Magnetic Circuits

LessonAbout 30 min

Magnetic circuit is a closed path in which magnetic induction or flux flows.

1. Force on a Conductor Carrying Current in a Magnetic Field

Current-carrying conductor in magnetic field experiences mechanical force
Direction determined by right-hand rule or Fleming’s left-hand rule

Force on a Conductor

\[ F = \beta I L \sin\alpha \]

Where: \(\alpha\) = angle between conductor axis and magnetic field

Example. A wire \(12\ \text{cm}\) long and carrying a current of \(30\ \text{A}\) is placed in between the pole face of a magnet whose magnetic flux density is \(0.9\ \text{tesla}\). If the wire is inclined at an angle \(60^\circ\) from the plane of the magnetic field, what is the force exerted on the wire?

Solution.

\[ F = \beta I L \sin\alpha = (0.9)(30)(0.12) \sin 60^\circ = 2.8\ \text{N} \]

Answer: \(2.8\ \text{N}\)

2. Force Between Two Straight Parallel Wires Carrying Current

Like currents: attractive force between parallel conductors
Opposite currents: repulsive force between parallel conductors

Force Between Parallel Wires

\[ F = (2 \times 10^{-7}) I_1 I_2 \left(\dfrac{L}{d}\right) \]

Example. Two straight parallel wires \(2\ \text{m}\) long and \(3\ \text{mm}\) apart carry a current of \(8.0\ \text{A}\) in opposite directions. Calculate the force between these conductors.

Solution.

\[ F = (2 \times 10^{-7}) \left(\dfrac{I_1 I_2 L}{d}\right) = 2 \times 10^{-7} \left[\dfrac{(8)(8)(2)}{0.003}\right] = 8.53 \times 10^{-3}\ \text{N} \]

Answer: \(8.53 \times 10^{-3}\ \text{N}\)

3. Magnetomotive Force

Magnetomotive force is the force that sets up or tends to set up magnetic flux in a magnetic circuit.

Ampere-turn: SI unit of magnetomotive force (\(\text{AT}\))
Gilbert: cgs unit of magnetomotive force, named after William Gilbert

Magnetomotive Force

\[ \mathfrak{F} = NI \]

\[ \mathfrak{F} = 0.4\pi NI \]

Example. The coil in a magnetic contactor requires \(0.5\ \text{A}\) to provide a magnetizing force of \(500\ \text{AT}\). How many turns are necessary?

Solution.

\[ N = \dfrac{\mathfrak{F}}{I} = \dfrac{500}{0.5} = 1000\ \text{turns} \]

Answer: \(1000\ \text{turns}\)

4. Magnetic Flux

Magnetic flux is the number of magnetic lines of force in a magnetic field.

Weber: SI unit of magnetic flux equal to \(10^8\) lines, named after Wilhelm Weber
Maxwell: cgs unit of magnetic flux equal to one line of force, named after James Clerk Maxwell
Leakage flux: flux that does not follow intended path in magnetic circuit

Magnetic Flux Unit Relation

\[ 1\ \text{Wb} = 10^8\ \text{maxwells} \]

5. Magnetic Flux Density

Magnetic flux density is the number of lines of flux per unit area through any substance in a plane perpendicular to flux lines.

Tesla: SI unit equal to webers per square meter, named after Nikola Tesla
Gauss: cgs unit equal to maxwells per square centimeter, named after Johann Karl Friedrich Gauss

Magnetic Flux Density

\[ \beta = \dfrac{\Phi}{A} \]

Example. A coil with \(150\ \text{turns}\) is wound over a \(200\ \text{cm}\) long cylindrical iron core whose relative permeability is \(250\). If a current of \(2\ \text{A}\) passes through the coil, determine the flux density in the core.

Solution.

\[ \beta = \dfrac{\mu_0 \mu_r NI}{\ell} = \dfrac{(4\pi \times 10^{-7})(250)(150)(2)}{2} = 0.07\ \text{T} \]

Answer: \(0.07\ \text{T}\)

6. Magnetic Field Intensity

Magnetic field intensity is the MMF per unit length of path of magnetic flux, also called magnetizing force or magnetic potential gradient.

Ampere-turn per meter: SI unit of magnetic field intensity (\(\text{AT/m}\))
Oersted: cgs unit of magnetic field strength equal to gilbert per centimeter, named after Hans Christian Oersted

Magnetic Field Intensity

\[ H = \dfrac{NI}{\ell} \qquad\qquad H = \dfrac{\mathfrak{F}}{\ell} \]

Example. A solenoid \(30\ \text{cm}\) long is wound with \(300\ \text{turns}\). What is the value of its field strength inside the solenoid, when the coil is carrying a current of \(2\ \text{amperes}\)?

