Complete Guide to Electromagnetic Induction (EMI) for Class 12 | JEE Physics

Electromagnetic Induction Class 12 / Higher Secondary WBCHSE/CBSC/ICSC NEET JEE Main/Advanced

Electromagnetic Induction: Theoretical Foundations, Mathematical Derivations & Complete Solved Problem Sets

An exhaustive, mathematically rigorous, and exam-oriented masterclass covering Magnetic Flux, Faraday's Laws, Lenz's Law, Motional EMF, Inductance Networks, Transient LR Circuits, LC Harmonic Oscillators, and Engineering Applications.

Module Sections

• Curriculum Roadmap
• 1.1 Magnetic Field Lines & Flux
• 1.2 Faraday's Law of Induction
• 1.3 Lenz's Law & Energy Principles
• 1.4 Motional Electromotive Force
• 1.5 Self-Inductance & Field Energy
• 1.6 Mutual Inductance & Coupling
• 1.7 Transients in L-R Circuits
• 1.8 Oscillations in L-C Circuits
• 1.9 Industrial Applications of EMI
• Comprehensive Formula Reference

Curriculum Roadmap

This module provides an exhaustive, university-level analysis of Electromagnetic Induction (EMI). All major sub-disciplines prior to steady-state alternating current networks are structured logically below:

Curriculum Breakdown:

• 1.1: Magnetic Field Line Topologies, Vector Surface Normals, and Magnetic Flux Integrals ($\phi_B$).
• 1.2: Faraday's Law of Induction, Flux Variation Modes, and Time-Invariant Induced Charge Transfer ($\Delta q$).
• 1.3: Lenz's Law, Thermodynamic Energy Conservation, and Directional Right-Hand Rules.
• 1.4: Motional Electromotive Force: Microscopic Lorentz Balancing, Linear Translating Rods ($Bvl$), and Pivoted Rotating Bars ($\frac{1}{2}B\omega l^2$).
• 1.5: Self-Inductance ($L$), Solenoids, Inductor Voltage Drop Conventions, and Magnetic Field Energy Density.
• 1.6: Mutual Inductance ($M$), Reciprocity Theorem, Coupling Factor ($K$), and Inductor Combinations in Series and Parallel.
• 1.7: Transient Dynamics of Series $L\text{-}R$ Circuits: Calculus Derivations for Current Growth and Exponential Decay.
• 1.8: Undamped $L\text{-}C$ Harmonic Oscillations: Second-Order Differential Formulations and the Mass-Spring Analogy.
• 1.9: Industrial Applications: Eddy Current Damping, Motor Armature Back-EMF Dynamics, and the AC Dynamo Principle.
• Comprehensive Formula Reference: Master Equation Summary.

1.1 Magnetic Field Lines and Magnetic Flux

Magnetic field lines provide a spatial mapping of vector magnetic fields ($\mathbf{B}$). In classical electrodynamics, it is physically incorrect to designate them as "lines of force" because the magnetic Lorentz force acting on a point charge $q$ moving with velocity $\mathbf{v}$ is given by:

$$\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$$

The vector cross product dictates that $\mathbf{F}_m$ is mutually perpendicular to both $\mathbf{v}$ and $\mathbf{B}$. Consequently, the force acting on a moving charge acts strictly normal to the field lines rather than tangential to them.

Fundamental Properties of Magnetic Field Lines

• Tangential Field Direction: The tangent drawn to a field line at any spatial coordinate gives the direction of the magnetic induction vector $\mathbf{B}$ at that point.
• Spatial Density and Field Magnitude: The number of field lines crossing unit cross-sectional area normal to the lines is directly proportional to the magnitude of $\mathbf{B}$.
• Uniqueness and Non-Intersection: Field lines never intersect. If two lines crossed, the magnetic field at the point of intersection would have two distinct directions simultaneously, which is physically impossible.
• Continuous Closed Loops: Unlike electrostatic field lines (which originate on positive charges and terminate on negative charges), magnetic field lines form continuous closed loops without start or end points. This is the geometric manifestation of Gauss's Law for Magnetism ($\oint \mathbf{B} \cdot d\mathbf{s} = 0$), confirming the absence of isolated magnetic monopoles.

Mathematical Definition of Magnetic Flux

Magnetic flux ($\phi_B$) measures the total normal magnetic field passing through a bounded surface. For an arbitrary curved surface divided into differential vector area elements $d\mathbf{s} = \mathbf{\hat{n}} \, ds$ (where $\mathbf{\hat{n}}$ is the outward unit normal):

$$d\phi_B = \mathbf{B} \cdot d\mathbf{s} = B \, ds \cos\theta \implies \phi_B = \iint_S \mathbf{B} \cdot d\mathbf{s} = \iint_S B \cos\theta \, ds$$

For a planar surface of total area $S$ immersed in a uniform magnetic field $\mathbf{B}$:

$$\phi_B = B S \cos\theta$$

Where $\theta$ is the angle between the magnetic field vector $\mathbf{B}$ and the surface normal area vector $\mathbf{\hat{n}}$. If $\mathbf{B}$ is oriented perpendicular to the surface plane, $\theta = 0^\circ \implies \cos(0^\circ) = 1$, yielding maximum flux: $\phi_B = BS$.

Figure 1.1: Differential surface area vector element $d\mathbf{s}$ inclined at angle $\theta$ to a uniform magnetic field $\mathbf{B}$.

Flux Units and Dimensions:

• Scalar Character: Magnetic flux is a true scalar quantity resulting from the scalar product of two vectors.
• SI Unit: The SI unit is the weber ($\text{Wb}$). $1\text{ Wb} = 1\text{ T}\cdot\text{m}^2 = 1\text{ N}\cdot\text{m/A} = 1\text{ V}\cdot\text{s} = 1\text{ J/A}$.
• CGS Unit: The CGS unit is the maxwell ($\text{Mx}$). $1\text{ Wb} = 10^8\text{ Mx}$.
• Dimensional Formula: $[\phi_B] = [\text{M}^1 \text{L}^2 \text{T}^{-2} \text{I}^{-1}]$.

1.2 Faraday's Law of Electromagnetic Induction

Faraday's law of induction is the governing mathematical formulation of dynamic electromagnetism:

"The magnitude of electromotive force induced in any closed conducting loop is directly proportional to the time rate of change of magnetic flux linked through that circuit."

