MAGNETIC FIELD OF A MOVING PLATE

T. A. Lukovskaya

(Presented by the Department of Higher Mathematics of the Tomsk Polytechnic Institute)

The solution of the symmetric problem for an infinitely thin nonmagnetic plate was obtained by Yu. K. Fedosenko [1] in 1965. The problem was first posed by Steidinger [2], and was later considered by A. B. Sapozhnikov [3], but no solutions suitable for computation were obtained.

We shall consider a ferromagnetic plate whose cross section forms the region \(\left(-\dfrac{\delta}{2} \le z \le +\dfrac{\delta}{2}\right)\), with magnetic permeability \(\mu\), conductivity \(\sigma\), moving with constant velocity \(v\) in the positive direction of the \(OX\) axis.

Consider the following three cases.

§ 1. A metallic plate moves in the constant field of two like poles located at a distance \(h\) on both sides of the surface of the plate, Fig. 1. Denote by \(u=A_y\) the component of the vector potential \(\mathbf A\) along the \(OY\) axis in the region \(-\dfrac{\delta}{2}<z<\dfrac{\delta}{2}\); by \(u_0^+\) the component of the primary field in the region \(z>\dfrac{\delta}{2}\); and by \(u_0^-\) the component of the primary field in the region \(z<-\dfrac{\delta}{2}\). We denote the component of the secondary field by:

\[ u^+ \text{— in the region } z>\frac{\delta}{2}, \]

\[ u^- \text{— in the region } z<-\frac{\delta}{2}. \]

In the case of a linear magnet:

\[ u_0^+ = m \int_0^\infty e^{-\left|h+\frac{\delta}{2}-z\right|\alpha}\frac{\sin \alpha x}{\alpha}\,d\alpha; \]

\[ u_0^- = m \int_0^\infty e^{-\left|h+\frac{\delta}{2}+z\right|\alpha}\frac{\sin \alpha x}{\alpha}\,d\alpha. \]

In Chapter 22 of [3] it is shown that the problem reduces to solving the equations:

\[ \frac{\partial^{2}u^{+}}{\partial x^{2}}+\frac{\partial^{2}u^{+}}{\partial z^{2}}=0, \qquad z>\frac{\delta}{2} \tag{1} \]

\[ \frac{\partial^{2}u}{\partial x^{2}}-\omega\frac{\partial u}{\partial x} +\frac{\partial^{2}u}{\partial z^{2}}=0, \qquad -\frac{\delta}{2}<z<\frac{\delta}{2} \tag{2} \]

\[ \frac{\partial^{2}u^{-}}{\partial x^{2}}+\frac{\partial^{2}u^{-}}{\partial z^{2}}=0, \qquad z<-\frac{\delta}{2}, \tag{3} \]

where \(\omega=4\pi\sigma\mu v\cdot 10^{-9}\).

Let us assume that the external medium with respect to the plate is air \((\mu_{0}=1)\).

Fig. 1

At the boundaries of the adjacent media the following conditions hold:

\[ u\left(x,\frac{\delta}{2}\right) = u^{+}\left(x,\frac{\delta}{2}\right) + u_{0}^{+}\left(x,\frac{\delta}{2}\right); \]

\[ \frac{1}{\mu}\, \frac{\partial u\left(x,\frac{\delta}{2}\right)}{\partial z} = \frac{\partial u^{+}\left(x,\frac{\delta}{2}\right)}{\partial z} + \frac{\partial u_{0}^{+}\left(x,\frac{\delta}{2}\right)}{\partial z}, \tag{4} \]

\[ u\left(x,-\frac{\delta}{2}\right) = u^{-}\left(x,-\frac{\delta}{2}\right) + u_{0}^{-}\left(x,-\frac{\delta}{2}\right); \]

\[ \frac{1}{\mu}\, \frac{\partial u\left(x,-\frac{\delta}{2}\right)}{\partial z} = \frac{\partial u^{-}\left(x,-\frac{\delta}{2}\right)}{\partial z} + \frac{\partial u_{0}^{-}\left(x,-\frac{\delta}{2}\right)}{\partial z}. \tag{5} \]

