Mastering Magnetic Circuits: Conquer Fundamentals with Solved Problems

Problem-1

A mild-steel ring having a cross-sectional area of \(500 \mathrm{~mm}^{2}\) and a mean circumference of \(400 \mathrm{~mm}\) has a coil of 200 turns wound uniformly around it. Calculate:

the reluctance of the ring;
the current required to produce a flux of \(800 \mu \mathrm{Wb}\) in the ring.

Given: The relative permeability of mild steel for a flux density of \(1.6 \mathrm{~T}\) is about \(380 .\)

Solution-1

\[\frac{800 \times 10^{-6}[\mathrm{~Wb}]}{500 \times 10^{-6}\left[\mathrm{~m}^{-2}\right]}=1.6 \mathrm{~T}\]
Flux density in ring is
\[\frac{0.4}{380 \times 4 \pi \times 10^{-7} \times 5 \times 10^{-4}}=1.68 \times 10^{6} \mathrm{~A} / \mathrm{Wb}\]
The reluctance of ring is

\[\begin{aligned} \dfrac{F}{\Phi}&=\dfrac{l}{\mu_{0} \mu_{\mathrm{r}} A}=S \\ \Phi&=\frac{F}{S}\\ &800 \times 10^{-6}=\frac{\mathrm{m} . \mathrm{m} . \mathrm{f}}{1.677 \times 10^{6}}\\ \mathrm{mmf}~ F&=1342 \mathrm{~A} \end{aligned}\]
MMF
\[\begin{aligned} F&=N I\\ \dfrac{F}{N}&=\dfrac{1342}{200}=6.7 \mathrm{~A} \end{aligned}\]
magnetizing current

Alternatively:

\[\begin{array}{c} H=\dfrac{B}{\mu_{0}}=\dfrac{B}{4 \pi \times 10^{-7}} \\ H=\dfrac{B}{\mu_{\mathrm{r}} \mu_{0}}=\dfrac{1.6}{380 \times 4 \pi \times 10^{-7}}=3350 \mathrm{~A} / \mathrm{m} \\ \therefore \quad \mathrm{m} . \mathrm{m} . \mathrm{f} .=3350 \times 0.4=1340 \mathrm{~A} \end{array}\]
MMF
\[\dfrac{1340}{200}=6.7 \mathrm{~A}\]
Magnetising Current

Problem-2

A ring of mild steel stampings having a mean circumference of \(400 \mathrm{~mm}\) and a cross-sectional area of \(500 \mathrm{~mm}^{2}\) is wound with 200 turns.

Calculate the inductance of the coil corresponding to a reversal of a magnetizing current of \(2 \mathrm{~A}\)

Given: Corresponding flux density \(=1.13 \mathrm{~T}\)

Solution-2

\[\Phi=B A=1.13[\mathrm{~T}] \times 0.0005\left[\mathrm{~m}^{2}\right]=0.000565 \mathrm{~Wb}\]
Total Flux
\[L=\frac{N \Phi}{I}=(0.000565 \times 200) / 2=56.6 \mathrm{mH}\]
Inductance

Problem-3

An iron ring of circular cross-sectional area of \(3.0 \mathrm{~cm}^{2}\) and mean diameter of \(20 \mathrm{~cm}\) is wound with 500 turms of wire and carries a current of \(2.09 \mathrm{~A}\) to produce the magnetic flux of \(0.5 \mathrm{~m}\) Wb in the ring.

Determine the permeability of the material.

Solution-3

\[\begin{aligned} \mathrm{a}&=3 \mathrm{~cm}^{2}=3 \times 10^{-4} \mathrm{~m}^{2}\\ \mathrm{~d}&=20 \mathrm{~cm}, \quad \mathrm{~N}=500, \quad \mathrm{I}=2 \mathrm{~A}, \quad \phi=0.5 \mathrm{~mWb} \\ l&=\pi \times \mathrm{d}=\pi \times 20=62.8318 \mathrm{~cm}=0.628318 \mathrm{~m}\\ &\mathrm{S}=\frac{l}{\mu_{0} \mu_{\mathrm{r}} \mathrm{a}}=\frac{0.628313}{4 \pi \times 10^{-7} \times \mu_{\mathrm{r}} \times 3 \times 10^{-4}}\\ &=\frac{1.6667 \times 10^{9}}{\mu_{\mathrm{r}}} \quad \cdots~(1) \end{aligned}\]

\[\begin{aligned} \text{flux} & = \dfrac{\text{mmf} }{\text{reluctance} } = \dfrac{NI}{S}\\ \mathrm{S}&=\frac{\mathrm{N} \mathrm{I}}{\phi}=\frac{500 \times 2}{0.5 \times 10^{-3}} \\ &=2 \times 10^{6} \mathrm{AT} / \mathrm{Wb} \quad \cdots~(2) \end{aligned}\]

Equating (1) and (2),

Problem-4

A solenoid of \(100 \mathrm{~cm}\) is wound on a brass tube. If the current through the coil is \(0.5 A,\)

calculate the number of turns necessary over the solenoid to produce a field strength of \(500 \mathrm{AT} / \mathrm{m}\) at the centre of the coil.

Solution-4

The field strength on the axis of a long solenoid is given by

\[\begin{aligned} \mathrm{H} &=\frac{\mathrm{NI}}{l} \mathrm{AT} / \mathrm{m} \\ l &=\text { Length of coil }=100 \mathrm{~cm}=1 \mathrm{~m} \\ \mathrm{I} &=\text { Current }=0.5 \mathrm{~A} \\ 500 &=\frac{\mathrm{N} \times 0.5}{1} \\ \mathrm{~N} &=1000 \end{aligned}\]

Problem-5

The magnetic circuit has dimensions \(A_{c}=A_{g}=9\) \(\mathrm{cm}^{2}, g=0.050 \mathrm{~cm}, l_{c}=30 \mathrm{~cm},\) and \(N=500\) turns. Assume \(\mu_{\mathrm{r}}=\) 70,000 for core material.

Find the reluctances \(R_{c}\) and \(R_{g}\)
For the condition that the magnetic circuit is operating with \(B_{c}=\) \(1.0 \mathrm{~T},\) find
- flux and current

Solution-5

\[\begin{aligned} \mathcal{R}_{\mathrm{c}}&=\frac{l_{\mathrm{c}}}{\mu_{\mathrm{r}} \mu_{0} A_{\mathrm{c}}}=\frac{0.3}{70,000\left(4 \pi \times 10^{-7}\right)\left(9 \times 10^{-4}\right)}=3.79 \times 10^{3} \quad \text{AT/Wb} \\ \mathcal{R}_{\mathrm{g}}&=\frac{g}{\mu_{0} A_{\mathrm{g}}}=\frac{5 \times 10^{-4}}{\left(4 \pi \times 10^{-7}\right)\left(9 \times 10^{-4}\right)}=4.42 \times 10^{5} \quad \text{AT/Wb} \\ \phi&=B_{\mathrm{c}} A_{\mathrm{c}}=1.0\left(9 \times 10^{-4}\right)=9 \times 10^{-4} \mathrm{~Wb} \\ i&=\frac{\mathcal{F}}{N}=\frac{\phi\left(\mathcal{R}_{\mathrm{c}}+\mathcal{R}_{\mathrm{g}}\right)}{N}=\frac{9 \times 10^{-4}\left(4.46 \times 10^{5}\right)}{500}=0.80 \mathrm{~A} \end{aligned}\]