length (\(\text{m}\)), \(v\) = velocity (\(\text{m/s}\)), \(\alpha\) = angle of motion relative to field, \(t\) = time (\(\text{s}\))
Self-Induced Voltage	\(e = L \dfrac{di}{dt}\)	\(e\) = self-induced voltage (\(\text{V}\)), \(L\) = self-inductance (\(\text{H}\)), \(di/dt\) = rate of change of current (\(\text{A/s}\))
Self-Inductance Formulas	\(L = \dfrac{N\Phi}{I}\) \(L = \dfrac{N^2}{\mathfrak{R}}\) \(L = \dfrac{\mu_0 \mu_r A N^2}{\ell}\)	\(L\) = self-inductance (\(\text{H}\)), \(N\) = turn count, \(\Phi\) = magnetic flux (\(\text{Wb}\)), \(I\) = current (\(\text{A}\)), \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability, \(A\) = core area (\(\text{m}^2\)), \(\ell\) = mean path length (\(\text{m}\))
Inductance with Varying Reluctance	\(L = N \dfrac{d\Phi/dt}{di/dt}\)	\(L\) = inductance (\(\text{H}\)), \(N\) = number of turns, \(d\Phi/dt\) = flux change rate (\(\text{Wb/s}\)), \(di/dt\) = current change rate (\(\text{A/s}\))
Coaxial Cable Inductance	\(L = (2 \times 10^{-7}) \ln\left(\dfrac{D}{d}\right)\)	\(L\) = inductance per unit length (\(\text{H/m}\)), \(D\) = outer conductor diameter (\(\text{m}\)), \(d\) = inner conductor diameter (\(\text{m}\))
Parallel Conductors Inductance	\(L = (4 \times 10^{-7}) \ln\left(\dfrac{D}{\sqrt{r_1 r_2}}\right)\)	\(L\) = loop inductance per unit length (\(\text{H/m}\)), \(D\) = conductor spacing (\(\text{m}\)), \(r_1, r_2\) = conductor radii (\(\text{m}\))
Series Inductance	\(L_t = L_1 + L_2 + L_3 + \dots\)	\(L_t\) = total series inductance (\(\text{H}\)), \(L_1, L_2, L_3\) = individual inductances (\(\text{H}\))
Parallel Inductance	\(\dfrac{1}{L_t} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \dfrac{1}{L_3} + \dots\) \(L_t = \dfrac{L_1 L_2}{L_1 + L_2}\)	\(L_t\) = total parallel inductance (\(\text{H}\)), \(L_1, L_2, L_3\) = branch inductances (\(\text{H}\))
Mutual Inductance Formulas	\(M = k \sqrt{L_1 L_2}\) \(M = \dfrac{N_1 N_2}{\mathfrak{R}}\) \(M = \dfrac{\mu_0 \mu_r A N_1 N_2}{\ell}\) \(k = \dfrac{\Phi_{\text{link}}}{\Phi_{\text{total}}}\)	\(M\) = mutual inductance (\(\text{H}\)), \(k\) = coupling coefficient (\(0 \le k \le 1\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(N_1, N_2\) = turn counts, \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability, \(A\) = core area (\(\text{m}^2\)), \(\ell\) = path length (\(\text{m}\))
Series Coils with Mutual Inductance	\(L_{ta} = L_1 + L_2 + 2M\) \(L_{to} = L_1 + L_2 - 2M\) \(M = \dfrac{L_{ta} - L_{to}}{4}\)	\(L_{ta}\) = series aiding total inductance (\(\text{H}\)), \(L_{to}\) = series opposing total inductance (\(\text{H}\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(M\) = mutual inductance (\(\text{H}\))
Parallel Coils with Mutual Inductance	\(L_{ta} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}\) \(L_{to} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 + 2M}\)	\(L_{ta}\) = parallel aiding total inductance (\(\text{H}\)), \(L_{to}\) = parallel opposing total inductance (\(\text{H}\)), \(L_1, L_2\) = self-inductances (\(\text{H}\)), \(M\) = mutual inductance (\(\text{H}\))
Energy Stored in an Inductor	\(W = \dfrac{1}{2} L I^2\)	\(W\) = stored magnetic energy (\(\text{J}\)), \(L\) = inductance (\(\text{H}\)), \(I\) = current (\(\text{A}\))