Solution.

\[ H = \dfrac{NI}{\ell} = \dfrac{300(2)}{0.30} = 2000\ \text{AT/m} \]

Answer: \(2000\ \text{AT/m}\)

7. Permeability

Permeability is the ability of a material to conduct magnetic flux through it.

Free space permeability: \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\) in SI units (\(1\) in cgs units)
Relative permeability: ratio of material permeability to air or vacuum permeability
Reluctivity: reciprocal of permeability

Permeability and Relative Permeability

\[ \mu = \dfrac{\beta}{H} \]

\[ \mu_r = \dfrac{\mu}{\mu_0} \]

Example. Calculate the magnetic field intensity to produce a flux density of \(10 \times 10^{-3}\ \text{tesla}\) at the center of a long straight solenoid in air.

Solution.

\[ H = \dfrac{\beta}{\mu_0 \mu_r} = \dfrac{10 \times 10^{-3}}{(4\pi \times 10^{-7})(1)} = 7958\ \text{AT/m} \]

Answer: \(7958\ \text{AT/m}\)

8. Magnetic Field of an Infinitely Long Straight Wire

Concentric circular field lines around current-carrying conductor
Field strength varies inversely with perpendicular distance \(r\)

Field of Long Straight Wire

\[ H = \dfrac{I}{2\pi r} \]

\[ \beta = \dfrac{\mu_0 \mu_r I}{2\pi r} \]

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Example. If a current of \(5\ \text{A}\) flows through a long wire of radius \(0.004\ \text{meter}\), what is the flux density (\(H\)) produced \(0.02\ \text{meter}\) away from the surface of the wire?

Solution.

\[ H = \dfrac{I}{2\pi r} = \dfrac{5}{2\pi(0.02)} = 39.8\ \text{AT/m} \]

Answer: \(39.8\ \text{AT/m}\)

9. Reluctance

Reluctance is the property of a material that opposes flux flow, equal to the ratio of MMF to magnetic flux.

Reciprocal henry: SI unit of reluctance (\(\text{H}^{-1}\))
Gilbert per maxwell: cgs unit of reluctance

Reluctance

\[ \mathfrak{R} = \dfrac{\ell}{\mu_0 \mu_r A} \qquad\qquad \mathfrak{R} = v \dfrac{\ell}{A} \]

Where: \(v\) = reluctivity

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Example. The reluctance of a non-magnetic circuit is \(12\ \text{units}\). How much flux will be set up if surrounded by a coil \(600\ \text{turns}\) carrying a current of \(3\ \text{A}\)?

Solution.

\[ \Phi = \dfrac{\mathfrak{F}}{\mathfrak{R}} = \dfrac{NI}{\mathfrak{R}} = \dfrac{600(3)}{12} = 150\ \text{Wb} \]

Answer: \(150\ \text{Wb}\)

10. Permeance

Permeance is the reciprocal of reluctance.

Permeance

\[ \mathcal{P} = \dfrac{1}{\mathfrak{R}} \qquad\qquad \mathcal{P} = \dfrac{\mu_0 \mu_r A}{\ell} \]

11. Magnetic Circuit

Magnetic circuit is a closed path in which magnetic induction or flux flows.

Analogous to electric Ohm’s law (\(\text{MMF} \leftrightarrow \text{EMF}\), \(\text{Flux} \leftrightarrow \text{Current}\), \(\text{Reluctance} \leftrightarrow \text{Resistance}\))

Ohm’s Law for Magnetic Circuit

\[ \Phi = \dfrac{\mathfrak{F}}{\mathfrak{R}} = \dfrac{NI}{\mathfrak{R}} \]

\[ \Phi = \dfrac{0.4\pi NI}{\mathfrak{R}} \]

Example. A circular ring of iron having a cross sectional area of \(5\ \text{cm}^2\), an average diameter of \(30\ \text{cm}\) is wound with a coil of \(1000\ \text{turns}\). If a current of \(3\ \text{A}\) is pass through the coil, determine the flux generated? Assume the relative permittivity of iron is \(250\).

