ECM · ENEE 154

Ch 6. Magnetic Circuit and Induction

3 hours · 3 marks

Magnetic circuits analyse flux paths using analogies to electric circuits. This short chapter also covers B–H behaviour, hysteresis and eddy losses, Faraday induction, and force on current-carrying conductors — all frequent short-answer / numerical topics.

6.1 Magnetic Circuit Types

Toroidal Magnetic Circuit with Air Gap Air gap (lg) N turns Φ Iron path length (li) | Cross-section (A) | μr = relative permeability
Figure 6.1 — Toroidal magnetic circuit with air gap

Basic quantities

MagneticSymbol / unitElectric analogy
MMF (magnetomotive force)\(\mathcal{F}=NI\) (AT)EMF \(E\)
Flux\(\Phi\) (Wb)Current \(I\)
Reluctance\(\mathcal{R}=l/(\mu A)\) (AT/Wb)Resistance \(R\)
Permeance\(\mathcal{P}=1/\mathcal{R}\)Conductance \(G\)
Flux density\(B=\Phi/A\) (T)—
Field intensity\(H=\mathcal{F}/l=NI/l\) (AT/m)—
\[ \Phi=\frac{\mathcal{F}}{\mathcal{R}}=\frac{NI}{\mathcal{R}},\qquad B=\mu H=\mu_0\mu_r H,\qquad \mu_0=4\pi\times 10^{-7}\,\text{H/m} \]

Series magnetic circuit

Same flux \(\Phi\) through each section (core + air gap, etc.); MMFs (reluctance drops) add:

\[ NI=\Phi(\mathcal{R}_1+\mathcal{R}_2+\cdots)=H_1\ell_1+H_2\ell_2+\cdots \]

Air gaps dominate because \(\mu_r\approx 1\) ⇒ large \(\mathcal{R}_g=\ell_g/(\mu_0 A_g)\). Fringing increases effective gap area slightly; often neglected in first-order exam problems.

Parallel magnetic circuit

Same MMF across parallel branches; fluxes add:

\[ \Phi_{\text{total}}=\Phi_a+\Phi_b,\qquad NI=H_a\ell_a=H_b\ell_b \]

Composite cores (window frames, multiple limbs) are solved like parallel–series resistor networks.

Magnetization notes

6.2 B–H Curve and Hysteresis (DC)

H (AT/m) B (T) Br Hc Bsat
Figure 6.2 — B-H hysteresis loop

Plotting \(B\) vs \(H\) for a ferromagnetic core while slowly increasing then decreasing magnetizing current yields a hysteresis loop.

Point / quantityMeaning
Residual flux density \(B_r\)\(B\) remaining when \(H=0\) after magnetization
Coercive force \(H_c\)\(H\) needed to bring \(B\) back to zero
Saturation \(B_{\text{sat}}\)maximum practical \(B\)
Loop areaenergy loss per cycle per unit volume (DC cycling)

For DC (or very slow) magnetization, domain alignment lags \(H\), producing the loop. Remanence and coercivity characterise soft vs hard materials.

6.3 Hysteresis (AC)

Under AC excitation the operating point traverses the hysteresis loop every cycle. Instantaneous \(B\) and \(H\) are not single-valued; the loop area still represents energy dissipated as heat per cycle.

At higher frequency the loop may widen slightly (dynamic effects), but the classical separation still treats hysteresis loss as proportional to frequency (once per cycle) and eddy loss as proportional to \(f^2\).

6.4 Hysteresis Loss and Eddy Current Loss

Hysteresis loss (Steinmetz)

\[ P_h=\eta\,B_m^{x}\,f\,V \]

Eddy current loss

Changing flux induces circulating currents in the core (Faraday). These \(I^2R\) losses are reduced by laminating the core (thin sheets insulated from each other):

\[ P_e=k_e\,B_m^{2}\,f^{2}\,t^{2}\,V \] where \(t\) is lamination thickness. Finer laminations ⇒ smaller \(P_e\).

Total core (iron) loss

\[ P_{\text{iron}}=P_h+P_e \]

Separation of losses: measure core loss at two frequencies (same \(B_m\)) or use the \(P/f\) vs \(f\) method.

6.5 Faraday’s Law — Statically and Dynamically Induced EMF

Faraday’s law: the induced EMF in a coil equals the rate of change of flux linkage:

\[ e=-N\frac{d\Phi}{dt}=-\frac{d\lambda}{dt} \]

Lenz’s law: the minus sign — induced current opposes the change causing it.

Statically induced EMF

Conductor / coil is stationary; flux through it changes with time (transformer action).

For sinusoidal flux \(\Phi=\Phi_m\sin\omega t\):

\[ E_{\text{rms}}=4.44\,f\,N\,\Phi_m \] (transformer EMF equation).

Dynamically induced EMF

Flux density is constant in time, but conductors move in the field (generator action):

\[ e=B\,\ell\,v\sin\theta \] where \(\theta\) is the angle between velocity and \(\mathbf{B}\) (maximum when motion is perpendicular to \(\mathbf{B}\) and to the conductor length).

6.6 Force on a Current-Carrying Conductor

A conductor of length \(\ell\) carrying current \(I\) in a magnetic field \(\mathbf{B}\) experiences

\[ \mathbf{F}=I\,(\boldsymbol{\ell}\times\mathbf{B}),\qquad F=B I \ell\sin\theta \]

Direction: Fleming’s left-hand rule (motors) — Forefinger (Field), Centre finger (Current), Thumb (Motion/Force).

For two long parallel conductors:

\[ F/l=\frac{\mu_0 I_1 I_2}{2\pi d} \] (attractive if currents same direction).

