Comprehensive Guide to Self Inductance, Inductors, and Magnetic Energy

Overview of Inductors and Inductance (20.2.1)

• Definition of an Inductor: Inductors consist of a conductor wound into a coil. When a current flows through this coil, energy is temporarily stored in the magnetic field generated within it.

• Electromagnetic Induction Principle: When the current flowing through an inductor changes, the time-varying magnetic field induces a voltage in the conductor. According to Faraday's law of electromagnetic induction, this induced voltage opposes the change in current that created it.

• Inductance (L): An inductor is characterized by its property called inductance. It is defined as the ratio of the induced voltage to the rate of change of current.

• Units and Range: The SI unit of inductance is the Henry ($H$). Typical inductance values for commercial components range from $1\,\mu H$ ($10^{-6}\,H$) to $1\,H$.

• Alternative Name: An inductor is also commonly referred to as a "Choke".

Physics of Self-Inductance (20.2.2)

• Definition of Self-Inductance: This phenomenon occurs when a changing magnetic flux through a circuit arises from the circuit itself rather than an external source. The sequence of causality is defined as: $\Delta I \rightarrow \Delta B \rightarrow \Delta \Phi \rightarrow \epsilon_{ind}$.

• Impact of Current Changes:     * Current Increase: As current ($I$) increases, the magnetic flux through the loops of the coil also increases. This induces an electromotive force (emf) that opposes the change in magnetic flux. As the magnitude of the current increases, the rate of increase gradually lessens, and the induced emf decreases, resulting in a gradual increase of current until it reaches a constant state ($I_{max}$).     * Current Decrease: If the current decreases, the direction of the induced emf is the same as the original current to oppose the drop.

• Direction of Induced EMF (Lenz's Law):     * When current increases: Induced emf direction is opposite to the direction of current $I$.     * When current decreases: Induced emf direction is the same as the direction of current $I$.

Mathematical Foundations and Formulas

• Magnetic Field in a Solenoid:     * $B = \mu_0 n I$     * Where $n = \frac{N}{l}$ (number of turns per unit length).

• Self-Inductance Constant ($L$): The self-induced emf is proportional to the time rate of change of the current.     * $\epsilon = -N \frac{d\Phi}{dt} = N \frac{d(B \cdot A)}{dt} = NA \frac{dB}{dt}$     * Substituting the solenoid field: $\epsilon = NA \left( \frac{\mu_0 N}{l} \right) \frac{dI}{dt}$     * Therefore, $\epsilon = -L \frac{dI}{dt}$

• Inductance for a Solenoid:     * $L = \frac{\mu_0 N^2 A}{l}$     * Where:         * $N$ is the number of coils (more coils increase inductance).         * $\mu_0$ is the permeability of the core ($1.2567 \times 10^{-6}$, approximately $4\pi \times 10^{-7}$).         * $A$ is the cross-sectional area of the coil (larger area increases inductance).         * $l$ is the length of the coil (a shorter coil increases inductance due to narrower/overlapping coils).

• Magnetic Flux Relationship:     * $\Phi_B = \frac{LI}{N}$     * $L = \frac{N \Phi_B}{I}$

• Back EMF: The self-induced emf is frequently called "back emf" because it opposes the changing current established by the circuit's battery.

Mutual Inductance and Transformers

• Mutual Inductance (M): Occurs when the changing current in one coil induces an emf in a second, adjacent coil.     * $\epsilon_1 = -M \frac{di_2}{dt}$     * $\epsilon_2 = -M \frac{di_1}{dt}$     * Relationship: $M_{12} = M_{21} = M$     * Formula for Mutual Inductance: $M = \frac{\mu_0 N_p N_s A_s}{l}$

• Transformers: Devices used to change voltage and current levels.     * Ratio: $\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}$     * Ideal Transformer Model: Assumes no energy loss, therefore input power equals output power: $V_p I_p = V_s I_s$.     * Power Loss and Efficiency: Power is lost primarily due to resistance in the windings: $P = VI = I^2 R$. Efficiency is measured as the ratio of $P_{output}$ to $P_{input}$.

Energy Stored in an Inductor (20.2.3)

• Theory of Energy Storage: The induced emf prevents a battery from establishing a current instantaneously. The battery must perform work to push charges against this back emf. This work is stored as potential energy in the inductor's magnetic field.

• Energy Formula:     * $U = \frac{1}{2} L I^2$

• Comparison with Capacitors:     * Geometric Dependence: Capacitor $C = \frac{\epsilon_0 A}{d}$ vs. Inductor $L = \frac{\mu_0 N^2 A}{l}$.     * Variables: Capacitor depends on Voltage ($C = \frac{Q}{V}$) vs. Inductor depends on Current ($L = \frac{N\Phi}{I}$).     * Energy Formula: Capacitor $U = \frac{1}{2} C V^2$ vs. Inductor $U = \frac{1}{2} L I^2$.

Practical Exercises and Quantitative Solutions

• Exercise 1: Derivation & Identification:     * Question: Derive the back emf equation and identify factors changing inductor value.     * Solution: Derivation follows $\epsilon = N \frac{d\Phi}{dt} \rightarrow \dots \rightarrow L \frac{dI}{dt}$. Factors include $N$ (turns), $\mu_0$ (core material), $A$ (area), and $l$ (length).

• Exercise 2: Calculating Induced EMF:     * Parameters: $L = 25\,mH$, $I_1 = 12\,A$, $I_2 = 27\,A$, $\Delta t = 125\,ms$.     * Calculation: $\Delta I = 27 - 12 = 15\,A$. $\epsilon = - (25 \times 10^{-3}) \left( \frac{15}{125 \times 10^{-3}} \right) = -3\,V$.     * Explanation: The magnitude is $3\,V$. Because the current increases, the direction of the induced emf is opposite to the original current. This reduces the change of original current.

• Exercise 3: Solenoid Properties:     * Parameters: $l = 50\,cm$, diameter = $10\,cm$ ($r = 0.05\,m$), $N = 700$.     * Calculation: $A = \pi \times 0.05^2$. $L = \frac{(1.2567 \times 10^{-6}) (700^2) (\pi \times 0.05^2)}{0.5} \approx 9.67 \times 10^{-3}\,H = 9.67\,mH$.

• Exercise 4: Current Change Rates:     * Parameters: $P = 20\,W$, $L = 300\,mH$, $I = 3.5\,A$.     * Calculation: $V = \frac{P}{I} = \frac{20}{3.5} = 5.71\,V$. $\frac{dI}{dt} = \frac{V}{L} = \frac{5.71}{0.3} = 19\,As^{-1}$.

• Exercise 5: Energy Storage:     * Parameters: Solenoid $N = 300$, $l = 0.25\,m$, $A = 4.0\,cm^2$ ($4.0 \times 10^{-4}\,m^2$), $I = 0.5\,mA$.     * Calculation:         1. $L = \frac{(1.2567 \times 10^{-6}) (300^2) (4 \times 10^{-4})}{0.25} = 1.81 \times 10^{-4}\,H$.         2. $E = \frac{1}{2} (1.81 \times 10^{-4}) (0.5 \times 10^{-3})^2 = 2.26 \times 10^{-11}\,J$.

Questions & Discussion

• The session concludes by inviting questions from participants before moving to the next topic.