The electric current passing on the conductor creates around it magnetic field. The proportionality coefficient between current in a contour and the magnetic flux created by this current is called inductance of the coil.
Instruction
1. Leaning on definition term inductance, it is easy to guess calculation of this size. The simplest formula for calculation of inductance of the solenoid looks so: L=F/I, where L – inductance of a contour, F - a magnetic flux of the magnetic field covering the coil, I-current in the coil. This formula is the inductance defining unit of measure: 1 Weber / 1 Ampere = 1 Henry or, for short, 1 Wb / 1 A = 1 Gn. Example 1. On the coil the current with a force of 2 flows And, around it magnetic field which magnetic flux makes 0.012 Wb was formed. Define inductance of this coil. Decision: L = 0.012 Wb / 2 A = 0.006 Gn = 6 mgn.
2. The inductance of a contour (L) depends on the sizes and a form of the coil, on magnetic properties of the environment in which there is this conductor with current. Proceeding from it, the inductance of the long coil (solenoid) can be determined by the formula specified in figure 1 where µ0 – the magnetic constant equal 12.6*(10) in-7 degrees of Gn/m; µ - relative magnetic permeability of the environment in which the coil with current (the table value specified in physical reference books) is located; N – the number of rounds in the coil, lkat is length of the coil, S is the area is one round. Example 2. To find inductance of the coil having characteristics: length – 0.02 m, the area of a round – 0.02 sq.m., number of rounds = 200. Decision: If Wednesday in which there is a solenoid is not specified, then air by default undertakes, the magnetic permeability of air is equal to unit. Therefore, L = 12.6*(10) in-7 degrees * 1*(40000/0.02)*0.02=50.4*(10) in-3 degrees of Gn = 50.4 mgn.
3. Also it is possible to calculate magnetic induction of the solenoid, leaning on a formula of energy of magnetic field of current (see figure 2). From it it is visible that induction can be calculated, knowing energy of the field and current in the coil: L = 2W / (I) in a square. Example 3. The coil in which current 1 A flows creates around itself magnetic field energy 5 J. Opredelita inductance of such coil. Decision: L = 2 * 5/1 = 10 Gn.