Single Loop: $$e = -\frac{d\phi_B}{dt}$$ $N$-Turn Concentric Coil: $$e = -N \frac{d\phi_B}{dt}$$

Mechanisms Inducing Flux Variations ($\phi_B = B S \cos\theta$)

1. Time-Varying Magnetic Field ($B = B(t)$): The circuit geometry remains stationary while the external magnetic field changes over time.
2. Time-Varying Loop Area ($S = S(t)$): The magnetic field is static, but the loop moves across magnetic boundaries or changes physical area.
3. Time-Varying Orientation ($\theta = \theta(t)$): The coil rotates continuously inside a static magnetic field ($\theta(t) = \omega t$), generating sinusoidal alternating emf.

Induced Current and Displaced Charge ($\Delta q$)

For a closed loop with total resistance $R$, the instantaneous induced current is:

$$i(t) = \frac{|e(t)|}{R} = \frac{N}{R} \left|\frac{d\phi_B}{dt}\right|$$

The total charge $\Delta q$ passing through the circuit cross-section during finite time interval $\Delta t$ is:

$$\Delta q = \int i(t) \, dt = \frac{N}{R} \int_{\phi_i}^{\phi_f} |d\phi_B| = \frac{N |\Delta\phi_B|}{R}$$

Fundamental Principle: While induced emf ($e$) and current ($i$) are inversely proportional to time interval $\Delta t$, the total charge transferred ($\Delta q$) is strictly independent of the duration of flux change. It depends purely on the net change in flux linkage ($N|\Delta\phi_B|$) and circuit resistance ($R$).

Problem 1 Time-Dependent Flux

A stationary single-turn coil is placed in a region where the magnetic flux variation is described by the relation $\phi(t) = (10t^2 + 5t + 1)\text{ mWb}$. Determine the magnitude of the induced electromotive force in the coil at time $t = 5.0\text{ s}$.

Analytical Solution

Step 1: Convert flux relation to SI base units:
$$\phi(t) = (10t^2 + 5t + 1) \times 10^{-3}\text{ Wb}$$

Step 2: Differentiate with respect to time to obtain induced EMF:
$$|e(t)| = \frac{d\phi}{dt} = \frac{d}{dt}\left[(10t^2 + 5t + 1)\times 10^{-3}\right] = (20t + 5)\times 10^{-3}\text{ V}$$

Step 3: Evaluate at the specified instant $t = 5.0\text{ s}$: $$|e(5)| = [20(5.0) + 5] \times 10^{-3} = (100 + 5) \times 10^{-3} = 105 \times 10^{-3}\text{ V} = \mathbf{0.105\text{ V}} \text{ (or } 105\text{ mV)}$$

Problem 2 Spatial Field Decay

A planar square conducting loop of side $10.0\text{ cm}$ and electrical resistance $0.50\,\Omega$ is oriented vertically in the East-West plane. A uniform magnetic field of magnitude $0.10\text{ T}$ is directed North-East. The field is reduced to zero at a constant rate over a time interval of $0.70\text{ s}$. Calculate the magnitude of the induced current in the loop.

Figure 1.2: Square loop in the vertical E-W plane with a North-East directed magnetic field.

Analytical Solution

Step 1: Calculate geometric area and normal orientation:
Loop area: $A = (0.10\text{ m})^2 = 10^{-2}\text{ m}^2$.
Since the loop lies in the East-West vertical plane, its normal vector $\mathbf{\hat{n}}$ points North. The magnetic field vector $\mathbf{B}$ points North-East. Thus, the inclination angle is $\theta = 45^\circ$.

Step 2: Determine initial and final magnetic flux:
$$\phi_i = B A \cos(45^\circ) = (0.10\text{ T})(10^{-2}\text{ m}^2)\left(\frac{1}{\sqrt{2}}\right) = \frac{10^{-3}}{\sqrt{2}}\text{ Wb}$$ $$\phi_f = 0\text{ Wb}$$

Step 3: Compute induced electromotive force:
$$|e| = \frac{|\phi_f - \phi_i|}{\Delta t} = \frac{10^{-3} / \sqrt{2}}{0.70} = \frac{10^{-3}}{0.70 \times 1.4142} \approx 1.01 \times 10^{-3}\text{ V} \approx 1.0\text{ mV}$$

Step 4: Compute induced current using Ohm's Law:
$$i = \frac{|e|}{R} = \frac{1.0 \times 10^{-3}\text{ V}}{0.50\,\Omega} = 2.0 \times 10^{-3}\text{ A} = \mathbf{2.0\text{ mA}}$$

Problem 3 Charge vs Time Invariance

A square loop of area $20.0\text{ cm}^2$ and resistance $5.0\,\Omega$ is oriented with its plane perpendicular to a uniform magnetic field $B = 2.0\text{ T}$. The loop is rotated through $180^\circ$ about a coplanar axis. Find the induced electromotive force $e$, induced current $i$, and total transferred charge $\Delta q$ when the rotation takes place in: (a) $\Delta t = 0.010\text{ s}$, and (b) $\Delta t = 0.020\text{ s}$.

Analytical Solution

Step 1: Compute total net change in magnetic flux:
Initial state: $\theta_i = 0^\circ \implies \phi_i = B A \cos(0^\circ) = +BA$.
Final state after $180^\circ$ rotation: $\theta_f = 180^\circ \implies \phi_f = B A \cos(180^\circ) = -BA$.
$$|\Delta\phi| = |\phi_f - \phi_i| = |-BA - (+BA)| = 2BA$$ $$|\Delta\phi| = 2 \times (2.0\text{ T}) \times (20.0 \times 10^{-4}\text{ m}^2) = 8.0 \times 10^{-3}\text{ Wb}$$

Step 2: Case (a) Rotation duration $\Delta t = 0.010\text{ s}$: $$|e| = \frac{|\Delta\phi|}{\Delta t} = \frac{8.0 \times 10^{-3}\text{ Wb}}{0.010\text{ s}} = \mathbf{0.80\text{ V}}$$ $$i = \frac{|e|}{R} = \frac{0.80\text{ V}}{5.0\,\Omega} = \mathbf{0.16\text{ A}}$$ $$\Delta q = i \times \Delta t = (0.16\text{ A})(0.010\text{ s}) = \mathbf{1.6 \times 10^{-3}\text{ C}} \text{ (or } 1.6\text{ mC)}$$

Step 3: Case (b) Rotation duration $\Delta t = 0.020\text{ s}$: $$|e| = \frac{|\Delta\phi|}{\Delta t} = \frac{8.0 \times 10^{-3}\text{ Wb}}{0.020\text{ s}} = \mathbf{0.40\text{ V}}$$ $$i = \frac{|e|}{R} = \frac{0.40\text{ V}}{5.0\,\Omega} = \mathbf{0.080\text{ A}}$$ $$\Delta q = i \times \Delta t = (0.080\text{ A})(0.020\text{ s}) = \mathbf{1.6 \times 10^{-3}\text{ C}} \text{ (or } 1.6\text{ mC)}$$ Conclusion: Doubling the time interval halves the induced emf and current, but the total transferred charge $\Delta q$ remains exactly constant.

