Calculation of the magnetic field generated by a current-carrying conductor in a rectangular slot
Abstract Generated abstract
This paper addresses the engineering problem of calculating the two-dimensional magnetic field of a rectilinear current-carrying conductor located in a rectangular slot, as encountered in transformer windings and electric machine slots. Assuming infinitely permeable ferromagnetic boundaries, it formulates the field through Poisson’s equation for the magnetic vector potential and applies Kantorovich’s direct variational method with a single product approximation. The derivation yields an analytical expression for the vector potential, with coefficients determined from an approximated current-density distribution and boundary conditions. The resulting formula is presented as a convenient means for computing magnetic induction components at points in the slot, with a note on improving accuracy for the radial component by interchanging the prescribed and computed trial functions.
Full Text
CALCULATION OF THE MAGNETIC FIELD CREATED BY A CURRENT-CARRYING CONDUCTOR IN A RECTANGULAR SLOT
V. A. Lukutin
(Presented by the scientific seminar of the Department of Theoretical Foundations of Electrical Engineering)
The determination of electrodynamic forces in the windings of transformers and electrical machines, as well as the calculation of the leakage inductances of these windings, ultimately reduces to the calculation of the magnetic field created by a rectilinear current-carrying conductor placed in a slot of rectangular cross section. Such calculations are very common in engineering practice; therefore this question has often been discussed in the literature [2, 3, 4].
Usually the field is reduced to a two-dimensional one; for this purpose the cylindrical winding of transformers is developed into a rectilinear strip, and it is assumed that the surfaces of the ferromagnetic cores are planes bounding the computational region on three sides. Figure 1 shows a single-phase transformer with two windings; in the right half of this figure the computational region is shown, the boundaries of which are hatched. A winding is placed in the region, its cross section
Fig. 1.
which is represented by a rectangle with sides \(2a\) and \((d-c)\), and whose magnetomotive force is equal to \(Iw\).
Let us note that the same region is obtained when studying the field of current-carrying conductors placed in the slots of the rotor or stator of an electric machine. In this case the ferromagnetic slot walls are straightened, and we neglect the bulging of the flux.
Taking into account the influence of ferromagnetic boundaries forces one to use the method of mirror images, which leads to a cumbersome solution in the form of an infinite series, very inconvenient for engineering purposes [2]. In this connection approximate solutions are used, substantially simplifying the problem but at the same time losing accuracy [3].
Below we show the possibility of obtaining an approximate analytical expression for the vector potential in the region shown in Fig. 1, using Kantorovich’s direct variational method [1].
It is known that the field in the region satisfies Poisson’s equation
\[ \frac{\partial^2 A}{\partial x^2}+\frac{\partial^2 A}{\partial y^2} =-\mu\delta(x,y). \tag{1} \]
Here \(\delta\) is the current density in the cross section of the winding,
\(A\) is the magnetic vector potential.
If we assume that the ferromagnetic medium surrounding the region has \(\mu=\infty\), then the boundary conditions will be as follows:
\[ B_x=\frac{\partial A}{\partial y}=0,\quad \text{for } y=n \tag{2} \]
\[ B_y=-\frac{\partial A}{\partial x}=0,\quad \text{for } y=0 \text{ and } x=\pm m. \]
We shall seek the solution of the problem in the form of a single term, representing the product of two functions, one of which we choose arbitrarily. This function must satisfy the boundary conditions (2) and be, at least, twice differentiable. Taking into account the symmetry of the field, one may write the following solution of the stated problem:
\[ A(x,y)=\varphi(x)f(y), \tag{3} \]
where
\[ \varphi(x)=m^2-x^2+\frac{x^4}{2m^2} \]
is an arbitrarily chosen function, and \(f(y)\) is the unknown function to be determined.
