The toroidal core of Figure 8.26(a) has ρo = 10 cm and a circular cross section with a = 1 cm. If the core is made of steel (μ = 1000 μo) and has a coil with 200 turns, calculate the amount of current that will produce a flux of 0.5 mWb in the core.

Question

The toroidal core of Figure 8.26(a) has [latex]\rho_{o}=10cm[/latex] and a circular cross section with [latex]a=1 cm[/latex]. If the core is made of steel [latex](\mu=1000\mu_{o})[/latex] and has a coil with 200 turns, calculate the amount of current that will produce a flux of [latex]0.5mWb[/latex] in the core.

Accepted Answer

This problem can be solved in two different ways: by using the magnetic field approach (direct) or by using the electric circuit analog (indirect).

Method 1: Since [latex] ho_{o}[/latex] is large compared with a, from Example 7.6

[latex]B=\frac{\mu NI}{\ell}=\frac{\mu_{o}\mu_{r}NI}{2\pi ho_{o}}[/latex]

Hence

[latex]\Psi=BS=\frac{\mu_{o}\mu_{r}NI\pi a^{2}}{2\pi ho_{o}}[/latex]

or

[latex]I=\frac{2 ho_{o}\Psi}{\mu_{o}\mu_{r}Na^{2}}=\frac{2(10 imes10^{-2})(0.5 imes10^{-3})}{4\pi imes10^{-7}(1000)(200)(1 imes10^{-4})}=\frac{100}{8\pi}=3.979 A[/latex]

Method 2: The toroidal core in Figure 8.26(a) is analogous to the electric circuit of Figure 8.26(b). From the circuit and Table 8.4

Electric	Magnetic
Conductivity [latex]\sigma[/latex]	Permeability [latex]\mu[/latex]
Field intensity E	Field intensity H
Current [latex]I=\int J\cdot dS[/latex]	Magnetic flux [latex]\Psi=\int B\cdot dS[/latex]
Current density [latex]J=\frac{I}{S}=\sigma E[/latex]	Flux density [latex]B=\frac{\Psi}{S}=\mu H[/latex]
Electromotive force (emf) V	Magnetomotive force (mmf) [latex]?[/latex]
Resistance R	Reluctance [latex]\Re[/latex]
Conductance [latex]G=\frac{1}{R}[/latex]	Permeance [latex]?=\frac{1}{\Re}[/latex]
Ohm’s law [latex]R=\frac{V}{I}=\frac{\ell}{\sigma S}[/latex] or [latex]V=E\ell=IR[/latex]	Ohm’s law [latex]\Re=\frac{?}{\Psi}=\frac{\ell}{\mu S}[/latex] or [latex]?=H\ell=\Psi\Re=NI[/latex]
Kirchhoff ’s laws: [latex]\sum I=0\\sum V-\sum RI=0[/latex]	Kirchhoff ’s laws: [latex]\sum{\Psi=0}\\sum{?}-\sum{\Re\Psi =0} [/latex]

[latex]?=NI=\Psi ℜ=\Psi\frac{\ell}{\mu S}=\Psi \frac{2\pi ho _{o}}{\mu_{o}\mu_{r}\pi a^{2}}[/latex]

or

[latex]I=\frac{2 ho_{o}\Psi}{\mu_{o}\mu_{r}Na^{2}}=3.979A[/latex]

as obtained with Method 1.

Electric	Magnetic
Conductivity $\sigma$	Permeability $\mu$
Field intensity E	Field intensity H
Current $I=\int J\cdot dS$	Magnetic flux $\Psi=\int B\cdot dS$
Current density $J=\frac{I}{S}=\sigma E$	Flux density $B=\frac{\Psi}{S}=\mu H$
Electromotive force (emf) V	Magnetomotive force (mmf) $?$
Resistance R	Reluctance $\Re$
Conductance $G=\frac{1}{R}$	Permeance $?=\frac{1}{\Re}$
Ohm’s law $R=\frac{V}{I}=\frac{\ell}{\sigma S}$ or $V=E\ell=IR$	Ohm’s law $\Re=\frac{?}{\Psi}=\frac{\ell}{\mu S}$ or $?=H\ell=\Psi\Re=NI$
Kirchhoff ’s laws: $\sum I=0\\\sum V-\sum RI=0$	Kirchhoff ’s laws: $\sum{\Psi=0}\\\sum{?}-\sum{\Re\Psi =0}$

Question 8.14: The toroidal core of Figure 8.26(a) has ρo = 10 cm and a cir...

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