1.3 Lenz's Law and Thermodynamic Conservation

Lenz's Law establishes the direction of induced current in a circuit:

"The direction of any induced electromotive force or current is always such as to oppose the specific mechanical or magnetic change producing it."

Energy Conservation Consistency

Lenz's Law is a direct manifestation of the First Law of Thermodynamics. If the induced magnetic field aided the external flux change, bringing a magnetic pole toward a loop would induce an opposite pole on the facing surface, attracting the magnet. The magnet would accelerate toward the loop without any external mechanical work. This would create energy spontaneously from nothing, violating the Conservation of Energy.

Because the induced field opposes the approaching pole, an external mechanical force must do positive work to move the magnet against magnetic repulsion. This mechanical work is exactly converted into electrical energy and dissipated as Joule heating ($i^2R$) within the circuit resistance.

Direction Rules:

• RIN Rule: For a loop placed to the Right of a straight wire with current Increasing, the induced magnetic polarity is North ($\implies$ Anti-clockwise current).
• $\otimes\text{IN}$ Rule: If an external magnetic field is directed Inwards ($\otimes$) and is Increasing, the induced field must point Outwards ($\odot$), producing a North polarity ($\implies$ Anti-clockwise current).

Figure 1.3: Right-hand sign conventions establishing current directions for increasing versus decreasing magnetic flux.

Problem 4 Gravitational Dynamics

A permanent bar magnet is released from rest and falls vertically along the central axis of a fixed horizontal conducting copper ring. Analyze whether the instantaneous downward acceleration $a$ of the magnet during its descent is greater than, equal to, or less than $g$.

Figure 1.4: Bar magnet falling toward a closed conducting loop.

Analytical Solution

Step 1: Identify the source of induction:
As the magnet approaches the loop with its North pole directed downward, the downward magnetic flux passing through the loop increases.

Step 2: Apply Lenz's Law to determine opposing magnetic forces:
The loop induces an anti-clockwise current that establishes an upward magnetic dipole field, making the top face of the loop a North pole to repel the descending magnet.

Step 3: Analyze the net equation of motion:
The upward repulsive magnetic force $F_m$ opposes the downward gravitational force:
$$m g - F_m = m a \implies a = g - \frac{F_m}{m}$$ Because $F_m > 0$, the downward acceleration satisfies $\mathbf{a < g}$. (When the magnet passes below the loop and recedes, the loop develops an upper South pole to attract it upward, so $a < g$ remains true throughout).

Problem 5 Solenoid Interaction

A bar magnet is translated coaxially toward a closed solenoid and subsequently retracted away. Detail the mechanical force interactions experienced by the magnet during both stages.

Analytical Solution

Step 1: Approaching Stage:
Moving the North pole toward the solenoid increases flux linkage. By Lenz's Law, the facing end of the solenoid develops a North magnetic pole $\implies$ Mechanical Repulsion.

Step 2: Retracting Stage:
Withdrawing the North pole decreases flux linkage. By Lenz's Law, the facing end of the solenoid develops an attractive South magnetic pole $\implies$ Mechanical Attraction.

Problem 6 Field Decay Sense

A circular conducting loop lies in the plane of the paper to the right of a long vertical wire carrying an upward current. If the current in the wire decreases with time, determine the direction of the induced current in the loop.

Figure 1.5: Conducting loop adjacent to a straight wire carrying decreasing current.

Analytical Solution

Step 1: Establish initial magnetic field orientation:
Using the Right-Hand Grip Rule for the vertical wire, the magnetic field to its right points perpendicularly into the plane of the paper ($\otimes$).

Step 2: Identify flux variation:
Because current $I$ is decreasing, the inward flux ($\otimes$) through the loop is decreasing over time.

Step 3: Apply Lenz's Law:
The loop must induce a current whose magnetic field reinforces the decaying field (i.e., directed into the page $\otimes$). By the right-hand rule, an inward field requires a Clockwise induced current.

Problem 7 Coplanar Field Symmetry

A straight wire carrying current directed perpendicularly into the paper passes through the geometric centroid of a flat triangular conducting loop lying in the plane of the paper. If the current in the wire increases steadily, calculate the magnitude of the induced current in the triangular loop.

Figure 1.6: Circular magnetic field lines coplanar with a triangular loop.

Analytical Solution

Step 1: Analyze field geometry:
The magnetic field lines produced by the straight wire form concentric circles that lie entirely within the plane of the triangular loop.

Step 2: Compute flux linkage integral:
Because the field lines are parallel to the loop's surface, the angle between the field vector $\mathbf{B}$ and the normal vector $\mathbf{\hat{n}}$ is everywhere $\theta = 90^\circ$: $$\phi_B = \iint B \cos(90^\circ) \, ds = 0$$

Step 3: Conclusion:
Since $\phi_B = 0$ continuously, $d\phi_B/dt = 0$. The induced current in the triangular coil is identically Zero.

1.4 Motional Electromotive Force

Motional emf is the electromotive force induced across a conductor translating or rotating through a constant, time-invariant magnetic field.

Microscopic Derivation for a Translating Conductor

Consider a straight conducting rod of length $l$ moving with constant velocity $\mathbf{v}$ through a uniform magnetic field $\mathbf{B}$ directed into the page. Conduction electrons inside the rod move at velocity $\mathbf{v}$, experiencing a magnetic Lorentz force:

$$\mathbf{F}_m = -e(\mathbf{v} \times \mathbf{B})$$

By the cross product rule, $(\mathbf{v} \times \mathbf{B})$ points upward; therefore, electrons experience a downward force toward terminal $b$. This charge separation establishes an internal electrostatic field $\mathbf{E}$ pointing from $a$ to $b$. Equilibrium is reached when the electrostatic force $eE$ balances the magnetic Lorentz force $evB$:

$$eE = evB \implies E = vB$$

Integrating along the length $l$ gives the steady-state potential difference:

$$\Delta V = V_a - V_b = \int_0^l E \, dl = B v l$$

Figure 1.7: Conductor of length $l$ translating across a magnetic field, forming an equivalent battery of emf $e = Bvl$.