Solved Example — Toroidal Ring with Air Gap

Classic numerical

A circular magnetic ring has mean circumference (mean length of iron path) \(0.5\,\text{m}\), cross-section \(A=100\,\text{cm}^2\), air gap \(4\,\text{mm}\), flux \(\Phi=2\,\text{mWb}\), relative permeability of iron \(\mu_r=800\), turns \(N=200\). Find the MMF (and magnetizing current).

Given data (SI):

Flux density (assume same \(A\) in gap, neglect fringing):

\[ B=\frac{\Phi}{A}=\frac{0.002}{0.01}=0.2\,\text{T} \]

Field in iron:

\[ H_i=\frac{B}{\mu_0\mu_r}=\frac{0.2}{4\pi\times 10^{-7}\times 800} =\frac{0.2}{1.0053\times 10^{-3}}\approx 199\,\text{AT/m} \] \[ \text{AT for iron}=H_i\ell_i\approx 199\times 0.5\approx 99.5\,\text{AT} \]

Field in air gap (\(\mu_r=1\)):

\[ H_g=\frac{B}{\mu_0}=\frac{0.2}{4\pi\times 10^{-7}}\approx 1.59\times 10^{5}\,\text{AT/m} \] \[ \text{AT for gap}=H_g\ell_g\approx 1.59\times 10^{5}\times 0.004\approx 636\,\text{AT} \]

Total MMF:

\[ NI=H_i\ell_i+H_g\ell_g\approx 99.5+636=735.5\,\text{AT} \] \[ I=\frac{NI}{N}=\frac{735.5}{200}\approx 3.68\,\text{A} \]

Observation: the tiny air gap accounts for most of the MMF (\(\approx 86\%\)). This is why air-gap design dominates machine/magnetics calculations.

Reluctance check:

\[ \mathcal{R}_i=\frac{\ell_i}{\mu_0\mu_r A},\quad \mathcal{R}_g=\frac{\ell_g}{\mu_0 A},\quad NI=\Phi(\mathcal{R}_i+\mathcal{R}_g) \] yields the same numerical result.

Series vs parallel — quick comparison

SeriesParallel
FluxSame \(\Phi\) in each partSplits among branches
MMFDrops add (\(NI=\sum H\ell\))Same \(NI\) on each branch
Typical structureRing + one air gapMultiple limbs / leakage paths

Most Probable Questions

Q1. Define magnetic circuit. Compare series and parallel magnetic circuits. 3 marks
A magnetic circuit is a closed path for flux, analysed with MMF, flux and reluctance (analogous to \(E\), \(I\), \(R\)). Series: same \(\Phi\), MMFs add. Parallel: same MMF across branches, fluxes add.
Q2. A circular ring has mean length \(0.5\) m, \(A=100\) cm\(^2\), air gap \(4\) mm, \(\Phi=2\) mWb, \(\mu_r=800\), \(N=200\). Find MMF. 3 marks
Worked example above: \(NI\approx 736\) AT (iron \(\approx 100\) AT, gap \(\approx 636\) AT). \(I\approx 3.68\) A.
Q3. Explain the B–H curve and hysteresis loop for DC magnetization. Define \(B_r\) and \(H_c\). 3 marks
\(B\) vs \(H\) is nonlinear and multi-valued under cyclic magnetization. \(B_r\): remanent flux density at \(H=0\). \(H_c\): coercivity — field to demagnetize to \(B=0\). Loop area ∝ energy loss per cycle.
Q4. Distinguish hysteresis loss and eddy current loss. How are eddy losses reduced? 3 marks
Hysteresis: domain friction, \(P_h\propto B_m^x f\). Eddy: induced circulating currents, \(P_e\propto B_m^2 f^2 t^2\). Reduce eddy loss by thin laminations (small \(t\)) and high-resistivity core steel.
Q5. State Faraday’s law. Differentiate statically and dynamically induced EMF. 3 marks
\(e=-N\,d\Phi/dt\). Static: conductor fixed, flux changes (transformer; self/mutual). Dynamic: conductor moves in (usually steady) field — \(e=B\ell v\sin\theta\) (generator).
Q6. Write the expression for force on a current-carrying conductor in a magnetic field. State Fleming’s left-hand rule. 3 marks
\(F=BI\ell\sin\theta\). Left hand: Forefinger = Field (\(B\)), Centre = Current (\(I\)), Thumb = Force/Motion — used for motors.
Q7. Why does a small air gap require a large MMF? 2 marks
Air has \(\mu_r\approx 1\), so reluctance \(\ell_g/(\mu_0 A)\) is huge compared with iron (\(\mu_r\sim 10^2\text{–}10^3\)). Most ampere-turns drop across the gap.
Q8. Write Steinmetz’s formula for hysteresis loss and the eddy-current loss formula. 3 marks
\(P_h=\eta B_m^x f V\); \(P_e=k_e B_m^2 f^2 t^2 V\). Total iron loss \(P_h+P_e\).
Q9. Derive / state the transformer EMF equation \(E=4.44 f N\Phi_m\). 3 marks
\(\Phi=\Phi_m\sin\omega t\) ⇒ \(e=N\omega\Phi_m\cos\omega t\) ⇒ \(E_{\text{max}}=2\pi f N\Phi_m\) ⇒ \(E_{\text{rms}}=2\pi f N\Phi_m/\sqrt{2}=4.44 f N\Phi_m\).
Q10. What is meant by saturation of a magnetic material? 2 marks
Beyond the knee of the B–H curve, further increase in \(H\) produces little rise in \(B\) because domains are nearly fully aligned; effective \(\mu_r\) drops sharply.
← Chapter 5 Next: Chapter 7 →