The system below is a contactor The system is lying down and is not affected by the force of gravity. Furthermore, the two parts of the core are symmetric, that is, the dimensions of the fixed part (lower) are equal to those of the mobile part (upper). Hy = 500 P=2cm 1-1 1,5cm 1cm 1cm 2cm 2cm 2cm 1cm a) Obtain the inductance of the circuit as a function of the position x of the moving part of the core. b) Calculate the magnetic force as a function of position x and current i. c) Obtain the current differential equation. d) Obtain the differential equation for the position x. e) Obtain the differential equation of velocity v. Note: As expected, to simulate the system dynamics, we should obtain three differential equations. To ob- tain the current differential equation, we must obtain it from the manipulation of the electrical circuit mesh equation in the figure below. + R Circuito Mag In this circuit, Vin is a constant voltage that will feed the coil and is the induced voltage of the coil. To obtain the mechanical equations, that is, the differential equations of x and v, we must draw on our knowledge of basic physics. ma = 2 Forces Where m and a are the mass and acceleration of the moving part. Consider three forces in the system, the magnetic force, the spring force, and the aerodynamic friction force (-bv, where b is the coeffici- ent of friction). Remember that the equations must be in Function of x, v and i. That is, the a cannot appear. The diffe- rential equation of x is straightforward, that is, it is a general definition of kinetics. Use bold letters for the parameters below: Moving part mass m; Aerodynamic friction coefficient b; Number of turns N; Coil Resistance R; Spring force constant k; Condition in which the spring is not stressed xr; Supply voltage Vin The system below is a contactor The system is lying down and is not affected by the force of gravity. Furthermore, the two parts of the core are symmetric, that is, the dimensions of the fixed part (lower) are equal to those of the mobile part (upper). Hy = 500 P=2cm 1-1 1,5cm 1cm 1cm 2cm 2cm 2cm 1cm a) Obtain the inductance of the circuit as a function of the position x of the moving part of the core. b) Calculate the magnetic force as a function of position x and current i. c) Obtain the current differential equation. d) Obtain the differential equation for the position x. e) Obtain the differential equation of velocity v. Note: As expected, to simulate the system dynamics, we should obtain three differential equations. To ob- tain the current differential equation, we must obtain it from the manipulation of the electrical circuit mesh equation in the figure below. + R Circuito Mag In this circuit, Vin is a constant voltage that will feed the coil and is the induced voltage of the coil. To obtain the mechanical equations, that is, the differential equations of x and v, we must draw on our knowledge of basic physics. ma = 2 Forces Where m and a are the mass and acceleration of the moving part. Consider three forces in the system, the magnetic force, the spring force, and the aerodynamic friction force (-bv, where b is the coeffici- ent of friction). Remember that the equations must be in Function of x, v and i. That is, the a cannot appear. The diffe- rential equation of x is straightforward, that is, it is a general definition of kinetics. Use bold letters for the parameters below: Moving part mass m; Aerodynamic friction coefficient b; Number of turns N; Coil Resistance R; Spring force constant k; Condition in which the spring is not stressed xr; Supply voltage Vin