Energy Conservation and Power Balance

If the rod slides along a U-shaped track of circuit resistance $R$ (and internal rod resistance $r$), the induced current is $i = \frac{Bvl}{R+r}$. The moving rod experiences a retarding magnetic force $F_m = i l B = \frac{B^2 l^2 v}{R+r}$. An external mechanical agent must exert an equal and opposite force $F_{\text{ext}} = F_m$ to maintain constant velocity. The mechanical input power equals the electrical power dissipated as Joule heat:

$$P_{\text{mech}} = F_{\text{ext}} v = \frac{B^2 l^2 v^2}{R+r} = i^2 (R+r) = P_{\text{elec}}$$

Rotational Motional EMF

For a conducting rod of length $l$ pivoted at one end $O$ and rotating with uniform angular velocity $\omega$ in a perpendicular magnetic field $\mathbf{B}$:

$$de = B v(r) \, dr = B (\omega r) \, dr \implies e = \int_0^l B \omega r \, dr = \frac{1}{2} B \omega l^2$$

Problem 8 Multi-Source Rail Network

Two parallel horizontal conducting rails separated by $10.0\text{ cm}$ are bridged by a fixed central resistor $R = 5.0\,\Omega$. Two conducting rods with internal resistances $r_1 = 10.0\,\Omega$ and $r_2 = 15.0\,\Omega$ slide away from the central resistor in opposite directions at constant speeds $v_1 = 4.00\text{ m/s}$ (moving left) and $v_2 = 2.00\text{ m/s}$ (moving right) through a uniform vertical field $B = 0.010\text{ T}$ directed into the page. Determine the magnitude and direction of the current flowing through the central resistor.

Figure 1.8: Parallel rail network with two moving rods generating opposing loop EMFs.

Analytical Solution

Step 1: Compute motional EMF magnitudes for both sliding rods:
Left Rod ($e_1$): $e_1 = B v_1 l = (0.010\text{ T})(4.00\text{ m/s})(0.10\text{ m}) = 4.0 \times 10^{-3}\text{ V} = 4.0\text{ mV}$.
Right Rod ($e_2$): $e_2 = B v_2 l = (0.010\text{ T})(2.00\text{ m/s})(0.10\text{ m}) = 2.0 \times 10^{-3}\text{ V} = 2.0\text{ mV}$.

Step 2: Determine branch polarities using $\mathbf{v} \times \mathbf{B}$: * Left rod (moving left): $\mathbf{v}\times\mathbf{B}$ points downward $\implies$ lower terminal positive, driving current UPWARD through central resistor $R$.
* Right rod (moving right): $\mathbf{v}\times\mathbf{B}$ points upward $\implies$ upper terminal positive, driving current DOWNWARD through central resistor $R$.

Step 3: Apply the Principle of Superposition:
Due to $e_1$ alone: Branch 1 ($r_1 = 10\,\Omega$) connects in series with the parallel combination of $R = 5\,\Omega$ and $r_2 = 15\,\Omega$.
$$R_{p1} = \frac{5 \times 15}{5 + 15} = 3.75\,\Omega \implies R_{\text{eq,1}} = 10 + 3.75 = 13.75\,\Omega = \frac{55}{4}\,\Omega$$ $$I_{\text{total,1}} = \frac{e_1}{R_{\text{eq,1}}} = \frac{4\text{ mV}}{55/4} = \frac{16}{55}\text{ mA}$$ $$i_{1,R} = I_{\text{total,1}} \times \frac{r_2}{R + r_2} = \frac{16}{55} \times \frac{15}{20} = \frac{12}{55}\text{ mA} \text{ (Directed Upward)}$$
Due to $e_2$ alone: Branch 2 ($r_2 = 15\,\Omega$) connects in series with the parallel combination of $R = 5\,\Omega$ and $r_1 = 10\,\Omega$.
$$R_{p2} = \frac{5 \times 10}{5 + 10} = \frac{10}{3}\,\Omega \implies R_{\text{eq,2}} = 15 + \frac{10}{3} = \frac{55}{3}\,\Omega$$ $$I_{\text{total,2}} = \frac{e_2}{R_{\text{eq,2}}} = \frac{2\text{ mV}}{55/3} = \frac{6}{55}\text{ mA}$$ $$i_{2,R} = I_{\text{total,2}} \times \frac{r_1}{R + r_1} = \frac{6}{55} \times \frac{10}{15} = \frac{4}{55}\text{ mA} \text{ (Directed Downward)}$$

Step 4: Superpose the branch currents:
$$i_{\text{net}} = i_{1,R} - i_{2,R} = \frac{12}{55} - \frac{4}{55} = \mathbf{\frac{8}{55}\text{ mA}} \approx \mathbf{0.145\text{ mA}} \text{ (Directed Upward)}$$

1.5 Self-Inductance and Inductors

Self-inductance ($L$) is the electrical analogue of mechanical mass, representing the circuit's opposition to variations in current.

Dual Formal Definitions of Self-Inductance

1. Magnetic Flux Linkage Definition: The total flux linkage $N\phi_B$ is proportional to current $i$:
   $$N\phi_B = L i \implies L = \frac{N\phi_B}{i}$$
2. Induced Electromotive Force Definition: The back-emf $e$ is proportional to the rate of current change:
   $$e = -L \frac{di}{dt} \implies L = \left|\frac{e}{di/dt}\right|$$

The SI unit of self-inductance is the Henry ($\text{H}$): $1\text{ H} = 1\text{ Wb/A} = 1\text{ V}\cdot\text{s/A} = 1\,\Omega\cdot\text{s}$.

Potential Difference Across an Inductor

When traversing an inductor in the direction of assumed current, the potential difference is $V_{ab} = V_a - V_b = L \frac{di}{dt}$:

Figure 1.9: Inductor potential drop polarities for steady, increasing, and decreasing currents.

Problem 9 Back-EMF Calculation

An inductor with inductance $L = 0.54\text{ H}$ carries a current decreasing at a uniform rate $\frac{di}{dt} = -0.030\text{ A/s}$. Calculate the self-induced electromotive force.

Analytical Solution

Step 1: Apply the fundamental self-induction relation:
$$e = -L \frac{di}{dt} = -(0.54\text{ H})(-0.030\text{ A/s}) = \mathbf{+1.62 \times 10^{-2}\text{ V}} \text{ (or } 16.2\text{ mV)}$$

Problem 10 KVL Branch Equation

In an active circuit branch, a resistor $R = 10.0\,\Omega$, an inductor $L = 5.0\text{ H}$, and a DC source $E = 20.0\text{ V}$ are connected in series from terminal $a$ to terminal $b$. At a given instant, current $i = 2.0\text{ A}$ flows from $a$ to $b$ and is decreasing at a rate of $-1.0\text{ A/s}$. Find the potential difference $V_{ab} = V_a - V_b$.

Figure 1.10: Active branch schematic for KVL traversal.

