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Chapter 6: Electromagnetic Induction

In earlier chapters, we discovered that moving electric charges (current) produce magnetic fields (Oersted & Ampere). In 1830, Michael Faraday in England and Joseph Henry in the USA demonstrated the converse: changing magnetic fields generate electric currents. This phenomenon, known as Electromagnetic Induction (EMI), is the foundation of modern electric power generators, dynamos, and transformers!

1. Magnetic Flux ($\Phi_B$)

Definition: The total number of magnetic field lines passing normally (perpendicularly) through a given surface area is defined as the Magnetic Flux linked with that surface.

Figure 6.4 & 6.5: Magnetic Flux through Surface Elements in Magnetic Field
Figure 6.4 & 6.5: Magnetic flux through planar surface and area elements \(d\vec{A}_i\) in magnetic field.
Mathematical Formula & Properties

For a uniform magnetic field $\vec{B}$ through a flat surface of area vector $\vec{A}$:

$$\Phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta$$

Where $\theta$ is the angle between the magnetic field $\vec{B}$ and the Area Vector $\vec{A}$ (perpendicular to the plane).

For a non-uniform magnetic field or curved surface:

$$\Phi_B = \int \vec{B} \cdot d\vec{A} = \sum_{\text{all}} \vec{B}_i \cdot d\vec{A}_i$$

Significance of the Angle $\theta$:

2. The Experiments of Faraday and Henry

Faraday and Henry performed three landmark experiments that established the foundations of Electromagnetic Induction:

Experiment 6.1: Current Induced by Relative Motion of a Magnet

Figure 6.1: Magnet and Coil Experiment
Figure 6.1: When the bar magnet is pushed towards the coil, the pointer in galvanometer G deflects.

Experiment 6.2: Current Induced by Motion of a Current-Carrying Coil

Figure 6.2: Current induced by motion of current carrying coil
Figure 6.2: Current is induced in coil \(C_1\) due to motion of the current-carrying coil \(C_2\).
Example 6.1 Question: Consider Experiment 6.2. (a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer?
Solution:
(a) To obtain a larger deflection:
  1. Insert a soft iron core inside coil $C_2$ $(\text{increases relative permeability } \mu_r)$.
  2. Connect coil $C_2$ to a higher voltage battery $(\text{increases magnetic field } B)$.
  3. Move coil $C_2$ very rapidly towards or away from coil $C_1$ $\left(\text{increases } \frac{d\Phi_B}{dt}\right)$.
(b) Replace the galvanometer with a small torch light bulb. The induced current from rapid relative motion will make the bulb glow visibly.

Experiment 6.3: Current Induced by Varying Current (No Relative Motion)

Figure 6.3: Stationary coils with tapping key
Figure 6.3: Experimental set-up for stationary coils with tapping key.

3. Faraday's Laws of Electromagnetic Induction

Faraday's Laws

First Law (Qualitative): Whenever the magnetic flux linked with a closed circuit changes with time, an electromotive force (EMF) is induced in the circuit. The induced EMF lasts only as long as the change in magnetic flux continues.

Second Law (Quantitative): The magnitude of the induced EMF in a circuit is directly proportional to the time rate of change of magnetic flux linked with the circuit.

$$\epsilon = -\frac{d\Phi_B}{dt}$$

For a closely wound coil of $N$ turns:

$$\epsilon = -N \frac{d\Phi_B}{dt}$$

(The negative sign represents the direction of induced EMF according to Lenz's Law).

Three Methods to Induce an EMF:

Since $\Phi_B = BA\cos\theta$, an EMF can be induced by changing any of the three variables:

  1. By changing the magnetic field magnitude ($B$): Done by moving a magnet, changing current in an adjacent coil, or using AC.
  2. By changing the effective area ($A$): Done by stretching, shrinking, or moving a coil into/out of a magnetic field (Motional EMF).
  3. By changing the orientation angle ($\theta$): Done by rotating the coil relative to the magnetic field (Principle of AC Generator).
Key Formulae: Induced Current & Induced Charge

