A magnetic flux of 0.015 Vb occurs in the coil with a current strength in the turns of 5.0 A.

A magnetic flux of 0.015 Vb occurs in the coil with a current strength in the turns of 5.0 A. How many turns does the coil contain if its inductance is 60 mH.

To find the number of turns contained in the specified coil, we apply the formula: L = N * F / I, whence we express: N = I * L / F.

Variables: I – current in turns (I = 5 A); L – inductance of the coil (L = 60 mH = 0.06 H); Ф – the arisen magnetic flux (Ф = 0.015 Wb).

Let’s perform the calculation: N = I * L / F = 5 * 0.06 / 0.015 = 20 pcs.

Answer: The indicated coil contains 20 turns.