We shall seek the solution of the problem posed in the form of Fourier integrals:

\[ u^{+}(x,z)=\int\limits_{0}^{\infty}\left[A^{+}(\alpha,z)\cos\alpha x+B^{+}(\alpha,z)\sin\alpha x\right]\,d\alpha,\qquad z>\frac{\delta}{2} \tag{6} \]

\[ u(x,z)=\int\limits_{0}^{\infty}\left[A(\alpha,z)\cos\alpha x+B(\alpha,z)\sin\alpha x\right]\,d\alpha,\qquad -\frac{\delta}{2}<z<\frac{\delta}{2} \tag{7} \]

\[ u^{-}(x,z)=\int\limits_{0}^{\infty}\left[A^{-}(x,z)\cos\alpha x+B^{-}(\alpha,z)\sin\alpha x\right]\,d\alpha,\qquad z<-\frac{\delta}{2}. \tag{8} \]

Substituting the integrals into equations (1)—(3) and using the uniqueness of representation by a Fourier integral (see [4]), we obtain the equations:

\[ \frac{d^{2}A^{+}}{dz^{2}}-\alpha^{2}A^{+}=0,\qquad \frac{d^{2}B^{+}}{dz^{2}}-\alpha^{2}B^{+}=0, \tag{9} \]

\[ \frac{d^{2}A^{-}}{dz^{2}}-\alpha^{2}A^{-}=0,\qquad \frac{d^{2}B^{-}}{dz^{2}}-\alpha^{2}B^{-}=0, \tag{10} \]

\[ \frac{d^{2}A}{dz^{2}}-\alpha^{2}A-\alpha\omega B=0,\qquad \frac{d^{2}B}{dz^{2}}-\alpha^{2}B+\alpha\omega A=0. \tag{11} \]

Particular solutions of equations (9) and (10) are the expressions \(\exp(+\alpha z)\) and \(\exp(-\alpha z)\), but for the upper medium \(z>\frac{\delta}{2}\) only the second gives a bounded solution, and for the lower medium \(z<-\frac{\delta}{2}\) only the first; thus we shall have:

\[ A^{+}(\alpha,z)=c_{1}^{+}(\alpha)e^{-\alpha z},\qquad B^{+}(\alpha,z)=c_{2}^{+}(\alpha)e^{-\alpha z}, \tag{12} \]

\[ A^{-}(\alpha,z)=c_{1}^{-}(\alpha)e^{\alpha z},\qquad B^{-}(\alpha,z)=c_{2}^{-}(\alpha)e^{\alpha z}. \tag{13} \]

System (11) reduces to a fourth-order equation

\[ \frac{d^{4}A}{dz^{4}}-2\alpha^{2}\frac{d^{2}A}{dz^{2}}+\alpha^{2}(\omega^{2}+\alpha^{2})A=0. \tag{14} \]

The general solution of this equation may be written in the form

\[ A(\alpha,z)=c_{1}e^{\kappa_{1}z}+c_{2}e^{\kappa_{2}z}+c_{3}e^{-\kappa_{1}z}+c_{4}e^{-\kappa_{2}z}, \tag{15} \]

where

\[ K_{1,2}=a\pm bi,\quad a=\frac{1}{\sqrt{2}}\sqrt{\alpha^{2}+\sqrt{\alpha^{2}\omega^{2}+\alpha^{4}}} =\frac{\alpha}{\sqrt{2}}\sqrt{1+\sqrt{1+\frac{\omega^{2}}{\alpha^{2}}}},\quad b=\frac{\alpha\omega}{2a} \tag{16} \]

or

\[ A(\alpha,z)=e^{az}(C_{1}\cos bz+C_{2}\sin bz)+e^{-az}(C_{3}\cos bz+C_{4}\sin bz), \tag{17} \]

if we set \(c_{1}+c_{2}=C_{1}\), \(c_{3}+c_{4}=C_{3}\), \(i(c_{1}-c_{2})=C_{2}\), \(i(c_{4}-c_{3})=C_{4}\).

