Example

Question:

A solenoid has a core of material with relative permeability 400. The windings are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) \(H\), (b) \(M\), (c) \(B\) and (d) the magnetising current \(I_m\).

Solution:

(a) Field \(H\): \[ H = nI = 1000 \times 2.0 = 2 \times 10^3\,\text{A/m} \] (b) Magnetic field \(B\): \[ B = \mu_r \mu_0 H = 400 \times 4\pi \times 10^{-7} \times 2 \times 10^3 = 1.0\,\text{T} \] (c) Magnetisation \(M\): \[ M = \frac{B - \mu_0 H}{\mu_0} = (\mu_r - 1) H = 399 \times H \approx 8 \times 10^5\,\text{A/m} \] (d) Magnetising current \(I_m\) (additional current for same \(B\) in absence of core):
\[ B = \mu_r n (I + I_m) \] Using \(I = 2\,\text{A}\), \(B = 1\,\text{T}\): \[ I_m = 794\,\text{A} \]