Analytical Solution

Step 1: Write Kirchhoff's Voltage Law from terminal $a$ to terminal $b$:
$$V_a - iR - L\frac{di}{dt} - E = V_b$$

Step 2: Substitute values with appropriate algebraic signs:
$$V_a - (2.0\text{ A})(10.0\,\Omega) - (5.0\text{ H})(-1.0\text{ A/s}) - 20.0\text{ V} = V_b$$ $$V_a - 20.0 + 5.0 - 20.0 = V_b \implies V_a - 35.0 = V_b$$

Step 3: Compute $V_{ab}$:
$$V_{ab} = V_a - V_b = \mathbf{35.0\text{ V}}$$

Self-Inductance of a Long Solenoid

For a long solenoid of length $l$, area $S$, and turns $N$ ($n = N/l$), the axial field is $B = \mu_0 n i$. The total flux linkage is $N\phi_B = N(BS) = \left(\frac{\mu_0 N^2 S}{l}\right)i$, yielding:

$$L = \frac{\mu_0 N^2 S}{l} = \mu_0 n^2 S l = \mu_0 n^2 V_{\text{solenoid}}$$

Problem 11 Coaxial Search Coil

A long solenoid with $200\text{ turns/cm}$ carries a current of $1.50\text{ A}$. Located at its center is an inner secondary coil of 100 turns and cross-sectional area $3.14 \times 10^{-4}\text{ m}^2$. If the solenoid current is reversed in direction in $0.050\text{ s}$, calculate the induced electromotive force in the inner coil.

Analytical Solution

Step 1: Compute magnetic field of the primary solenoid:
$n = 200\text{ turns/cm} = 20,000\text{ turns/m}$.
$$B = \mu_0 n i = (4\pi \times 10^{-7}\text{ T}\cdot\text{m/A})(20,000\text{ m}^{-1})(1.50\text{ A}) = 1.2\pi \times 10^{-2}\text{ T} \approx 3.77 \times 10^{-2}\text{ T}$$

Step 2: Calculate initial flux linkage:
$$\Phi_i = N_2 B A = 100 \times (3.77 \times 10^{-2}\text{ T}) \times (3.14 \times 10^{-4}\text{ m}^2) \approx 1.184 \times 10^{-3}\text{ Wb}$$

Step 3: Calculate flux change upon reversal:
$$\Delta \Phi = \Phi_f - \Phi_i = (-\Phi_i) - \Phi_i = -2\Phi_i = -2.368 \times 10^{-3}\text{ Wb}$$

Step 4: Compute induced EMF:
$$|e| = \frac{|\Delta\Phi|}{\Delta t} = \frac{2.368 \times 10^{-3}\text{ Wb}}{0.050\text{ s}} \approx \mathbf{0.048\text{ V}} \text{ (or } 48\text{ mV)}$$

Problem 12 Toroidal Induction

(a) An air-core toroidal solenoid has mean radius $r = 15.0\text{ cm}$, cross-sectional area $A = 12.0\text{ cm}^2$, and $N_1 = 1200\text{ turns}$. Find its self-inductance.
(b) A secondary winding of $N_2 = 300\text{ turns}$ is wound closely on the toroid. If primary current increases from 0 to $2.0\text{ A}$ in $0.050\text{ s}$, calculate the induced emf in the secondary.

Analytical Solution

Step 1: Part (a) Toroid Self-Inductance ($l = 2\pi r = 0.3\pi\text{ m}$):
$$L = \frac{\mu_0 N_1^2 A}{l} = \frac{(4\pi \times 10^{-7})(1200)^2 (12.0 \times 10^{-4}\text{ m}^2)}{0.3\pi\text{ m}} = \mathbf{2.304 \times 10^{-3}\text{ H}} = \mathbf{2.304\text{ mH}}$$

Step 2: Part (b) Induced EMF in secondary:
$$M = \frac{\mu_0 N_1 N_2 A}{l} = \frac{(4\pi \times 10^{-7})(1200)(300)(12.0 \times 10^{-4})}{0.3\pi} = 5.76 \times 10^{-4}\text{ H}$$ $$|e_2| = M \frac{\Delta I}{\Delta t} = (5.76 \times 10^{-4}\text{ H})\left(\frac{2.0\text{ A} - 0}{0.050\text{ s}}\right) = (5.76 \times 10^{-4})(40) = \mathbf{0.023\text{ V}} \text{ (or } 23\text{ mV)}$$

Problem 13 Scaling Relation

How does the self-inductance of a solenoid change if its total number of turns and its length are both doubled while its cross-sectional area remains constant?

Analytical Solution

Step 1: Apply the functional scaling law:
$$L = \frac{\mu_0 N^2 A}{l} \propto \frac{N^2}{l}$$

Step 2: Substitute scaled parameters ($N' = 2N$ and $l' = 2l$):
$$L' = \frac{\mu_0 (2N)^2 A}{2l} = \frac{4 N^2}{2 l} \mu_0 A = 2 \left(\frac{\mu_0 N^2 A}{l}\right) = \mathbf{2L}$$ Conclusion: The self-inductance is doubled.

Problem 14 Solenoid Parameters

(i) Calculate the self-inductance of an air-core solenoid containing 300 turns with length $25.0\text{ cm}$ and area $4.00\text{ cm}^2$.
(ii) Calculate the induced emf if the current decreases at a uniform rate of $50.0\text{ A/s}$.

Analytical Solution

Step 1: Part (i) Self-inductance calculation:
$$L = \frac{\mu_0 N^2 A}{l} = \frac{(4\pi \times 10^{-7}\text{ T}\cdot\text{m/A})(300)^2 (4.00 \times 10^{-4}\text{ m}^2)}{0.25\text{ m}} = \mathbf{1.81 \times 10^{-4}\text{ H}} \text{ (or } 0.181\text{ mH)}$$

Step 2: Part (ii) Induced EMF:
$$|e| = L \left|\frac{di}{dt}\right| = (1.81 \times 10^{-4}\text{ H})(50.0\text{ A/s}) = \mathbf{9.05 \times 10^{-3}\text{ V}} = \mathbf{9.05\text{ mV}}$$

Problem 15 Magnetic Energy Capacity

Determine the self-inductance required to store $1.0\text{ kWh}$ of magnetic energy in an inductor carrying a steady current of $200\text{ A}$. ($1.0\text{ kWh} = 3.6 \times 10^6\text{ J}$).

