Question 3.5.1: Consider the magnetic circuit shown in Fig. 3.28. Determine ...
Consider the magnetic circuit shown in Fig. 3.28. Determine the flux density in the gap, and the inductance of the coil. Assume that the core has a permeability μ >> μ_{0} and that there is no leakage flux.
We determine the field in the gap first. Choose an integration path along the axis of the circuit as indicated by the dotted line. Apply Eq.(3.138) to the path assuming that H is parallel to it. This gives
Eq.(3.138): \oint_{c}H· dl = I_{tot},
\oint_{c} H · dl = ni,
or
H_{c}l_{c} + H_{g} l_{g} = ni, (3.155)
where l_{c} and l_{g} are the lengths of the path in the core and gap, respectively, and H_{c} and H_{g} are the fields in these sections. Notice that I_{tot}, = ni. Next, apply Eq. (3.139) at the core-gap interface. As there is no leakage flux, all the flux that passes through the core also passes through the gap Φ_{c} = Φ_{g},
Eq. (3.139): \oint_{s} B· ds = 0.
B_{c}A_{c} = B_{g} A_{g} (no fringing flux at gap), (3.156)
where A_{c} and A_{g} are the cross-sectional areas of the core and gap. Substitute Eq.(3.156) into Eq. (3.155), making use of B_{c} = μH_{c} and B_{g} = μ_{o}H_{g}. This gives
B_{g} = \frac{ni}{(A_{g}/A_{c})(l_{c} /μ)+(l_{g}/μ_{o})}
= \frac{μ_{o} ni}{(A_{g}/A_{c})(μ_{o}/μ) l_{c} + l_{g}} (3.157)
If we assume that A_{g}/A_{c} ≈ 1 and μ_{o}/μ << 1, then Eq. (3.157) reduces to B_{g} = μ_{o}ni/l_{g}. It is instructive to rederive Eq. (3.157) using Eq. (3.144). Specifically,
Eq.(3.144): Φ \sum\limits_{i = 1}^{m} \mathscr{R}_{i} = I_{tot}.
Φ = \frac{ni}{\mathscr{R}_{c} + \mathscr{R}_{g}}, (3.158)
where \mathscr{R}_{c} = l_{c}/ μA_{c} and \mathscr{R}_{g} = l_{g}/ μ_{0}A_{g}. If we substitute \mathscr{R}_{c}, \mathscr{R}_{g}, and Φ = B_{g}A_{g} into Eq. (3.158) we obtain Eq. (3.157). When \mathscr{R}_{g} >>\mathscr{R}_{core}, we obtain
Φ = \frac{μ_{o} ni A_{g}}{l_{g}} \quad (μ >> μ_{0}). (3.159)
Inductance: Next, we determine the inductance. From Eqs. (3.156) and (3.157)
we have
B_{c} = \frac{μ_{o} ni}{(μ_{o}/μ)l_{c} + (A_{c}/A_{g})l_{g}}. (3.160)
Thus, the flux through the core is
Φ_{c} = B_{c} A_{c}
= \frac{μ_{o} ni A_{c}}{(μ_{o}/μ)l_{c} + (A_{c}/A_{g})l_{g}} (3.161)
Because this flux passes through each turn of the coil, the flux linkage is
∧ = \frac{μ_{o} n^{2}i A_{c}}{(μ_{o}/μ)l_{c} + (A_{c}/A_{g})l_{g}}. (3.162)
Finally, the inductance of the coil follows from Eq. (3.78),
Eq. (3.78): L = \frac{∧}{I}
L = \frac{∧}{i}
= \frac{μ_{o} n^{2} A_{c}}{(μ_{o}/μ)l_{c} +( A_{c}/A_{g})l_{g}}. (3.163)
If there is no gap, l_{g} = 0 and we obtain
L = \frac{μn^{2} A_{c}}{l_{c}} (no gap). (3.164)
On the other hand, if there is a gap, and the permeability of the core is high
(μ >> μ_{0}), then L can be approximated by
L = \frac{μ_{o}n^{2} A_{g}}{l_{g}}. (3.165)
Calculations: We demonstrate the theory with some sample calculations. Consider a magnetic circuit with the following parameters: n = 110, i = 1 A,
l_{c} = 10 cm, l_{g} = 1 mm, μ = 1000 μ_{0}, and A_{c}= A_{g} = 1 cm^{2}. The fields and flux in the core and the gap are as follows:
| \begin{matrix} & Core & Gap \\ H(A/m) & 100 & 100,000 \\ B (T) & 4π × 10^{-2} & 4π × 10^{-2} \\ Φ(Wb) & 4π × 10^{-6} & 4π × 10^{-6}\end{matrix} |
If the circuit has no gap (l_{g} = 0), then we find that
| \begin{matrix} & Core \\ H(A/m) & 1100 \\ B (T) & 1.38 \\Φ(Wb) & 1.38× 10^{-4}\end{matrix} |