From system (11) we have:

\[ B(\alpha,z)=\frac{1}{\alpha\omega}\left(\frac{d^{2}A}{dz^{2}}-\alpha^{2}A\right). \]

And, substituting the value of \(A(\alpha,z)\), we obtain:

\[ B(\alpha,z)=i\left(c_{1}e^{\kappa_{1}z}-c_{2}e^{\kappa_{2}z}+c_{3}e^{-\kappa_{1}z}-c_{4}e^{-\kappa_{2}z}\right) \]

or

\[ B(a,z)=e^{az}(C_2\cos bz-C_1\sin bz)+e^{-az}(C_3\sin bz-C_4\cos bz). \tag{18} \]

Subjecting the solutions (6)—(8) to the boundary conditions (4)—(5), we obtain a system of eight equations for the unknowns \(c_1^+\), \(c_2^+\), \(c_1\), \(c_2\), \(c_3\), \(c_4\), \(c_1^-\), \(c_2^-\):

\[ c_1 e^{\frac{\kappa_1\delta}{2}}+c_2 e^{\frac{\kappa_2\delta}{2}}+c_3 e^{-\frac{\kappa_1\delta}{2}}+c_4 e^{-\frac{\kappa_2\delta}{2}} = c_1^+ e^{-\frac{\alpha\delta}{2}}, \]

\[ c_1 e^{-\frac{\kappa_1\delta}{2}}+c_2 e^{-\frac{\kappa_2\delta}{2}}+c_3 e^{\frac{\kappa_1\delta}{2}}+c_4 e^{\frac{\kappa_2\delta}{2}} = c_1^- e^{-\frac{\alpha\delta}{2}}, \]

\[ \left[\kappa_1\left(c_1e^{\frac{\kappa_1\delta}{2}}-c_3e^{-\frac{\kappa_1\delta}{2}}\right) +\kappa_2\left(c_2e^{\frac{\kappa_2\delta}{2}}-c_4e^{-\frac{\kappa_2\delta}{2}}\right)\right] = -\mu\alpha c_1^+ e^{-\frac{\alpha\delta}{2}}, \]

\[ \left[\kappa_1\left(c_1e^{-\frac{\kappa_1\delta}{2}}-c_3e^{\frac{\kappa_1\delta}{2}}\right) +\kappa_2\left(c_2e^{-\frac{\kappa_2\delta}{2}}-c_4e^{\frac{\kappa_2\delta}{2}}\right)\right] = \mu\alpha c_1^- e^{-\frac{\alpha\delta}{2}}, \]

\[ i\left(c_1e^{\frac{\kappa_1\delta}{2}}-c_2e^{\frac{\kappa_2\delta}{2}}+c_3e^{-\frac{\kappa_1\delta}{2}}-c_4e^{-\frac{\kappa_2\delta}{2}}\right) = c_2^+e^{-\frac{\alpha\delta}{2}}+\frac{m}{\alpha}e^{-\alpha h}, \tag{19} \]

\[ i\left(c_1e^{-\frac{\kappa_1\delta}{2}}-c_2e^{-\frac{\kappa_2\delta}{2}}+c_3e^{\frac{\kappa_1\delta}{2}}-c_4e^{\frac{\kappa_2\delta}{2}}\right) = c_2^-e^{-\frac{\alpha\delta}{2}}+\frac{m}{\alpha}e^{-\alpha h}, \]

\[ i\kappa_1\left(c_1e^{\frac{\kappa_1\delta}{2}}-c_3e^{-\frac{\kappa_1\delta}{2}}\right) -i\kappa_2\left(c_2e^{\frac{\kappa_2\delta}{2}}-c_4e^{-\frac{\kappa_2\delta}{2}}\right) = -\mu\alpha c_2^+e^{-\frac{\alpha\delta}{2}}+m\mu e^{-\alpha h}, \]

\[ i\kappa_1\left(c_1e^{-\frac{\kappa_1\delta}{2}}-c_3e^{\frac{\kappa_1\delta}{2}}\right) -i\kappa_2\left(c_2e^{-\frac{\kappa_2\delta}{2}}-c_4e^{\frac{\kappa_2\delta}{2}}\right) = \mu\alpha c_2^-e^{-\frac{\alpha\delta}{2}}-m\mu e^{-\alpha h}. \]

