Instruction

1

Based on the definition of inductance, it is easy to guess about the calculation of this value. The simplest formula to calculate the inductance of the solenoid is: L=f/I where L is inductance, f is the magnetic flux embracing the coil of the magnetic field, I is the current in the coil. This formula is defining the unit of inductance: 1 Weber / 1 Ampere = 1 Henry or, for short, 1 WB, 1 And = 1 GN.

Example 1. Through the coil flows a current of 2 A, formed around it a magnetic field, the magnetic flux which is 0,012 WB. Determine the inductance of this coil. Solution: L= 0,012 WB / 2 (A = 0,006 H = 6 mH.

Example 1. Through the coil flows a current of 2 A, formed around it a magnetic field, the magnetic flux which is 0,012 WB. Determine the inductance of this coil. Solution: L= 0,012 WB / 2 (A = 0,006 H = 6 mH.

2

Inductance (L) depends on the size and shape of the coils, from the magnetic properties of the environment in which the current-carrying conductor. Accordingly, the inductance of a long coil (solenoid) can be determined according to the formula given in figure 1, where µ0 – magnetic constant, equal to 12.6*(10) in -7 degree GN/m; µ - relative magnetic permeability of the medium, which is a coil current (tabular value specified in physical books); N is the number of turns in the coil, lкат – coil length, S – area of one loop.

Example 2. Find the inductance of the coil having the characteristics: the length is 0.02 m, the area round to 0.02 sq. m. number of turns = 200. Solution: If the environment in which the solenoid is not specified, the default is taken to air, the magnetic permeability of air is unity. Therefore, L = 12,6*(10) in -7 degree *1*(40000/0,02)*0,02=50,4*(10) in -3 degree H = 50,4 mH.

Example 2. Find the inductance of the coil having the characteristics: the length is 0.02 m, the area round to 0.02 sq. m. number of turns = 200. Solution: If the environment in which the solenoid is not specified, the default is taken to air, the magnetic permeability of air is unity. Therefore, L = 12,6*(10) in -7 degree *1*(40000/0,02)*0,02=50,4*(10) in -3 degree H = 50,4 mH.

3

Also calculate the magnetic induction of the solenoid is possible, based on the formula of magnetic field energy current (see figure 2). It can be seen that the induction can be calculated by knowing the energy of the field and the current in the coil is: L = 2W/(I) squared.

Example 3. Coil in which is flowing a current of 1 A, creates a magnetic field around itself with the energy of 5 joules. Determine the inductance of a coil. Solution: L = 2* 5/1 = 10 GN.

Example 3. Coil in which is flowing a current of 1 A, creates a magnetic field around itself with the energy of 5 joules. Determine the inductance of a coil. Solution: L = 2* 5/1 = 10 GN.