1. Induced Current ($I$): If the circuit has total resistance $R$, the induced current is:

$$I = \frac{|\epsilon|}{R} = \frac{N}{R} \left| \frac{d\Phi_B}{dt} \right|$$

2. Induced Charge ($\Delta q$): Since $I = \frac{dq}{dt}$, we have:

$$\frac{dq}{dt} = \frac{N}{R} \frac{d\Phi_B}{dt} \implies \Delta q = \frac{N}{R} \Delta \Phi_B = \frac{|\Phi_{\text{final}} - \Phi_{\text{initial}}|}{R_{\text{total}}}$$

Crucial Fact for Exams: Induced EMF and Current depend on the rate of flux change (time-dependent), but the total induced charge ($\Delta q$) is completely independent of time and only depends on the net change in flux and circuit resistance!

NCERT Example 6.2 Question: A square loop of side $10\text{ cm}$ and resistance $0.5\,\Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\text{ T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.70\text{ s}$ at a steady rate. Determine the magnitudes of induced emf and current during this time interval.
Solution:
Area of square loop: $$A = (10\text{ cm})^2 = (0.1\text{ m})^2 = 10^{-2}\text{ m}^2$$ The normal to the loop faces East (or West). The field is North-East, so angle $\theta = 45^\circ$.
Initial magnetic flux: $$\Phi_{\text{initial}} = BA\cos 45^\circ = (0.10\text{ T}) \times (10^{-2}\text{ m}^2) \times \frac{1}{\sqrt{2}} = \frac{10^{-3}}{\sqrt{2}}\text{ Wb}$$ Final flux $\Phi_{\text{final}} = 0$, in time $\Delta t = 0.70\text{ s}$.
Magnitude of induced EMF: $$\begin{aligned} |\epsilon| &= \frac{|\Delta \Phi|}{\Delta t} = \frac{10^{-3}}{\sqrt{2} \times 0.70} \\ &= \frac{10^{-3}}{1.414 \times 0.70} \approx \mathbf{1.0\text{ mV}} = 1.0 \times 10^{-3}\text{ V} \end{aligned}$$ Magnitude of induced current: $$I = \frac{\epsilon}{R} = \frac{1.0 \times 10^{-3}\text{ V}}{0.5\,\Omega} = \mathbf{2.0\text{ mA}}$$
NCERT Example 6.3 Question: A circular coil of radius $10\text{ cm}$, $500\text{ turns}$, and resistance $2\,\Omega$ is placed with its plane perpendicular to the horizontal component of Earth's magnetic field ($B_H = 3.0 \times 10^{-5}\text{ T}$). It is rotated about its vertical diameter through $180^\circ$ in $0.25\text{ s}$. Estimate the magnitudes of the emf and current induced in the coil.
Solution:
Area $A = \pi r^2 = \pi \times (0.1\text{ m})^2 = \pi \times 10^{-2}\text{ m}^2$.
Initial flux per turn: $$\Phi_{\text{initial}} = B A \cos 0^\circ = 3.0 \times 10^{-5} \times \pi \times 10^{-2} = 3\pi \times 10^{-7}\text{ Wb}$$ Final flux per turn: $$\Phi_{\text{final}} = B A \cos 180^\circ = -3\pi \times 10^{-7}\text{ Wb}$$ Change in flux per turn: $$\Delta\Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = -6\pi \times 10^{-7}\text{ Wb}$$ Total induced EMF: $$\begin{aligned} \epsilon &= N \frac{|\Delta\Phi|}{\Delta t} = 500 \times \frac{6\pi \times 10^{-7}}{0.25} \\ &= 2000 \times 6\pi \times 10^{-7} = 1.2\pi \times 10^{-3}\text{ V} \approx \mathbf{3.8\text{ mV}} \end{aligned}$$ Induced current: $$I = \frac{\epsilon}{R} = \frac{3.8 \times 10^{-3}\text{ V}}{2\,\Omega} = \mathbf{1.9\text{ mA}}$$

4. Lenz's Law and Conservation of Energy

Statement: The polarity of the induced electromotive force is always such that it tends to produce an electric current which opposes the change in magnetic flux that produced it.

