Magnetic Circuits Basics (Kankon)

eee-1109 EEE-1109 Basic Electrical Engineering

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Part 1: Magnetic Concepts and Fundamental Laws

Magnetic Concepts and Circuits:

• Magnetic Fields: Any region in space in which a current-carrying conductor is acted upon by a force is said to be a magnetic field.

• Magnetic Flux Density: The forces which act upon the sides of the exploring coil, or upon any current-carrying conductor in a magnetic field, depend upon a characteristic of the field called magnetic flux density. $F=BLI$ $⟹B=\frac{F}{LI}$ where: $F$ is force (Newton-N) $B$ is magnetic flux density ($N{A}^{-1}{m}^{-1}$ or webers per square meter) $L$ is the length of a conductor (m) $I$ is current (Ampere)

• $\to$ The right hand rule: $B=\frac{\Phi }{A}$ where: $\Phi$ = magnetic flux (Weber - Wb) $A$ = area through it passes ($m^2$)

• forces between current-carrying conductors: $F=\frac{\mu L{I}_1{I}_2}{2\pi S}$ Where, $F$ is force in newtons $\mu$ is permeability, of the medium $L$ is parallel length of the conductors $I_1$ is the current in 1st conductor $I_2$ is the current in 2nd conductor $S$ is the separation of the conductors

• Permeability: The property of space and air and iron which influences the force between current carrying conductors is called permeability. In free space, $\mu =4\pi \times 10^{-7}$ Now, $\mu ={\mu }_r{\mu }_0$ $\to$ free space permeability $\to$ Relative permeability

Part 2: Biot-Savart Law and Ampere’s Law

• Biot-Savart Law: Flux density due to a long straight current-carrying conductor. $F=BL{I}_2$ $\to$ Force acting on second conductor $\to$ Reaction force of first conductor We know, $F=\frac{\mu L{I}_1{I}_2}{2\pi S}$ … (1) Equating these two expressions, $BL{I}_2=\frac{\mu L{I}_1{I}_2}{2\pi S}$ $B=\frac{\mu I_1}{2\pi S}$ webers per sq. m. $\to$ This is known as Biot-Savart Law

• Ampere’s Law: (Magnetic Field Intensity) $B=\frac{\mu I}{2\pi S}$ … (III) By another concept, $B=\mu H$ … (IV) where $B$ be the flux density $\mu$ is permeability $H$ is magnetic field intensity Substituting (III) and (IV) equation, $\mu H=\frac{\mu I}{2\pi S}$ $\therefore H=\frac{I}{2\pi S}$ $\to$ Ampere’s Law

Part 3: Reluctance and Magnetic Circuits

• Reluctance and the magnetic circuit:

• Magnetic Circuit: When a magnetic field is practically confined to a definite region is what we called a magnetic circuit.

• The reluctance equation, $R=\frac{\mathcal{F}}{\Phi }$ where: $R$ is reluctance $\mathcal{F}$ is magnetic force $\Phi$ is magnetic flux

• Also, we know, $\mathcal{F}=HL$ … (V) $\Phi =BA$ … (VI)

• Substituting these equations, $R=\frac{HL}{\mu HA}$ $\therefore R=\frac{L}{\mu A}$ $\to$ Reluctance equation where, $R$ is reluctance $\mu$ is permeability of the material $L$ is the length of magnetic circuit

• We can also compare this equation, $R=\frac{\rho L}{A}\to \text{Resistance}R=\frac{L}{\mu A}\to \text{Reluctance}$ $R=\frac{V}{I}\to \text{Resistance}R=\frac{\mathcal{F}}{\Phi }\to \text{Reluctance}$

Part 4: Applied Examples (MMF Calculations)

Example 1: Cast-Steel Ring

• *****Find the MMF necessary to establish a flux of $0.0006$ weber in a cast-steel ring of circular cross section. The outside diameter is 8 inch and inside is 6 inch.
  • Soln: Here, Magnetic Flux, $\Phi =0.0006$ weber
    • $=0.0006\times 10^5$ kilolines
    • $=60$ kilolines
  • Cross-sectional area, $A=\pi (0.5{)}^2=0.785$ sq. inch
  • Flux density, $B=\frac{\Phi }{A}=\frac{60}{0.785}=76.4$ kilolines per sq. in.
  • Magnetic field intensity, $H(\text{from curve})=27.5$ ampere-turns per in
  • $l(\text{mean length of path})=\pi \times 7=22$ inch
  • MMF, $\mathcal{F}=\sum H\Delta l=Hl=27.5\times 22=605$ ampere-turns

Example 2: Composite Magnetic Circuit (Wrought Iron and Cast Iron)

• *****A magnetic circuit consists of a wrought-iron rod, $1.5$ inch. in diameter and $30$ inch long bent into a semicircle, and a cast iron slab, $2$ inch. thick and $4$ inch wide as in Fig-02. A perfect contact between the slab and the ends of the rod is assumed. How many ampere-turns will be required to establish a flux of $0.002$ weber?

  • Soln: Magnetic flux, $\Phi =0.002\times 10^5=200$ kilolines

  • For wrought-iron rod:

    • Cross-sectional area $A=\pi (\frac{1.5}{2}{)}^2=1.77$ sq inch
    • Flux Density, $B=\frac{\Phi }{A}=\frac{200}{1.77}=113$ kilolines per sq inch
    • $H(\text{from curve})=80$ ampere-turns per inch
    • MMF, ${\mathcal{F}}_{WI}=Hl=80\times 30=2400$ ampere-turns

  • For cast-iron slab:

    • Cross-sectional area $=2\times 4=8$ sq. inch
    • $B=\frac{\Phi }{A}=\frac{200}{8}=25$ kilolines per sq. in.
    • $H(\text{from curve})=16$ ampere-turns per inch
    • The length of the path in the slab is the diameter of the semi-circle formed by the rod and is given by,
    • $2\pi rl=2\times 30$
    • $⟹2l=\frac{2\times 30}{\pi }$
    • $\therefore 2l=19.1$ inch
    • MMF, ${\mathcal{F}}_{CI}=16\times 19.1=306$ ampere-turns

  • Total MMF, ${\mathcal{F}}_{net}=2400+306=2706$ ampere-turns

Example 3: Air Gap Calculation

• *****A magnetic circuit contains an air gap $0.1$ inch long and having a cross-sectional area which may be considered the same as the cross section of the steel faces between which it lies. These areas are each $4$ sq. inch. How many ampere-turns are required to establish a flux of $400$ kilolines across the gap?

  • Soln: Here,
  • Reluctance, $R=\frac{L}{{\mu }_{0}A}=\frac{(0.1/39.37)}{4\pi \times 10^{-7}\times (4/39.37{)}^2}$ $=7.81\times 10^5\text{ mks units}$
  • Flux, $\Phi =400\times 10^5$ weber
  • MMF, $\mathcal{F}=Hl=\Phi R=(400\times 10^5)(7.81\times 10^5)$ $=3120\text{ ampere-turns}$