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Fig. 15 shows magnetic field produced by current in a solenoid.

Let H = magnetizing force along the axis of the solenoid.

Assumptions :

  • H remains constant through the length l of the solenoid.
  • The value of H outside the solenoid is negligible

Let a unit N-pole is placed at a point A (outside the solenoid) and is taken once round the completed path (shown dotted in Fig. 15) in a direction opposite to H.

Now work done in one round = H × l joules

The ampere-turns linked with this path = NI

Where N = number of turns of solenoid, and

I = current (A) passing through the solenoid.

As per work law, H × l = N × I






Example 3. A solenoid of 300 turns is wound on a continuous ring of magnetic material of relative permeability 1000. If the flux density in the magnetic material of the core in the solenoid is 1.3 T, what is the current in the solenoid? Mean diameter of the solenoid is 100 mm.

Solution. Number of turns of the solenoid, N = 300

Relative permeability, µr = 1000

Flux density B = 1.3 T

Mean diameter of the solenoid, d = 100 mm = 0.1 m