Solution.

\[ \beta = \dfrac{\mu NI}{\ell} = \dfrac{(4\pi \times 10^{-7})(250)(1000)(3)}{\pi(0.30)} = 1\ \text{T} \]

\[ \Phi = \beta A = (1)\left(5\ \text{cm}^2 \times \dfrac{1\ \text{m}^2}{10000\ \text{cm}^2}\right) = 5 \times 10^{-4}\ \text{Wb} \]

Answer: \(5 \times 10^{-4}\ \text{Wb}\)

12. Reluctance in Series

Total reluctance is the sum of individual series reluctances
Magnetic flux remains constant through series path

Series Reluctance

\[ \mathfrak{R}_t = \mathfrak{R}_1 + \mathfrak{R}_2 + \mathfrak{R}_3 + \dots \]

\[ \Phi_t = \Phi_1 = \Phi_2 = \Phi_3 = \dots \]

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Example. The relative permeability of a certain silicon steel is \(4500\). A certain magnetic loop consists of a silicon steel of \(10\ \text{cm}\) square, \(20\ \text{cm}\) long and an air gap of \(1/4\ \text{cm}\). What is the reluctance of the magnetic circuit?

Solution.

\[ \mathfrak{R} = \mathfrak{R}_{\text{core}} + \mathfrak{R}_{\text{air}} = \dfrac{\ell_{\text{core}}}{\mu_r \mu_0 A_{\text{core}}} + \dfrac{\ell_{\text{air}}}{\mu_0 A_{\text{air}}} \]

\[ \mathfrak{R} = \dfrac{20}{4500(1)(100)} + \dfrac{0.25}{1(100)} = 2.54 \times 10^{-3}\ \text{cgs units} \]

Answer: \(2.54 \times 10^{-3}\ \text{cgs units}\)

13. Reluctance in Parallel

Reciprocal of total reluctance equals sum of reciprocals of branch reluctances
Total flux divides among parallel branches

Parallel Reluctance

\[ \dfrac{1}{\mathfrak{R}_t} = \dfrac{1}{\mathfrak{R}_1} + \dfrac{1}{\mathfrak{R}_2} + \dfrac{1}{\mathfrak{R}_3} + \dots \]

\[ \Phi_t = \Phi_1 + \Phi_2 + \Phi_3 + \dots \]

14. Ampere-Turns for Air Gap

Air gap reluctance requires significant portion of total MMF

Air Gap MMF

\[ \mathfrak{F}_{\text{air}} = \dfrac{\beta d}{\mu_0} \]

Example. A magnetic ring (relative permittivity = \(800\)) has a mean radius of \(10\ \text{cm}\) and a cross sectional area of \(5\ \text{cm}^2\). An air gap measuring \(1.5\ \text{mm}\) is cut in the ring. Determine the required mmf in order to produce a flux of \(0.25\ \text{mWb}\) in the air gap.

Solution.

\[ \beta = \dfrac{\Phi}{A} = \dfrac{0.25 \times 10^{-3}}{5 \times 10^{-4}} = 0.5\ \text{T} \]

\[ \mathfrak{F}_{\text{air}} = \dfrac{\beta \ell_{\text{air}}}{\mu_0} = \dfrac{0.5(0.0015)}{4\pi \times 10^{-7}} = 596.83\ \text{AT} \]

\[ \mathfrak{F}_{\text{core}} = \dfrac{\beta \ell_{\text{core}}}{\mu_0 \mu_r} = \dfrac{0.5(2\pi \times 0.1)}{(4\pi \times 10^{-7})(800)} = 312.5\ \text{AT} \implies \mathfrak{F}_{\text{total}} = 596.83 + 312.5 = 909\ \text{AT} \]

Answer: \(909\ \text{AT}\)

15. Energy Stored in Magnetic Circuit

Energy stored in magnetic field volume of core or air gap

Stored Magnetic Energy

\[ W_m = \dfrac{1}{2} \mathfrak{R} \Phi^2 \]

\[ W_m = \dfrac{1}{2} \ell A \left(\dfrac{\beta^2}{\mu}\right) \]

Example. A magnetic ring with a mean diameter of \(25\ \text{cm}\) and a cross sectional area of \(5\ \text{cm}^2\) is wound with a coil of \(600\ \text{turns}\). An air gap of \(4\ \text{mm}\) is made by cutting a section of the ring. A current of \(10\ \text{A}\) is passed through the coil. Determine the energy stored in the air gap. Assume relative permittivity of the ring to be \(1000\).

