Assuming that the current flowing in the coil is 0.55A, and the windings in the coil are evenly wound on the iron core, the number of turns is 200. We see that the cross-section of the iron core is circular, its outer ring radius Rw=25mm, inner ring radius Rn=20mm, so the core cross-section diameter is 25-20=5mm. Let us set the relative permeability of the iron core [formula].
The first step: Let us find the magnetic induction intensity B at the outer ring and inner ring of the iron core when the air gap is not opened, and the magnetic flux [formula] in the iron core.
We know that from the magnetic field distribution of the densely wound current-carrying spiral loop and the Ampere loop theorem, the magnetic induction intensity of the densely wound current-carrying spiral loop is:
[formula]
But we can see from the picture that there is an iron core here. So of course the permeability has also changed. The permeability in the expression should be multiplied by the relative permeability of the iron core. Now we can get the value of magnetic induction intensity:
The radius of the outer ring of the iron core is 25, so the magnetic induction intensity Bw at the outer ring:
[formula]
The radius of the inner ring of the iron core is 20, so the magnetic induction intensity Bn at the inner ring:
[formula]
What about the middle of the core? Of course it is the average of the two, that is:
[formula]
Now let's find the magnetic flux. Since the magnetic field in the ring section is uniform, the average magnetic flux is calculated by the magnetic induction intensity B value in the middle of the ring. The calculation formula is: [formula]. S here is the area of the core section. So there are:
[formula]
Step 2: Let us find the magnetic induction intensity and magnetic flux after the iron core is slotted.
It can be seen from the figure that the width of the groove, that is, the width of the air gap, is 2.5 mm.
Now, there is an important law about to emerge, which is Kirchhoff's second law of magnetic circuits. The law is as follows:
[formula]
Note that the left side of this formula is called magnetomotive force, which is also called magnetomotive force. It is equal to the product of the current and the number of winding turns N; the right side of the formula is the algebraic sum of the magnetic voltage drops in each section of the magnetic circuit.
At the same time, we also know that the magnetic voltage drop is equal to the product of magnetic flux and magnetic resistance, that is: [formula]
The [formula] here is magnetic flux, and the [formula] is magnetoresistance. So the expression of Kirchhoff's second law of magnetic circuit becomes:
[formula]
With the calculation basis, let's consider the problem:
When the core ring is not slotted, according to Kirchhoff's second law, we have:
[Formula], we call this formula as formula 1.
Now the core is slotted. We know that the magnetic circuit is divided into two sections, one is the air gap magnetic circuit, and the other is the iron core magnetic circuit. These two magnetic circuits are connected in series. According to Kirchhoff's second law, we have:
[Formula], we call this formula as formula 2.
In Equation 2, the first term on the right is the magnetic flux and magnetic resistance of the air gap magnetic circuit, and the second term is the iron core magnetic flux and magnetic resistance.
Since the two magnetic circuits are connected in series, the magnetic fluxes in them are of course equal, and both are equal to [formula]. So Equation 2 is changed to:
[Formula], we call this formula as formula 3.
Now, we divide the left and right sides of Equation 3 by the left and right sides of Equation 1, thereby obtaining the ratio of the magnetic flux before and after the air gap is opened:
[formula]
If we look at the coil diagram, we will find that the width of the air gap is almost negligible for the circumference of the core ring, so there is: [formula], substituting the above formula, we get:
[Formula], we call this formula as formula 4
So what is the magnetic resistance Rj of the core ring? Its value is related to the cross-sectional area S of the iron core ring, the center line circumference L of the iron core ring, and the magnetic permeability of the iron core ring. Note that the radius of the outer ring of the iron core is Rw=25, and the radius of the inner ring is Rn=20, so the centerline radius is (Rw+Rn)/2. Let's look at its expression:
[formula]
What about [formula]? Its value is related to the air gap width parameter [formula] and the cross-sectional area S of the core ring, and its expression is:
[formula]
We substitute these two parameters into Equation 4, simplify and bring in specific values to obtain:
[formula]
Now, we finally get the result: the magnetic flux after the air gap is opened in the core ring is only 2.75% of the previous magnetic flux.
If we want to maintain the magnetic flux of the core ring after the air gap is opened, the magnetic flux does not change compared to the original one. From the [formula] we can see that because the current I is proportional to the air gap magnetic flux [formula], the current needs to increase. The multiple that the current increases is equal to the multiple that the magnetic flux decreases.
Now, let's answer and calculate the last question: If we want to keep the magnetic flux unchanged before and after, then the current I after opening the air gap must be how many times the iron core ring current [formula] when the air gap is not opened before?
[formula]
That is to say, for our example, after the air gap is opened, the current must be increased to 36.4 times the original, so as to keep the magnetic flux in the iron core ring basically unchanged.
Above, we take the toroidal core as an example to illustrate the problem.
For relays and contactors, their iron cores are generally U-shaped or E-shaped, and the magnetic flux before attraction (that is, the air gap flux) is about only the magnetic flux after attraction (that is, pure ferromagnetic flux). About 4%, after taking the reciprocal, the ratio of the pull-in current to the holding current is about 25 times.
We can see from the ABB contactor parameters given earlier. If the voltage before and after the contactor is closed, it can be seen from the table that the power consumption of the coil is 50VA when the contactor is closed, and the power consumption of the coil is 2.2VA when it is held. The ratio of the two is about 22.7 times. This is also the ratio of the pull-in current to the holding current of this type of contactor coil.
