Question

An electromagnet consists of an iron core with an air gap as shown below. The core is square shaped with a side length of $L_{1}$ and has a relative permeability of $\mu _{r}$ and a cross-sectional area of A. The air gap has a length of $L_{g}$, and = ($L_{1}$ - $L_{g}$)/2. The coil has N turns. Derive a formula for the magnitude of the magnetic flux density B and the magnitude of the magnetic field H inside the air gap.

Answer 1

Given that There is a electromagnet is given And it consists of an iron core. Side length = L_1 Relative permeability = mu_r Cross sectional Area = A Air Gap length = L_1 - L_g/2 The coil has N turns Now we have to derive the formula for the magnetic flux density B and the magnitude of the magnetic field H. For calculating this we have MMF = NI, Rl_core = l/mu A = 3L-1 + 2L_2/mu_0 mu_r A 3L_1 + 2L_2 is the total length of core Rl_air = l/mu A = L_g/mu_0 A Rl_core = 3L_1 + 2L_2/mu_0 mu_r A = 3L_1 + L-1 - L_g/mu_0 mu_r A \r\n Rl_core = 4L_1 - L_g/mu_0 mu_r A Then, MMF = Phi(Rl_cora + Rl_air) where, Phi = MMF/Rl_core + Rl_air = NI/4L_1 - L_g/mu_0 mu_r A + L_g/mu_0 A = NI/4L_1 L_g + L_g(mu_r - 1)/mu_0 mu_r A Phi = NI mu_0 mu_r A/4L_1 + L_g(mu_r - 1) Then, we have also Phi = integral_s B. ds we have B = Phi/A = NI mu_0 m_r A/4L-1 + L_g(mu_0 - 1) A B = N I mu_0 mu_r/4L_1 + L_g(mu_0 - 1) tesla This is the required formula for the