Solution.

\[ \mathfrak{F}_{\text{total}} = NI = 600(10) = 6000\ \text{AT} \]

\[ \mathfrak{F}_{\text{air}} + \mathfrak{F}_{\text{core}} = 3183.098\beta + 625\beta = 3808.098\beta = 6000 \implies \beta = 1.57\ \text{T} \]

\[ W_m = \dfrac{\beta^2 V}{2\mu_0} = \dfrac{(1.57)^2 (5 \times 10^{-4})(0.004)}{2(4\pi \times 10^{-7})} = 1.96\ \text{J} \]

Answer: \(1.96\ \text{J}\)

16. Tractive Force of a Magnet

Tractive force is the mechanical lifting force exerted at magnetic contact surfaces.

Tractive Force

\[ F = \dfrac{\beta^2 A}{2\mu_0} \]

17. Steinmetz Formula

According to experiments by Charles Proteus Steinmetz, the heat energy due to hysteresis released per cycle per unit volume of iron is proportional to flux density raised to 1.6 exponent.

Steinmetz Law

\[ W_h = \eta \beta_m^{1.6} f V \]

Where: \(\eta\) = hysteresis coefficient, \(\beta_m\) = maximum flux density, \(V\) = core volume

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Example. A small single-phase transformer has \(10.2\ \text{watts}\) at no load. The core has a volume of \(750\ \text{cubic cm}\). The maximum flux density is \(10{,}000\ \text{gauss}\) and the hysterisis constant of the core is \(5 \times 10^{-4}\). Using the Steinmetz law to find the hysterisis, determine the eddy current loss.

Solution.

\[ W_h = \eta \beta_m^{1.6} f V = (5 \times 10^{-4})(10000)^{1.6}(60)(750) = 56.5174 \times 10^6\ \text{erg/s} = 5.65\ \text{W} \]

\[ P_e = P_{\text{core}} - P_h = 10.2 - 5.65 = 4.55\ \text{W} \]

Answer: \(4.55\ \text{W}\)