It follows from this system that \(c_3=c_1\), \(c_4=c_2\), \(c_1^-=c_1^+\), \(c_2^-=c_2^+\). Then also \(C_3=C_1\), and \(C_4=-C_2\). And from (17) and (18) we shall have:

\[ A(a,z)=e^{az}(C_1\cos bz+C_2\sin bz)+e^{-az}(C_1\cos bz-C_2\sin bz), \]

or

\[ A(a,z)=2C_1\cos bz\,\operatorname{ch}az+2C_2\sin bz\,\operatorname{sh}az, \]

\[ B(a,z)=e^{az}(C_2\cos bz-C_1\sin bz)+e^{-az}(C_2\cos bz+C_1\sin bz) \]

or

\[ B(a,z)=2C_2\cos bz\,\operatorname{ch}az-2C_1\sin bz\,\operatorname{sh}az. \]

Now, instead of system (19), we shall have a system of four equations with four unknowns \(c_1^+\), \(C_1\), \(C_2\), \(c_2^+\):

\[ 2C_1\cos\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +2C_2\sin\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} = c_1^+e^{-\frac{\alpha\delta}{2}}, \]

\[ 2C_1\left(a\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2}\right) + \]

\[ +2C_2\left(a\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2}\right) = -\alpha\mu c_1^+e^{-\frac{\alpha\delta}{2}}, \tag{20} \]

\[ -2C_1\sin\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} +2C_2\cos\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} = c_2^+e^{-\frac{\alpha\delta}{2}}+\frac{m}{\alpha}e^{-\alpha h}, \]

\[ -2C_1\left(a\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2}\right) + \]

\[ +2C_2\left(a\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2}\right) = -\alpha\mu c_2^+e^{-\frac{\alpha\delta}{2}}+m\mu e^{-\alpha h}. \]

Solving system (20), we obtain:

\[ c_1=-\frac{m\mu p}{p^2+q^2}e^{-\alpha h},\qquad c_2=\frac{m\mu q}{p^2+q^2}e^{-\alpha h}, \]

\[ c_1^+=-\frac{m\mu}{p^2+q^2}(a\sin b\delta+b\sinh a\delta)e^{-\alpha\left(h-\frac{\delta}{2}\right)}, \]

\[ c_2^+=\frac{m\mu}{p^2+q^2}\left[ a\sinh a\delta-b\sin b\delta +2\alpha\mu\left(\sin^2\frac{b\delta}{2}\sinh^2\frac{a\delta}{2} +\cos^2\frac{b\delta}{2}\cosh^2\frac{a\delta}{2}\right) \right]e^{-\alpha\left(h-\frac{\delta}{2}\right)} -\frac{m}{\alpha}e^{-\alpha\left(h-\frac{\delta}{2}\right)}, \]

where

\[ p=\alpha\mu\sin\frac{b\delta}{2}\sinh\frac{a\delta}{2} +a\sin\frac{b\delta}{2}\cosh\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\sinh\frac{a\delta}{2} \]

\[ q=\alpha\mu\cos\frac{b\delta}{2}\cosh\frac{a\delta}{2} +a\cos\frac{b\delta}{2}\sinh\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\cosh\frac{a\delta}{2}. \tag{21} \]

Thus, for the values found for \(C_1, C_2, c_1^+\), and \(c_2^+\), the solution is expressed in the form of integrals:

\[ u^+(x,z)=\int_0^\infty [c_1^+(\alpha)\cos\alpha x+c_2^+(\alpha)\sin\alpha x]e^{-\alpha z}\,d\alpha \]

\[ u^-(x,z)=\int_0^\infty [c_1^+(\alpha)\cos\alpha x+c_2^+(\alpha)\sin\alpha x]e^{\alpha z}\,d\alpha, \qquad u^-(x,-z)=u^-(x,z), \]

\[ u(x,z)=2\int_0^\infty\left[(C_1(\alpha)\cos bz\cosh az+C_2(\alpha)\sin bz\sinh az)\cos\alpha x+\right. \]

\[ \left. +(C_2(\alpha)\cos bz\cosh az-C_1(\alpha)\sin bz\sinh az)\sin\alpha x \right]d\alpha. \]

We note that the field is symmetric with respect to the plane \(XOY\).