Figure 6.6: Illustration of Lenz's law
Figure 6.6: (a) Magnet moving towards loop induces CCW current (N-pole). (b) Magnet moving away induces CW current (S-pole).

Lenz's Law as a Consequence of Energy Conservation:

Lenz's law is a direct manifestation of the Law of Conservation of Energy:

Rule for Direction (End-Face Rule)
Figure 6.7: Planar loops in magnetic field
Figure 6.7: Planar loops of different shapes entering and leaving a magnetic field region.
Example 6.4 Question: Predict the direction of induced current in planar loops of different shapes moving into/out of a perpendicular magnetic field pointing into the page (Fig. 6.7):
Solution:
  1. Rectangular loop entering field: Magnetic flux into the page is increasing. To oppose this, the loop must generate an outward magnetic field ($\odot$). By the right-hand grip rule, current must flow anti-clockwise ($b \to c \to d \to a \to b$).
  2. Triangular loop leaving field: Inward flux is decreasing. The loop must support the decreasing field by generating an inward field ($\otimes$). Current flows clockwise ($b \to a \to c \to b$).
  3. Irregular loop leaving field: Inward flux is decreasing. To oppose the decrease, current flows clockwise ($c \to d \to a \to b \to c$).
Note: Induced current exists only while a loop is entering or leaving the boundary. When fully inside or fully outside, $\frac{d\Phi}{dt} = 0$, so induced current is zero!
Example 6.5 Question: (a) A closed loop is held stationary between the poles of two very strong permanent magnets. Can we generate current?
(b) A closed loop moves normal to the constant electric field between capacitor plates. Is a current induced?
(c) A rectangular loop and a circular loop move out of a uniform magnetic field with constant velocity $v$ (Fig. 6.8). In which loop is the induced EMF constant?
(d) Predict the polarity of capacitor plates A and B in Fig. 6.9 when North poles of two magnets approach from opposite sides.
Figure 6.8 & 6.9: Exiting loops and capacitor polarity
Figure 6.8 & 6.9: (c) Rectangular and circular loops moving out of magnetic field, and (d) Capacitor connected to loop with approaching magnets.
Solution:
(a) No. Strong magnets create a large magnetic flux, but since the loop and magnets are stationary, $\frac{d\Phi_B}{dt} = 0$, so no EMF is induced.
(b) No. Changing electric field/flux does not induce current in a conducting loop in this manner (EMI requires changing magnetic flux).
(c) The induced EMF is constant only in the rectangular loop because its rate of change of area ($\frac{dA}{dt} = l v$) remains constant. For the circular loop, the width cutting the boundary changes continuously as it exits, so $\frac{dA}{dt}$ varies with time.
(d) As North poles approach from both sides, the magnetic flux through the loop increases. The induced current must oppose this by creating opposing magnetic fields. Current flows such that Plate A becomes Positive and Plate B becomes Negative.

5. Motional Electromotive Force

Definition: The electromotive force induced across the ends of a conductor due to its motion through a magnetic field is called Motional EMF.

Figure 6.10: Motional EMF
Figure 6.10: Arm PQ moving across U-shaped conductor in uniform magnetic field.

A. Derivation using Faraday's Flux Rule (Translational Motional EMF)

Let a straight conducting rod $PQ$ of length $l$ slide with velocity $v$ on parallel conducting rails separated by distance $l$, placed in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the rails (into the page $\otimes$).

Let $x$ be the position of the rod from the closed end $RS$. The area enclosed by loop $PQRS$ is $A = lx$.

Magnetic flux linked with the loop at instant $t$:

$$\Phi_B = BA = Blx$$

According to Faraday's Law, the induced EMF is:

$$\epsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) = -Bl\left(\frac{dx}{dt}\right)$$

Since the rod is moving to the left, $x$ decreases with time, so $\frac{dx}{dt} = -v$. Therefore:

Motional EMF Formula $$\epsilon = Blv$$

Where $B$, $l$, and $v$ are mutually perpendicular. In vector notation: $\epsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.

B. Microscopic Explanation using Lorentz Magnetic Force:

Consider a free charge carrier $q$ (electron) inside the conductor moving with velocity $\vec{v}$ through field $\vec{B}$.