Analytical Solution

Step 1: Relate magnetic energy to inductance and current:
$$U = \frac{1}{2} L i^2 \implies L = \frac{2U}{i^2}$$

Step 2: Compute numerical value:
$$L = \frac{2(3.6 \times 10^6\text{ J})}{(200\text{ A})^2} = \frac{7.2 \times 10^6}{40,000} = \mathbf{180\text{ H}}$$

1.6 Mutual Inductance and Coupled Circuits

Mutual induction occurs when a changing current $i_1$ in a primary coil induces an emf $e_2$ in an adjacent secondary coil:

$$N_2 \phi_{B_2} = M i_1 \quad \text{and} \quad e_2 = -M \frac{di_1}{dt}$$

Reciprocity Theorem: For any two stationary coils, $M_{12} = M_{21} = M$.

Coupling Coefficient ($K$)

$$M = K \sqrt{L_1 L_2} \quad (0 \le K \le 1)$$

Combinations of Inductors

• Series (without mutual coupling): $L_{\text{eq}} = L_1 + L_2 + L_3 + \dots$
• Series (with mutual coupling $M$): $L_{\text{eq}} = L_1 + L_2 \pm 2M$ ($+2M$ for aiding flux; $-2M$ for opposing flux).
• Parallel (zero mutual coupling): $\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} + \dots$

Problem 16 Single-Turn Flux

A coil with 200 turns has a self-inductance of $10.0\text{ mH}$. Calculate the magnetic flux through the cross-section of a single turn when the coil carries a current of $4.0\text{ mA}$.

Analytical Solution

Step 1: Compute total flux linkage ($\Phi = Li$):
$$\Phi_{\text{total}} = (10.0 \times 10^{-3}\text{ H})(4.0 \times 10^{-3}\text{ A}) = 4.0 \times 10^{-5}\text{ Wb}$$

Step 2: Find magnetic flux per individual turn ($\phi = \Phi / N$):
$$\phi = \frac{4.0 \times 10^{-5}\text{ Wb}}{200} = \mathbf{2.0 \times 10^{-7}\text{ Wb}}$$

Problem 17 Concentric Square Loops

A small square conducting loop of side $l$ is placed coplanarly and concentrically inside a much larger square loop of side $L$ ($L \gg l$). Derive an exact analytical expression for the mutual inductance $M$ of the system.

Figure 1.11: Concentric coplanar square loops with $L \gg l$.

Analytical Solution

Step 1: Compute central magnetic field of the large square loop carrying current $i$:
The large loop consists of 4 straight wires of length $L$ at perpendicular distance $d = L/2$ with $\alpha = \beta = 45^\circ$:
$$B_1 = \frac{\mu_0 i}{4\pi (L/2)} [\sin(45^\circ) + \sin(45^\circ)] = \frac{\mu_0 i}{2\pi L} \left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2}\mu_0 i}{\pi L}$$ Total central field from all 4 sides: $$B = 4 B_1 = \frac{4\sqrt{2}\mu_0 i}{\pi L}$$

Step 2: Calculate flux linked through the inner small loop ($A = l^2$):
Because $L \gg l$, the field is uniform across the small area $l^2$: $$\phi_2 = B A = \left(\frac{2\sqrt{2}\mu_0 i}{\pi L}\right) l^2$$

Step 3: Extract mutual inductance $M = \phi_2 / i$:
$$M = \mathbf{\frac{2\sqrt{2}\mu_0 l^2}{\pi L}}$$

Problem 18 Coaxial Windings

A primary solenoid with $n_1 = 50.0\text{ turns/cm}$ is surrounded by a secondary coil of $N_2 = 200\text{ turns}$. The cross-sectional area of the primary is $4.00\text{ cm}^2$. Determine the mutual inductance $M$.

Analytical Solution

Step 1: Express parameters in SI units:
$n_1 = 5000\text{ turns/m}$, $N_2 = 200$, $A = 4.00 \times 10^{-4}\text{ m}^2$.

Step 2: Apply mutual inductance formula:
$$M = \mu_0 n_1 N_2 A = (4\pi \times 10^{-7}\text{ T}\cdot\text{m/A})(5000\text{ m}^{-1})(200)(4.00 \times 10^{-4}\text{ m}^2)$$ $$M = 16\pi \times 10^{-5}\text{ H} \approx \mathbf{5.0 \times 10^{-4}\text{ H}} \text{ (or } 0.50\text{ mH)}$$

Problem 19 Inductor Network Algebra

The equivalent inductance of two uncoupled inductors is $2.40\text{ H}$ in parallel and $10.0\text{ H}$ in series. Find the individual inductances $L_1$ and $L_2$.

Analytical Solution

Step 1: Set up system of equations:
Series: $L_1 + L_2 = 10.0\text{ H}$.
Parallel: $\frac{L_1 L_2}{L_1 + L_2} = 2.40\text{ H} \implies L_1 L_2 = 2.40 \times 10.0 = 24.0\text{ H}^2$.

Step 2: Solve the quadratic system:
$$(L_1 - L_2)^2 = (L_1 + L_2)^2 - 4L_1 L_2 = (10.0)^2 - 4(24.0) = 4.0 \implies L_1 - L_2 = 2.0\text{ H}$$ Solving gives: $$\mathbf{L_1 = 6.0\text{ H}} \quad \text{and} \quad \mathbf{L_2 = 4.0\text{ H}}$$

Problem 20 Coupled Energy Ratios

Two distinct coils have self-inductances $L_1 = 8.0\text{ mH}$ and $L_2 = 2.0\text{ mH}$. At a given instant, the current in both coils increases at the same constant rate, and equal electrical power is delivered to both coils. Find the ratio of magnetic energies stored in the coils at that instant.

Analytical Solution

Step 1: Relate power to inductance and current:
$$P = e i = \left(L \frac{di}{dt}\right) i$$ Because $P_1 = P_2$ and $(di/dt)_1 = (di/dt)_2$: $$L_1 i_1 = L_2 i_2 \implies \frac{i_1}{i_2} = \frac{L_2}{L_1} = \frac{2.0}{8.0} = \frac{1}{4}$$

Step 2: Compute ratio of stored energies:
$$\frac{U_1}{U_2} = \frac{\frac{1}{2}L_1 i_1^2}{\frac{1}{2}L_2 i_2^2} = \left(\frac{L_1}{L_2}\right)\left(\frac{i_1}{i_2}\right)^2 = \left(\frac{8.0}{2.0}\right)\left(\frac{1}{4}\right)^2 = 4 \times \frac{1}{16} = \mathbf{\frac{1}{4}}$$

1.7 Transient Dynamics in L-R Circuits

In circuits containing resistance $R$ and inductance $L$, current cannot jump instantaneously because an infinite back-emf would be required ($e = -L\frac{di}{dt} \to \infty$).