§ 2. The plate moves in the field of unlike poles arranged as in the first case. The primary field will have the form

\[ u_0^+=m\int_0^\infty e^{-\left|h+\frac{\delta}{2}-z\right|\alpha}\frac{\sin\alpha x}{\alpha}\,d\alpha; \qquad u_0^-=-m\int_0^\infty e^{-\left|h+\frac{\delta}{2}+z\right|\alpha}\frac{\sin\alpha x}{\alpha}\,d\alpha. \]

The problem reduces to solving equations (1)—(3) with boundary conditions (4)—(5).

As before, we seek the solution of the problem in the form of Fourier integrals, using the uniqueness of the representation of this solution by a Fourier integral. In this way we obtain

\[ c_3=-c_1,\quad c_4=-c_2,\quad \bar c_1=-c_1^+,\quad \bar c_2=-c_2^+ \quad\text{and}\quad C_3=-C_1,\quad C_4=-C_2. \]

In this case the solution is expressed in the form of integrals:

\[ u^+(x,z)=\int_0^\infty [c_1^+(\alpha)\cos\alpha x+c_2^+(\alpha)\sin\alpha x]e^{-\alpha z}\,d\alpha, \]

\[ u^-(x,z)=-\int_0^\infty [c_1^+(\alpha)\cos\alpha x+c_2^+(\alpha)\sin\alpha x]e^{\alpha z}\,d\alpha, \]

\[ u(x,z)=2\int_0^\infty\left\{[C_1(\alpha)\cos bz\sinh az+C_2(\alpha)\sin bz\cosh az]\cos\alpha x+\right. \]

\[ \left. +[C_2(\alpha)\cos bz\sinh az-C_1(\alpha)\sin bz\cosh az]\sin\alpha x \right\}d\alpha, \]

where

\[ C_1(\alpha)=-\frac{m\mu p_1}{p_1^2+q_1^2}\,e^{-\alpha h}; \qquad C_2(\alpha)=-\frac{m\mu q_1}{p_1^2+q_1^2}\,e^{-\alpha h}; \]

\[ c_1^+(\alpha)=\frac{m\mu}{p_1^2+q_1^2}(a\sin b\delta-b\sinh a\delta)\, e^{-\alpha\left(h-\frac{\delta}{2}\right)}, \]

\[ c_2^+(\alpha)=\frac{m\mu}{p_1^2+q_1^2} \left[ a\sinh a\delta+b\sin b\delta+ \right. \]

\[ \left. +2a\mu\left(\sin^2\frac{b\delta}{2}\cosh^2\frac{a\delta}{2} +\cos^2\frac{b\delta}{2}\sinh^2\frac{a\delta}{2}\right) \right]e^{-\alpha\left(h-\frac{\delta}{2}\right)} -\frac{m}{\alpha}e^{-\alpha\left(h-\frac{\delta}{2}\right)}. \]

\[ p_1=\alpha\mu\sin\frac{b\delta}{2}\cosh\frac{a\delta}{2} +a\sin\frac{b\delta}{2}\sinh\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\cosh\frac{a\delta}{2}, \]

\[ q_1=\alpha\mu\cos\frac{b\delta}{2}\sinh\frac{a\delta}{2} +a\cos\frac{b\delta}{2}\cosh\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\sinh\frac{a\delta}{2}. \]

§ 3. The plate moves in the field of four magnetic poles arranged as shown in Fig. 2.

Fig. 2

Keeping the previous notation, we shall have:

\[ u_0^+=m\int_0^\infty e^{-\left|h+\frac{\delta}{2}-z\right|\alpha} \,[\sin(x+d)\alpha-\sin(x-d)\alpha]\, \frac{d\alpha}{\alpha} \]

or

\[ u_0^+=2m\int_0^\infty e^{-\left|h+\frac{\delta}{2}-z\right|\alpha} \cos xz\,\sin d\alpha\,\frac{d\alpha}{\alpha}; \]

\[ u_0^-=2m\int_0^\infty e^{-\left|h+\frac{\delta}{2}+z\right|\alpha} \cos x\alpha\,\sin\alpha d\,\frac{d\alpha}{\alpha}. \]

