Unit 04: Electromagnetic Induction • Chapter 06
Inductance & The AC Generator
Electrical inertia, solenoid self & mutual inductance, stored magnetic energy, and working principle of the alternating current generator.
Mechanics mein jaise mass $m$ linear inertia hota hai jo achanak speed badhane ya rokne ka virodh karta hai, theek waise hi electrical circuits mein Inductance ($L$) electrical inertia hai! Jab aap switch on karte hain, inductor ek opposing back-EMF ($\mathcal{E} = -L\frac{dI}{dt}$) paida karke current ko achanak badhne se rokta hai, aur switch off karte waqt current ko girne se rokta hai (spark ke roop mein). Aur jab aap ek rectangular coil ko magnetic field mein lagatar ghumate hain, toh har aadhi cycle mein area vector $\vec{A}$ aur $\vec{B}$ ka angle $0^\circ \to 90^\circ \to 180^\circ$ continuous change hota hai—isi mechanical rotation se paida hoti hai hamare gharon ki Alternating Current (AC)!
1. Self & Mutual Inductance
Flux linkage proportional to circuit currents:
SI Unit: $\text{Henry (H)} = \text{Wb/A} = \text{V}\cdot\text{s/A}$. Scalar quantity depending on coil geometry and core permeability.
2. AC Generator Equation & Energy
Sinusoidal voltage and magnetic field energy density:
Notice: When magnetic flux $\Phi = BA\cos\omega t$ is maximum ($\theta=0^\circ$), induced EMF $\mathcal{E}$ is zero ($d\Phi/dt = 0$)!
Mathematical Derivation of Alternating EMF Generated in a Rotating Coil
Fig 6.3A: Essential components of an AC generator: Armature coil $ABCD$, permanent magnet poles, slip rings, and carbon brushes.
1. Consider a rectangular coil of $N$ turns and cross-sectional area $A$ rotating with constant angular velocity $\omega$ in a uniform magnetic field $\vec{B}$.
2. At time $t = 0$, let the plane of the coil be perpendicular to $\vec{B}$ (i.e. normal area vector $\vec{A}$ is parallel to $\vec{B}$, so $\theta = 0^\circ$). At time $t$, angle between $\vec{A}$ and $\vec{B}$ is $\theta = \omega t$.
3. Instantaneous magnetic flux linked with the $N$-turn coil:
4. By Faraday's Law of induction, induced EMF is the negative rate of change of flux:
5. If external load circuit has resistance $R$, the induced alternating current:
Self-Inductance of a Long Uniform Solenoid
1. Consider a long solenoid of length $l$, cross-sectional area $A$, and total turns $N$ (turns per unit length $n = N/l$).
2. Magnetic field inside the solenoid when carrying current $I$:
3. Magnetic flux through each single turn: $\Phi_1 = B A = \mu_0 \left(\frac{N}{l}\right) I A$.
4. Total magnetic flux linkage across all $N$ turns of the solenoid:
5. By definition of self-inductance ($N\Phi_B = L I$):
Magnetic Core Insertion: If a core of relative permeability $\mu_r$ is placed inside: $L = \mu_r \mu_0 n^2 A l$.
Mutual Inductance of Two Long Coaxial Solenoids
1. Consider two long coaxial solenoids $S_1$ and $S_2$ of common length $l$. Inner solenoid $S_1$ has radius $r_1$, turns $N_1$ ($n_1 = N_1/l$). Outer solenoid $S_2$ has radius $r_2$ ($r_2 > r_1$), turns $N_2$ ($n_2 = N_2/l$).
2. Pass current $I_2$ through outer solenoid $S_2$. The uniform magnetic field produced inside $S_2$ is $B_2 = \mu_0 n_2 I_2$.
3. This field passes through inner solenoid $S_1$ of cross-sectional area $A_1 = \pi r_1^2$. Total flux linked with all $N_1$ turns of inner solenoid:
4. By definition of mutual inductance ($N_1\Phi_1 = M_{12} I_2$):
Reciprocity Theorem: Mutual inductance is symmetric ($M_{12} = M_{21}$) and depends strictly on the area of the smaller inner core ($A_1 = \pi r_1^2$)!