Growth of Current (Charging Circuit)

Applying Kirchhoff's loop law: $E - iR - L \frac{di}{dt} = 0 \implies \frac{di}{\frac{E}{R} - i} = \frac{R}{L} dt$. Integrating from $t = 0$ ($i = 0$) gives:

$$i(t) = i_0 \left(1 - e^{-t/\tau_L}\right) \quad \text{where } i_0 = \frac{E}{R}, \; \tau_L = \frac{L}{R}$$

At $t = \tau_L$, the current reaches $i(\tau_L) = i_0 (1 - e^{-1}) \approx \mathbf{0.632 \, i_0}$ (63.2% of maximum).

Decay of Current (Discharging Circuit)

When the source is bypassed: $iR + L\frac{di}{dt} = 0 \implies i(t) = i_0 e^{-t/\tau_L}$.

At $t = \tau_L$, the current drops to $i(\tau_L) = i_0 e^{-1} \approx \mathbf{0.368 \, i_0}$ (36.8% of initial value).

Problem 21 Transient Time Constant

An inductor $L = 20.0\text{ mH}$, a resistor $R = 100\,\Omega$, and a battery $E = 10.0\text{ V}$ are connected in series. How much time elapses before the current reaches 99.0% of its maximum steady-state value?

Analytical Solution

Step 1: Compute inductive time constant $\tau_L$:
$$\tau_L = \frac{L}{R} = \frac{20.0 \times 10^{-3}\text{ H}}{100\,\Omega} = 0.20 \times 10^{-3}\text{ s} = 0.20\text{ ms}$$

Step 2: Solve the growth relation for $i = 0.99 i_0$:
$$0.99 i_0 = i_0(1 - e^{-t/\tau_L}) \implies e^{-t/\tau_L} = 0.010 \implies -\frac{t}{\tau_L} = \ln(0.010) = -\ln(100) \approx -4.605$$ $$t = 4.605 \times \tau_L = 4.605 \times 0.20\text{ ms} = \mathbf{0.92\text{ ms}}$$

Problem 22 Thermal Dissipation

An $L\text{-}R$ circuit with $L = 4.0\text{ H}$, $R = 1.0\,\Omega$, and $E = 6.0\text{ V}$ is energized at $t = 0$. Determine the instantaneous rate of power dissipation in Joule heating across an equivalent dissipation stage at $t = 4.0\text{ s}$.

Analytical Solution

Step 1: Calculate circuit time constant $\tau_L$:
$$\tau_L = \frac{L}{R} = \frac{4.0\text{ H}}{1.0\,\Omega} = 4.0\text{ s}$$

Step 2: Evaluate instantaneous current at $t = 4.0\text{ s}$ ($t = \tau_L$):
$$i(4.0) = \frac{E}{R}(1 - e^{-1}) = \frac{6.0}{1.0}(1 - 0.368) = 6.0 \times 0.632 = 3.80\text{ A}$$

Step 3: Compute power dissipation ($P = i^2 R_{\text{load}}$ for $10\,\Omega$ reference stage):
$$P = (3.80\text{ A})^2 \times 10.0\,\Omega = 14.44 \times 10.0 \approx \mathbf{140\text{ W}}$$

Problem 23 Differential Current Derivative

A coil of resistance $20.0\,\Omega$ and inductance $0.50\text{ H}$ is connected across a $200\text{ V}$ DC supply. Calculate the rate of increase of current: (a) at the instant of switch closure ($t = 0$), (b) after one time constant ($t = \tau_L$), and (c) find the final steady-state current.

Analytical Solution

Step 1: Formulate rate of current rise equation:
$$E - iR - L\frac{di}{dt} = 0 \implies \frac{di}{dt} = \frac{E - iR}{L} = \frac{E}{L} e^{-t/\tau_L}$$

Step 2: Part (a) At $t = 0$ ($i = 0$):
$$\left.\frac{di}{dt}\right|_{t=0} = \frac{E}{L} = \frac{200\text{ V}}{0.50\text{ H}} = \mathbf{400\text{ A/s}}$$

Step 3: Part (b) At $t = \tau_L$:
$$\left.\frac{di}{dt}\right|_{t=\tau_L} = 400 \times e^{-1} = 400 \times 0.368 = \mathbf{148\text{ A/s}}$$

Step 4: Part (c) Steady-state current ($t \to \infty$):
$$i_0 = \frac{E}{R} = \frac{200\text{ V}}{20.0\,\Omega} = \mathbf{10.0\text{ A}}$$

1.8 Oscillations in L-C Circuits and Mechanical Analogy

In an ideal $L\text{-}C$ loop (zero resistance), energy oscillates back and forth between the capacitor's electric field and the inductor's magnetic field without dissipation.

Second-Order Differential Equation

$$\frac{q}{C} - L\frac{di}{dt} = 0 \implies \frac{d^2q}{dt^2} + \left(\frac{1}{LC}\right)q = 0$$ $$\omega = \frac{1}{\sqrt{LC}}, \quad f = \frac{1}{2\pi\sqrt{LC}}, \quad T = 2\pi\sqrt{LC}$$ $$q(t) = q_0 \cos(\omega t), \quad i(t) = -\frac{dq}{dt} = \omega q_0 \sin(\omega t) = i_0 \sin(\omega t)$$

Mechanical vs Electrical Analogy

Mechanical System (Mass-Spring)	Electrical System ($L\text{-}C$ Circuit)
Displacement ($x$)	Charge ($q$)
Velocity ($v = dx/dt$)	Current ($i = dq/dt$)
Mass / Inertia ($m$)	Inductance ($L$)
Spring Constant ($k$)	Reciprocal Capacitance ($1/C$)
Potential Energy: $U = \frac{1}{2}kx^2$	Electric Field Energy: $U_E = \frac{q^2}{2C}$
Kinetic Energy: $K = \frac{1}{2}mv^2$	Magnetic Field Energy: $U_B = \frac{1}{2}Li^2$
Total Energy: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{const}$	Total Energy: $E = \frac{1}{2}Li^2 + \frac{q^2}{2C} = \frac{q_0^2}{2C} = \text{const}$

Problem 24 Energy Exchange Verification

A $25.0\,\mu\text{F}$ capacitor is charged to an initial potential difference of $300\text{ V}$ and connected across an ideal inductor $L = 10.0\text{ mH}$.
(a) Calculate the natural oscillation frequency.
(b) Determine the potential difference across the capacitor and circuit current magnitude at $t = 1.20\text{ ms}$.
(c) Verify the conservation of total energy at $t = 0$ and $t = 1.20\text{ ms}$.

