Ejercicios Resueltos | Circuitos Magneticos

Antes de resolver ejercicios, recordemos las analogías clave entre circuitos eléctricos y magnéticos:

| Magnitud Eléctrica | Magnitud Magnética | Relación | |--------------------|--------------------|-----------| | Fuerza electromotriz (FEM, E) | Fuerza magnetomotriz (FMM, ℱ) | ℱ = N·I (Amperios-vuelta) | | Corriente (I) | Flujo magnético (Φ) | Unidad: Weber (Wb) | | Resistencia eléctrica (R) | Reluctancia (ℛ) | ℛ = l / (μ·A) | | Ley de Ohm: I = V/R | Ley de Hopkinson: Φ = ℱ / ℛ | |

Fórmulas críticas:

| Electric Circuit | Magnetic Circuit | |----------------|------------------| | Electromotive force (EMF), ( E ) (volts) | Magnetomotive force (MMF), ( \mathcalF = N I ) (ampere-turns) | | Current, ( I ) (amperes) | Magnetic flux, ( \Phi ) (webers) | | Resistance, ( R = \frac\rho lA ) (ohms) | Reluctance, ( \mathcalR = \fracl\mu A ) (A-turns/Wb) | | Conductivity, ( \sigma ) | Permeability, ( \mu = \mu_r \mu_0 ) | | Ohm’s law: ( I = E/R ) | Ohm’s law for magnetics: ( \Phi = \mathcalF / \mathcalR ) | | Kirchhoff’s voltage law (KVL) | Ampère’s law: ( \sum N I = \sum H l = \sum \Phi \mathcalR ) | | Kirchhoff’s current law (KCL) | Flux continuity: ( \sum \Phi = 0 ) at a node |

Key formulas:

Important note: In ferromagnetic materials, ( \mu_r ) is not constant (saturation, hysteresis). Many introductory solved exercises assume linearity (constant ( \mu_r )).

Datos:

Resultado: B_entrehierro ≈ 0.471 T; caída m.m.f.: hierro ≈ 112.5 A·v, aire ≈ 187.5 A·v.

Un circuito magnético tiene una sección transversal de 0,01 m² y una longitud de 0,5 m. La inducción magnética en el circuito es de 1,5 T. Calcular el flujo magnético. circuitos magneticos ejercicios resueltos

Solución

Φ = B * A = 1,5 T * 0,01 m² = 0,015 Wb

A well-structured set of "circuitos magnéticos ejercicios resueltos" should progressively build from simple toroids to complex three-leg cores with air gaps and non-linear materials. The best resources include:

If you are a student, focus on problems that require drawing the magnetic equivalent circuit first—that skill alone solves 80% of exam questions. For self-study, compare your solutions against solved examples that explain why a step is taken, not just the arithmetic. Important note: In ferromagnetic materials, ( \mu_r )

Problem:
A magnetic circuit consists of an iron core (mean length ( l_core = 0.3 ) m, ( A = 4 \times 10^-4 ) m², ( \mu_r = 1000 )) and an air gap of length ( l_g = 1 ) mm (( A_g = A )). The coil has ( N = 500 ) turns. Find the current ( I ) needed to produce a flux density ( B_g = 0.8 ) T in the air gap. Neglect fringing.

Solution:

Answer: ( I \approx 1.66 ) A.

(Note: The air gap dominates reluctance despite its small length.) If you are a student