Trap 1: Phase Difference Between Flux and Induced EMF
*Board Question:* In an AC generator, when the plane of the armature coil is perpendicular to magnetic field lines, what is the induced EMF?
• When plane $\perp \vec{B}$: Normal vector $\vec{A} \parallel \vec{B} \implies \Phi_B = NBA\cos 0^\circ = \mathbf{\Phi_{\max}}$.
• However, $\mathcal{E} = -d\Phi/dt = \mathcal{E}_0\sin 0^\circ = \mathbf{0}$!
Induced EMF is ZERO when flux is maximum, and maximum when flux is zero ($\Delta\phi = 90^\circ$ phase shift).
Trap 2: Dependence of Self-Inductance on Turns ($L \propto N^2$)
*Exam Trap:* If a wire of length $L_w$ is wound into a solenoid of $N$ turns, and then rewound into a solenoid of $2N$ turns with the same total wire length:
Because total wire length is constant, $2\pi r \times N = \text{const} \implies r \propto 1/N \implies A = \pi r^2 \propto 1/N^2$.
Length of solenoid $l \propto N$.
$$L = \frac{\mu_0 N^2 A}{l} \propto \frac{N^2 (1/N^2)}{N} \propto \frac{1}{N}$$
Always read whether the solenoid is *stretched* or *rewound with the same wire length*!
Trap 3: Area in Mutual Inductance of Unequal Solenoids
When deriving mutual inductance between outer solenoid of radius $r_2$ and inner solenoid of radius $r_1$ ($r_2 > r_1$):
Never use area of the outer solenoid $\pi r_2^2$! The field $B$ is only linked through the turns of the inner solenoid across its physical cross-section $\pi r_1^2$. The formula strictly uses $A_{\text{inner}} = \pi r_1^2$.
Trap 4: Why are Resistance Coils Doubly Wound?
Standard resistance boxes use wires folded back on themselves and then wound non-inductively (bifilar winding). Current flows in opposite directions in adjacent wires, so their magnetic fields cancel out everywhere ($\Phi_{\text{net}} = 0$). This reduces self-inductance to near zero ($L \approx 0$), preventing false inductive voltage spikes during switching!
Problem: An AC generator consists of a coil of $50\text{ turns}$ and area $2.5\text{ m}^2$ rotating at an angular speed of $60\text{ rad/s}$ in a uniform magnetic field $B = 0.30\text{ T}$ between two fixed poles. The resistance of the circuit including that of the coil is $500\text{ }\Omega$. (i) What is the maximum current drawn from the generator? (ii) What is the maximum power dissipation in the coil?
Solution:
1. Peak EMF ($\mathcal{E}_0$):
2. Maximum current drawn ($I_0$):
3. Maximum instantaneous power dissipation:
Note on Average Power: Over a complete cycle, average power dissipated is $P_{\text{avg}} = \frac{1}{2}\mathcal{E}_0 I_0 = 5062.5\text{ W}$.
Problem: A solenoid of length $50\text{ cm}$ with cross-sectional area $4.0\text{ cm}^2$ has $500\text{ turns}$. If a current of $2.0\text{ A}$ flowing through it is reversed in direction in $0.10\text{ s}$, find (i) self-inductance of the solenoid, (ii) magnitude of back-EMF induced during current reversal, and (iii) magnetic energy stored before reversal.
Solution:
1. Length $l = 0.50\text{ m}$, Area $A = 4.0 \times 10^{-4}\text{ m}^2$, $N = 500$:
2. Rate of change of current: Initial $I_1 = +2\text{ A}$, reversed $I_2 = -2\text{ A}$:
Induced back-EMF:
3. Stored magnetic energy $U$:
Self-Inductance
$L = \frac{\mu_0 N^2 A}{l}$
Mutual Inductance
$M = \frac{\mu_0 N_1 N_2 A_1}{l}$
AC Peak Voltage
$\mathcal{E}_0 = N B A \omega$
Stored Energy
$U = \frac{1}{2} L I^2$