Analytical Solution

Step 1: Part (a) Compute oscillation frequency:
$$f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(10.0 \times 10^{-3}\text{ H})(25.0 \times 10^{-6}\text{ F})}} = \frac{1}{2\pi (5.0 \times 10^{-4}\text{ s})} = \mathbf{318.3\text{ Hz}}$$ $$\omega = 2\pi f = 2000\text{ rad/s}$$

Step 2: Part (b) Voltage and Current at $t = 1.20\text{ ms}$: Maximum initial charge: $q_0 = C V_0 = (25.0 \times 10^{-6}\text{ F})(300\text{ V}) = 7.50 \times 10^{-3}\text{ C}$.
Phase angle: $\theta = \omega t = (2000\text{ rad/s})(1.20 \times 10^{-3}\text{ s}) = 2.40\text{ radians}$.
Charge: $q = q_0 \cos(2.40\text{ rad}) = (7.50 \times 10^{-3})(-0.7374) = -5.53 \times 10^{-3}\text{ C}$.
Potential Difference: $V = \frac{|q|}{C} = \frac{5.53 \times 10^{-3}\text{ C}}{25.0 \times 10^{-6}\text{ F}} = \mathbf{221.2\text{ V}}$.
Current magnitude: $|i| = \omega q_0 |\sin(2.40\text{ rad})| = (2000)(7.50 \times 10^{-3})(0.6755) = \mathbf{10.13\text{ A}}$.

Step 3: Part (c) Verify Conservation of Total Energy:
At $t = 0$: $i = 0 \implies U_B = 0$.
$$U_E = \frac{1}{2} C V_0^2 = \frac{1}{2}(25.0 \times 10^{-6})(300)^2 = \mathbf{1.125\text{ J}} \implies U_{\text{total}} = 1.125\text{ J}$$
At $t = 1.20\text{ ms}$:
$$U_B = \frac{1}{2} L i^2 = \frac{1}{2}(10.0 \times 10^{-3}\text{ H})(10.13\text{ A})^2 = \mathbf{0.513\text{ J}}$$ $$U_E = \frac{q^2}{2C} = \frac{(-5.53 \times 10^{-3}\text{ C})^2}{2(25.0 \times 10^{-6}\text{ F})} = \mathbf{0.612\text{ J}}$$ $$U_{\text{total}} = U_B + U_E = 0.513\text{ J} + 0.612\text{ J} = \mathbf{1.125\text{ J}} \quad (\text{Strictly Conserved}).$$

1.9 Industrial Applications of Electromagnetic Induction

(i) Eddy Currents (Foucault Currents)

When bulk conductors move through magnetic fields or experience changing magnetic flux, closed circulating loops of current are induced within the body of the metal. Because bulk conductors offer very low electrical resistance, eddy currents can be large, leading to significant ohmic heating ($P = i^2 R$).

• Electromagnetic Damping: A metallic pendulum plate swinging into a magnetic field experiences opposing Lorentz forces from eddy currents, quickly dissipating kinetic energy as thermal energy. Used in deadbeat moving-coil galvanometers and electromagnetic train brakes.
• Lamination of Magnetic Cores: In transformer and motor cores, solid iron blocks are replaced by thin laminated sheets coated with insulating varnish, interrupting conduction paths to minimize eddy current losses.

Figure 1.12: Eddy current loops and magnetic damping on a swinging copper pendulum plate.

(ii) Back EMF in Electric Motors

An electric motor converts electrical energy into mechanical work. As the armature coil rotates in the magnetic field, an opposing back emf ($e$) is generated. The current drawn by the motor armature is:

$$i = \frac{V_{\text{applied}} - e}{R_{\text{armature}}}$$

At startup ($t = 0$), the armature is stationary $\implies e = 0 \implies i_{\text{start}} = V/R$, which can cause a dangerously large inrush current. A temporary series starter resistor protects the motor until rotation establishes back emf.

(iii) AC Generator (Dynamo) Principle

A coil of $N$ turns and area $A$ rotating with uniform angular velocity $\omega$ inside a uniform magnetic field $B$ produces a sinusoidal flux $\phi_B(t) = N B A \cos(\omega t)$. The induced alternating electromotive force is:

$$e(t) = -\frac{d\phi}{dt} = N B A \omega \sin(\omega t) = e_0 \sin(\omega t) \quad \text{where } e_0 = N B A \omega$$

Problem 25 Dynamo Rotation Geometry

A circular planar coil of radius $a$ containing $n$ turns is placed in a uniform vertical magnetic field $\mathbf{B}$. Derive the expression for the induced electromotive force when the coil is rotated with uniform angular velocity $\omega$ about:
(i) An axis passing through its center and perpendicular to the plane of the loop.
(ii) An axis aligned along its diameter.

Figure 1.13: Coil rotation about normal central axis vs diametrical axis.

Analytical Solution

Step 1: Part (i) Rotation about perpendicular central axis:
The area vector $\mathbf{A}$ remains parallel to the vertical magnetic field $\mathbf{B}$ at all times ($\theta = 0^\circ$). The flux $\phi = n B (\pi a^2)$ is constant $\implies \frac{d\phi}{dt} = 0 \implies \mathbf{e = 0}$.

Step 2: Part (ii) Rotation about diametrical axis:
The angle between the surface normal and the field changes continuously as $\theta(t) = \omega t$.
$$\phi(t) = n B (\pi a^2) \cos(\omega t) \implies e(t) = -\frac{d\phi}{dt} = \mathbf{n \pi a^2 B \omega \sin(\omega t)}$$

Comprehensive Formula Reference

• Magnetic Flux: $\phi_B = \iint \mathbf{B}\cdot d\mathbf{s} = B S \cos\theta$.
• Faraday's Law: $e = -N \frac{d\phi_B}{dt}, \quad \Delta q = \frac{N|\Delta\phi_B|}{R}$.
• Translational Motional EMF: $e = Bvl$.
• Rotational Motional EMF: $e = \frac{1}{2}B\omega l^2$.
• Solenoid Self-Inductance: $L = \frac{\mu_0 N^2 S}{l} = \mu_0 n^2 S l = \mu_0 n^2 V_{\text{solenoid}}$.
• Magnetic Stored Energy: $U = \frac{1}{2}Li^2$.
• Mutual Inductance & Coupling: $M = K\sqrt{L_1 L_2}, \quad L_{\text{eq,series}} = L_1 + L_2 \pm 2M$.
• $L\text{-}R$ Transients: $i_{\text{growth}} = i_0(1 - e^{-t/\tau_L}), \quad i_{\text{decay}} = i_0 e^{-t/\tau_L}, \quad \tau_L = \frac{L}{R}$.
• $L\text{-}C$ Natural Frequency: $\omega = \frac{1}{\sqrt{LC}}, \quad f = \frac{1}{2\pi\sqrt{LC}}$.
• AC Dynamo Peak EMF: $e_0 = N B A \omega$.