Substitute the chosen solution into the functional [1]
\[ J=\iint\left[ \left(\frac{\partial A}{\partial x}\right)^2+ \left(\frac{\partial A}{\partial y}\right)^2 -2\mu\delta A \right]\,dx\,dy. \tag{4} \]
and carry out the integration with respect to one variable over the limits from \(x=-m\) to \(x=m\), taking into account that \(\delta=0\) for \(|x|>a\). Having performed all the necessary calculations, we obtain
\[ J=\int F(f',f,y)\,dy, \]
\[ F=0.609\,f^2(y)+1.237\,m^5 f'^{\,2}(y) -0.133\,\mu\delta(y)f(y) \left(30m^2a-10a^3\frac{3a^5}{m^2}\right). \tag{5} \]
Now let us use the Euler equation [1]
\[ \frac{\partial F}{\partial f}-\frac{\partial}{\partial y}\frac{\partial F}{\partial f'}=0 \tag{6} \]
and, substituting (5) into it, we obtain the following differential equation:
\[ f''(y)-\frac{0.493}{m^2}f(y) = -\mu \delta(y)\frac{0.539}{m^5} \left(30m^2a-10a^3+\frac{3a^5}{m^2}\right). \tag{7} \]
The general solution of (7) is written as
\[ f_1(y)=K_1e^{p_1y}+K_2e^{p_2y}, \tag{8} \]
where
\[ p_1=\frac{0.702}{m},\qquad p_2=-\frac{0.702}{m}. \]
On the right-hand side of equation (7) stands the current-density distribution function, which may be approximated by a power series
\[ \delta(y)=\frac{I\omega}{2a(d-c)} \left(D_1y+D_3y^3+\cdots+D_ky^k+\cdots\right). \]
If we restrict ourselves to the first three terms of the series, then the particular solution of equation (7) is written in the form
\[ f_2(y)=C_1y+C_3y^3+C_5y^5. \tag{9} \]
The integration constants \(K_1\) and \(K_2\) are determined by the choice of the initial line of force. Setting \(A=0\) at \(y=0\), from (8), (9) we obtain
\[ 0=f_1(0)+f_2(0)=K_1+K_2. \tag{10} \]
If we use the boundary condition \(\dfrac{\partial A}{\partial y}=0\) at \(y=n\), then it is possible to determine
\[ K_1=-K_2=-\frac{m\left(C_1+3C_3n^2+5C_5n^4\right)}{1.404\,\operatorname{Ch}p_1}. \tag{11} \]
The coefficients \(C_k\) are found by substituting the particular solution \(f_2(y)\) into differential equation (7):
\[ 6C_3y+20C_5y^3-\frac{0.493}{m^2} \left(C_1y+C_3y^3+C_5y^5\right)= \]
\[ =-\mu\delta_0^2\frac{0.0539}{m^5} \left(D_1y+D_3y^3+D_5y^5\right)\cdot b. \]
Equating the factors of equal powers of \(y\) in the last equality, we obtain formulas for finding \(C_k\):
\[ \begin{gathered} C_1=\frac{\mu \delta_0 b}{m^3}\left(0.109D_1+m^2 1.33D_3+m^4 53.9D_5\right);\\ C_3=\frac{\mu \delta_0 b}{m^3}\left(0.109D_3+m^2 4.43D_5\right);\\ C_5=\frac{\mu \delta_0 b}{m^3}D_5;\qquad \delta_0=\frac{I\omega}{2a(d-c)};\\ b=30m^2a-10a^3+\frac{3a^5}{m^2}. \end{gathered} \tag{12} \]
Now the coefficients found can be substituted into our solution. In general form it is
\[ A(x,y)=\left(m^2-x^2+\frac{x^4}{2m^2}\right) \left[ C_1y+C_3y^3+C_5y^5- \frac{m\left(C_1+3C_3n^2+5C_5n^4\right)} {0.702\,\operatorname{Ch}p_1} \cdot \operatorname{Sh}\frac{0.702y}{m} \right]. \tag{13} \]
This formula is convenient for calculations; it makes it possible to find easily the components of the magnetic induction at any point in space from the formulas
\[ B_x=\frac{\partial A}{\partial y};\qquad B_y=-\frac{\partial A}{\partial x}. \tag{14} \]
In a number of problems the radial \((B_y)\) component of the magnetic induction is of greatest interest. In this case it is advisable to prescribe in solution (3) the function \(f(y)\), and to compute \(\varphi(x)\) by the indicated method, which will make it possible to obtain more accurate results.
References
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L. V. Kantorovich, V. I. Krylov. Approximate Methods of Higher Analysis. Moscow, State Publishing House of Technical-Theoretical Literature, 1949.
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K. Binns, P. Lawrence. Analysis and Calculation of Electric and Magnetic Fields. “Energiya,” 1970.
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G. N. Petrov, I. S. Nayashkov. Electrodynamic forces in transformers. “Elektrichestvo,” 1955, No. 8.
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D. N. Morozov. Transformers. Moscow, Gosenergoizdat, 1962.