As in the preceding cases, the problem reduces to solving equations (1)—(3). At the boundaries of adjacent media the conditions (4)—(5) hold. We shall seek the solution of the problem in the form of Fourier integrals (6)—(8). Subjecting the solution (6)—(8) to the boundary conditions (4)—(5), we obtain a system of eight equations

\[ c_1e^{\frac{\varkappa_1\delta}{2}} +c_2e^{\frac{\varkappa_2\delta}{2}} +c_3e^{-\frac{\varkappa_1\delta}{2}} +c_4e^{-\frac{\varkappa_2\delta}{2}} = c_1^+e^{-\frac{\alpha\delta}{2}} +\frac{2m}{\alpha}\sin\alpha d\cdot e^{-\alpha h}, \]

\[ C_1 e^{-\frac{\kappa_1\delta}{2}}+C_2 e^{-\frac{\kappa_2\delta}{2}}+C_3 e^{\frac{\kappa_1\delta}{2}}+C_4 e^{\frac{\kappa_2\delta}{2}} = c_1^{-}e^{-\frac{\alpha\delta}{2}}+\frac{2m}{\alpha}\sin \alpha d\cdot e^{-\alpha h}, \tag{22} \]

\[ \kappa_1\left(C_1 e^{\frac{\kappa_1\delta}{2}}-C_3 e^{-\frac{\kappa_1\delta}{2}}\right) +\kappa_2\left(C_2 e^{\frac{\kappa_2\delta}{2}}-C_4 e^{-\frac{\kappa_2\delta}{2}}\right) = -\alpha\mu c_1^{+}e^{-\frac{\alpha\delta}{2}}+2m\mu\sin\alpha d\cdot e^{-\alpha h}. \]

\[ \kappa_1\left(C_1 e^{-\frac{\kappa_1\delta}{2}}-C_3 e^{\frac{\kappa_1\delta}{2}}\right) +\kappa_2\left(C_2 e^{-\frac{\kappa_2\delta}{2}}-C_4 e^{\frac{\kappa_2\delta}{2}}\right) = \alpha\mu c_1^{-}e^{-\frac{\alpha\delta}{2}}-2m\mu\sin\alpha d\cdot e^{-\alpha h}. \]

\[ i\left(C_1 e^{\frac{\kappa_1\delta}{2}}-C_2 e^{\frac{\kappa_2\delta}{2}}+C_3 e^{-\frac{\kappa_1\delta}{2}}-C_4 e^{-\frac{\kappa_2\delta}{2}}\right) = c_2^{+}e^{-\frac{\alpha\delta}{2}}. \]

\[ i\left(C_1 e^{-\frac{\kappa_1\delta}{2}}-C_2 e^{-\frac{\kappa_2\delta}{2}}+C_3 e^{\frac{\kappa_1\delta}{2}}-C_4 e^{\frac{\kappa_2\delta}{2}}\right) = c_2^{-}e^{-\frac{\alpha\delta}{2}}. \]

\[ i\kappa_1\left(C_1 e^{\frac{\kappa_1\delta}{2}}-C_3 e^{-\frac{\kappa_1\delta}{2}}\right) -i\kappa_2\left(C_2 e^{\frac{\kappa_2\delta}{2}}-C_4 e^{-\frac{\kappa_2\delta}{2}}\right) = -\alpha\mu c_2^{+}e^{-\frac{\alpha\delta}{2}}. \]

\[ i\kappa_1\left(C_1 e^{-\frac{\kappa_1\delta}{2}}-C_3 e^{\frac{\kappa_1\delta}{2}}\right) -i\kappa_2\left(C_2 e^{-\frac{\kappa_2\delta}{2}}-C_4 e^{\frac{\kappa_2\delta}{2}}\right) = \alpha\mu c_2^{-}e^{-\frac{\alpha\delta}{2}}. \]

Whence

\[ C_3=C_1,\qquad C_4=C_2,\qquad c_1^{-}=c_1^{+},\qquad c_2^{-}=c_2^{+} \]

and

\[ C_3=C_1,\qquad C_4=-C_2. \]