Formula Summary

Formula Name	Formula	Variables
Force on a Conductor	\(F = \beta I L \sin\alpha\)	\(F\) = force (\(\text{N}\)), \(\beta\) = flux density (\(\text{T}\)), \(I\) = current (\(\text{A}\)), \(L\) = conductor length (\(\text{m}\)), \(\alpha\) = angle between conductor axis and magnetic field
Force Between Parallel Wires	\(F = (2 \times 10^{-7}) I_1 I_2 \left(\dfrac{L}{d}\right)\)	\(F\) = force (\(\text{N}\)), \(I_1, I_2\) = currents in wires (\(\text{A}\)), \(L\) = wire length (\(\text{m}\)), \(d\) = distance between wires (\(\text{m}\))
Magnetomotive Force	\(\mathfrak{F} = NI\) \(\mathfrak{F} = 0.4\pi NI\)	\(\mathfrak{F}\) = magnetomotive force (\(\text{AT}\) or \(\text{gilberts}\)), \(N\) = number of turns, \(I\) = current (\(\text{A}\))
Magnetic Flux Unit Relation	\(1\ \text{Wb} = 10^8\ \text{maxwells}\)	\(\text{Wb}\) = weber (\(\text{SI unit}\)), \(\text{maxwell}\) = cgs unit of flux
Magnetic Flux Density	\(\beta = \dfrac{\Phi}{A}\)	\(\beta\) = flux density (\(\text{T}\) or \(\text{Wb/m}^2\)), \(\Phi\) = magnetic flux (\(\text{Wb}\)), \(A\) = cross-sectional area (\(\text{m}^2\))
Magnetic Field Intensity	\(H = \dfrac{NI}{\ell}\) \(H = \dfrac{\mathfrak{F}}{\ell}\)	\(H\) = field intensity (\(\text{AT/m}\)), \(N\) = turns, \(I\) = current (\(\text{A}\)), \(\ell\) = mean path length (\(\text{m}\)), \(\mathfrak{F}\) = MMF (\(\text{AT}\))
Permeability and Relative Permeability	\(\mu = \dfrac{\beta}{H}\) \(\mu_r = \dfrac{\mu}{\mu_0}\)	\(\mu\) = permeability (\(\text{H/m}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\) (free space permeability), \(\mu_r\) = relative permeability, \(\beta\) = flux density (\(\text{T}\)), \(H\) = field intensity (\(\text{AT/m}\))
Field of Long Straight Wire	\(H = \dfrac{I}{2\pi r}\) \(\beta = \dfrac{\mu_0 \mu_r I}{2\pi r}\)	\(H\) = field intensity (\(\text{AT/m}\)), \(\beta\) = flux density (\(\text{T}\)), \(I\) = current (\(\text{A}\)), \(r\) = radial distance from wire (\(\text{m}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability
Reluctance	\(\mathfrak{R} = \dfrac{\ell}{\mu_0 \mu_r A}\) \(\mathfrak{R} = v \dfrac{\ell}{A}\)	\(\mathfrak{R}\) = reluctance (\(\text{AT/Wb}\) or \(\text{H}^{-1}\)), \(\ell\) = mean path length (\(\text{m}\)), \(A\) = area (\(\text{m}^2\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability, \(v\) = reluctivity
Permeance	\(\mathcal{P} = \dfrac{1}{\mathfrak{R}}\) \(\mathcal{P} = \dfrac{\mu_0 \mu_r A}{\ell}\)	\(\mathcal{P}\) = permeance (\(\text{H}\) or \(\text{Wb/AT}\)), \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\ell\) = length (\(\text{m}\)), \(A\) = area (\(\text{m}^2\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\), \(\mu_r\) = relative permeability
Ohm’s Law for Magnetic Circuit	\(\Phi = \dfrac{\mathfrak{F}}{\mathfrak{R}} = \dfrac{NI}{\mathfrak{R}}\) \(\Phi = \dfrac{0.4\pi NI}{\mathfrak{R}}\)	\(\Phi\) = magnetic flux (\(\text{Wb}\)), \(\mathfrak{F}\) = MMF (\(\text{AT}\) or \(\text{gilberts}\)), \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(N\) = turns, \(I\) = current (\(\text{A}\))
Series Reluctance	\(\mathfrak{R}_t = \mathfrak{R}_1 + \mathfrak{R}_2 + \mathfrak{R}_3 + \dots\) \(\Phi_t = \Phi_1 = \Phi_2 = \Phi_3 = \dots\)	\(\mathfrak{R}_t\) = total reluctance (\(\text{H}^{-1}\)), \(\mathfrak{R}_1, \mathfrak{R}_2\) = series reluctances (\(\text{H}^{-1}\)), \(\Phi_t\) = total flux (\(\text{Wb}\)), \(\Phi_1, \Phi_2\) = section fluxes (\(\text{Wb}\))
Parallel Reluctance	\(\dfrac{1}{\mathfrak{R}_t} = \dfrac{1}{\mathfrak{R}_1} + \dfrac{1}{\mathfrak{R}_2} + \dfrac{1}{\mathfrak{R}_3} + \dots\) \(\Phi_t = \Phi_1 + \Phi_2 + \Phi_3 + \dots\)	\(\mathfrak{R}_t\) = total reluctance (\(\text{H}^{-1}\)), \(\mathfrak{R}_1, \mathfrak{R}_2\) = parallel branch reluctances (\(\text{H}^{-1}\)), \(\Phi_t\) = total flux (\(\text{Wb}\)), \(\Phi_1, \Phi_2\) = branch fluxes (\(\text{Wb}\))
Air Gap MMF	\(\mathfrak{F}_{\text{air}} = \dfrac{\beta d}{\mu_0}\)	\(\mathfrak{F}_{\text{air}}\) = air gap MMF (\(\text{AT}\)), \(\beta\) = flux density (\(\text{T}\)), \(d\) = air gap length (\(\text{m}\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\)
Stored Magnetic Energy	\(W_m = \dfrac{1}{2} \mathfrak{R} \Phi^2\) \(W_m = \dfrac{1}{2} \ell A \left(\dfrac{\beta^2}{\mu}\right)\)	\(W_m\) = stored energy (\(\text{J}\)), \(\mathfrak{R}\) = reluctance (\(\text{H}^{-1}\)), \(\Phi\) = flux (\(\text{Wb}\)), \(\ell\) = length (\(\text{m}\)), \(A\) = area (\(\text{m}^2\)), \(\beta\) = flux density (\(\text{T}\)), \(\mu\) = permeability (\(\text{H/m}\))
Tractive Force	\(F = \dfrac{\beta^2 A}{2\mu_0}\)	\(F\) = tractive or lifting force (\(\text{N}\)), \(\beta\) = flux density (\(\text{T}\)), \(A\) = contact area (\(\text{m}^2\)), \(\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}\)
Steinmetz Law	\(W_h = \eta \beta_m^{1.6} f V\)	\(W_h\) = hysteresis power loss (\(\text{W}\)), \(\eta\) = Steinmetz hysteresis coefficient, \(\beta_m\) = maximum flux density (\(\text{T}\)), \(f\) = frequency (\(\text{Hz}\)), \(V\) = core volume (\(\text{m}^3\))