【正文】 dby an I current. The39。s distance x from the end of the coil.The magnetic field at some given point (see Fig. 3), maybe calculated according to the BiotSavarLaplace formula[ l l ] . Recall, that the magnetic field produced by a smallsegment of wire, dl , canying a current I (see Fig. 4a) isgiven byWhere u0 is the permeability of the free space and d l x r isthe vector product of vectors dl and r .Hence, the magnitude of the magnetic field beesThe magnetic field of a circular contour with an a radius, asshown in Fig. 4b, is given byNote, that from considerations of symmetry, the fieldponent perpendicular to the coil axis dB, must be zero onthe axis.In order to evaluate the field due to the many turns ( N )along the axis of the coil, let n be the number of turns permetre. Also, consider the solenoid given in Fig. 3 as a series ofequidistant circular contours at the mutual distances dx,canying the current n l d x . The total axial field from all turnsof the coil beesIntegrating Eq. (5) within the interval θ1 ≤θ≤θ2, giveswhich can be rewritten in the formHowever, the electromagnet is posed of many turnslayers with variable radius r1≤ r≤ r2. Using the foregoing, inconjunction with Eq. (7), the magnetic field beesAnd so, the total magnetic field is given byThe magnetic force on the metallic sphere can beexpressed as [ I I ]where S is the material surface crossed by the magnetic , the upwards force on the ball due to the field B isgiven byAnother approach to the determination of the force on theball due to the field is based on the fact that the force isproportional both to the induced dipole strength and field strength. Namely, the magnetic field induces a dipole in the sphere whose strength is proportional to the magnetic field and that means to the coil current. Of course, it is necessary to make the assumption, that the sphere is magnetized within the linear part of the magnetization curve and does not reach saturation. sphere whose strength is proportional to the magnetic field andThe location of thedipole within the sphere issuch that each pole is atthe center of mass of its respective hemisphere（see ）The forces on thesphere due to the magneticfield are an attractive forceon the North pole and arepulsive one on the South pole. Hence, the upwards force on the ball due to the magneticfield B in conjunction with Eq. (9) is given by and hencewhere G39。10% in qxtreme conditions.However, many studies dealing with the modeling of the magnetic levitation system based on model linearization using Taylor39。s position x [l], [14]. It has the largest value when the bearing ball is next to the coil, and decreases to a constant value as the bearing hall is enough removed. Some typical approximations for L(x) are given byNote that over the range of x considered in the experiments Xmon,≤ X≤ Xmax one can pick the parameters L1（∞），Li1, Xi0 ，i = l , 2 , 3 in Eqs. (18) to (20) for mutual approach purpose [7],[8].V. CONCLUSIONUsing the magnetic field to levitate a steel ball, the Magnetic Levitation System as a teaching aid enables thetheoretical study and practical investigation of basic and advanced approaches to control of nonlinear unstable equipment may operate in stand alone mode without the the digital mode system operates with MAT LAB^ /SMULNKso ftware. The scope of experimentation includes dynamics modeling, identification, analysis and various conlroller design using classical and modern methods. For the modeling of the consider