And instead of the system (21) we shall have a system of four equations with the unknowns \(c_1^{+}(\alpha)\), \(C_1(\alpha)\), \(C_2(\alpha)\), \(c_2^{+}(\alpha)\):

\[ 2C_1\cos\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +2C_2\sin\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} = c_1^{+}e^{-\frac{\alpha\delta}{2}}+\frac{2m}{\alpha}\sin\alpha d\cdot e^{-\alpha h}, \]

\[ 2C_1\left(a\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2}\right) +2C_2\left(a\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2}\right) = \]

\[ = -\alpha\mu c_1^{+}e^{-\frac{\alpha\delta}{2}}+2m\mu\sin\alpha d\cdot e^{-\alpha h}, \tag{23} \]

\[ -2C_1\sin\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} +2C_2\cos\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} = c_2^{+}e^{-\frac{\alpha\delta}{2}}, \]

\[ -2C_1\left(a\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2} +b\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2}\right) + \]

\[ +2C_2\left(a\cos\frac{b\delta}{2}\operatorname{sh}\frac{a\delta}{2} -b\sin\frac{b\delta}{2}\operatorname{ch}\frac{a\delta}{2}\right) = -\alpha\mu c_2^{+}e^{-\frac{\alpha\delta}{2}}. \]

Solving the system (23), we obtain

\[ C_1(\alpha)=\frac{2m\mu\sin\alpha d}{p^2+q^2}\cdot q e^{-\alpha h},\qquad C_2(\alpha)=\frac{2m\mu\sin\alpha d}{p^2+q^2}\cdot p e^{-\alpha h}, \]

\[ c_1^{+}(\alpha)=\frac{2m\mu\sin\alpha d}{p^2+q^2} \left[ 2\alpha\mu\left( \sin^2\frac{b\delta}{2}\operatorname{sh}^2\frac{a\delta}{2} +\cos^2\frac{b\delta}{2}\operatorname{ch}^2\frac{a\delta}{2} \right) +\right. \]

\[ \left. +a\operatorname{sh}a\delta-b\sin b\delta-\frac{p^2+q^2}{\alpha\mu} \right] e^{-\left(h-\frac{\delta}{2}\right)\alpha}, \]

\[ c_2^{+}(\alpha)=\frac{2m\mu\sin\alpha d}{p^2+q^2} \left(a\sin b\delta+b\operatorname{sh}a\delta\right) e^{-\alpha\left(h-\frac{\delta}{2}\right)}. \]

where \(q\) and \(p\) are determined by formulas (21). Then the solution is expressed in the form of integrals:

\[ u^{+}(x,z)=\int_{0}^{\infty}\left[c_{1}^{+}(\alpha)\cos\alpha x+c_{2}^{+}(\alpha)\sin\alpha x\right]e^{-\alpha z}\,d\alpha, \]

\[ u^{-}(x,z)=\int_{0}^{\infty}\left[c_{1}^{+}(\alpha)\cos\alpha x+c_{2}^{+}(\alpha)\sin\alpha x\right]e^{\alpha z}\,d\alpha, \]

\[ \begin{aligned} u(x,z)=2\int_{0}^{\infty}\{& \left[C_{1}(\alpha)\cos bz\,\operatorname{ch}az+C_{2}(\alpha)\sin bz\,\operatorname{sh}az\right]\cos\alpha x+\\ &+\left[C_{2}(\alpha)\cos bz\,\operatorname{ch}az-C_{1}(\alpha)\sin bz\,\operatorname{sh}az\right]\sin\alpha x\}\,d\alpha . \end{aligned} \]

Let us note that we can solve the problem as soon as the primary field is specified in the form of a Fourier integral. In particular, it is easy to obtain a solution for the case in which the primary field of line currents is specified. For the numerical analysis of the problem, it is best, using the relation \(\overline{B}=\operatorname{rot}\overline{A}\), to write expressions for \(H_x\) and \(H_z\)

\[ \mu H_x=-\frac{\partial u}{\partial z}, \qquad \mu H_z=\frac{\partial u}{\partial x} \]

and then to calculate their values on a computer at various points for various values of the parameters.

REFERENCES

1. Yu. K. Fedosenko. “Defectoscopy,” No. 4, 1965, p. 8.
2. W. Steidinger. Arch. für. El. XXII, 1953, 1929.
3. A. B. Sapozhnikov. Fundamentals of Electromagnetic Defectoscopy of Metallic Bodies. Dissertation, Tomsk, 1951, Ch. 22.
4. G. A. Byuler and T. A. Lukovskaya. — “Fizika,” No. 2, 1968.