Question 3.5.2: Determine the inductance of the actuator shown in Fig. 3.29....
Determine the inductance of the actuator shown in Fig. 3.29. Assume that the core has a permeability μ >> μ_{0}, and that there is no leakage flux.
We follow the same analysis as in the previous example. First, apply Eq. (3.138) to a path around the actuator circuit (dotted line) assuming that H is parallel to the path. This gives
Eq. (3.138): \oint_{c}H· dl = I_{tot},
\oint_{c} H · dl = ni
or
H_{c}l_{c} + 2H_{g} x = ni. (3.166)
Next, apply Eq. (3.139) at the core-gap interface assuming that 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), (3.167)
Substitute Eq. (3.167) into Eq. (3.166) with B_{c} = μH_{c} and B_{g} = μ_{o}H_{g}, and obtain
B_{g} = \frac{ni}{(A_{g}/A_{c})(l_{c} /μ)+ (2x/μ_{o})}
= \frac{μ_{o} ni}{(A_{g}/A_{c})(μ_{o}/μ) l_{c} + 2x}.
The flux through the circuit is Φ = B_{g}A_{g} . Therefore,
Φ = \frac{μ_{o} ni A_{g}}{(A_{g}/A_{c})(μ_{o}/μ) l_{c} + 2x}.
We are interested in the inductance of the coil. This can be determined from the flux linkage ∧. Because the flux Φ passes through each turn of the coil, the flux linkage is ∧ = nΦ The inductance follows from Eq. (3.78), and is given by
Eq. (3.78): L = \frac{∧}{I}
L = \frac{∧}{i}
= \frac{μ_{o} n^{2} A_{g}}{(A_{g}/A_{c})(μ_{o}/μ)l_{c} + 2x}. (3.169)
Notice that the inductance in a function of the gap spacing x, L = L(x). Because μ_{o}/μ << 1, we can estimate the inductance using
L(x) ≈ \frac{μ_{o} n^{2} A_{g}}{2x}