C. Power and Energy Conservation in Motional EMF:

If the circuit has total resistance $R$:

D. Rotational Motional EMF (Rotating Conductor / Disc)

Figure 6.11: Rotating metallic rod touching circular metallic ring in magnetic field
Figure 6.11: Rotating metallic rod hinged at center touching circular metallic ring in uniform magnetic field.
Derivation: Rotational Motional EMF

Method 1 (Integration of Small Elements):

Consider an infinitesimal element of length $dr$ at distance $r$ from the axis of rotation $O$. Linear speed of this element is $v = \omega r$.

Motional EMF induced across this small element $dr$ is:

$$d\epsilon = B v dr = B(\omega r) dr = B\omega r dr$$

Integrating from $r = 0$ (hinge) to $r = l$ (outer tip):

$$\epsilon = \int_0^l B\omega r dr = B\omega \left[ \frac{r^2}{2} \right]_0^l = \mathbf{\frac{1}{2} B \omega l^2} = \mathbf{\frac{1}{2} B \omega R^2}$$

Since $\omega = 2\pi\nu$, this can also be written as: $\epsilon = B \pi \nu l^2 = B A \nu$ (where $A = \pi l^2$ is the swept area per revolution).

Example 6.6 Question: A metallic rod of $1\text{ m}$ length is rotated with a frequency of $50\text{ rev/s}$, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $1\text{ m}$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field of $1.0\text{ T}$ parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?
Solution:
Given: $l = R = 1.0\text{ m}$, frequency $\nu = 50\text{ rev/s} \implies \omega = 2\pi\nu = 2\pi \times 50 = 100\pi\text{ rad/s}$, $B = 1.0\text{ T}$.
Using Rotational Motional EMF formula: $$\begin{aligned} \epsilon &= \frac{1}{2} B \omega R^2 \\ &= \frac{1}{2} \times 1.0 \times (100\pi) \times (1.0)^2 \\ &= 50\pi\text{ V} = 50 \times 3.1416 = \mathbf{157\text{ V}} \end{aligned}$$
Example 6.7 Question: A wheel with $10\text{ metallic spokes}$ each $0.5\text{ m}$ long is rotated with a speed of $120\text{ rev/min}$ in a plane normal to the horizontal component of Earth's magnetic field $B_H = 0.4\text{ G}$ at a place. What is the induced emf between the axle and the rim of the wheel? ($1\text{ G} = 10^{-4}\text{ T}$).
Solution:
Given: length of spoke $R = 0.5\text{ m}$, $B = 0.4\text{ G} = 0.4 \times 10^{-4}\text{ T}$.
Angular speed: $$\omega = \frac{120\text{ rev}}{60\text{ s}} \times 2\pi = 4\pi\text{ rad/s}$$ EMF induced across a single spoke: $$\begin{aligned} \epsilon &= \frac{1}{2} B \omega R^2 \\ &= \frac{1}{2} \times (0.4 \times 10^{-4}\text{ T}) \times (4\pi) \times (0.5)^2 \\ &= 0.2 \times 10^{-4} \times 4\pi \times 0.25 \\ &= \mathbf{6.28 \times 10^{-5}\text{ V}} \end{aligned}$$ Note on Number of Spokes: All 10 metallic spokes are connected in parallel between the central axle and outer rim. Just as identical cells connected in parallel provide the same terminal potential difference, the net EMF remains equal to the EMF of a single spoke ($\mathbf{6.28 \times 10^{-5}\text{ V}}$).

6. Inductance & Mutual Induction

Inductance: Inductance is the electrical property of a conductor/circuit by virtue of which it opposes any change in the electric current flowing through it. It acts as the electrical analogue of mass (inertia) in mechanics.

For a coil of $N$ turns, the total magnetic flux linked is proportional to the current $I$:

$$N\Phi_B \propto I \implies N\Phi_B = (\text{Inductance}) \times I$$
Figure 6.12: Two long co-axial solenoids illustrating mutual inductance
Figure 6.12: Two long co-axial solenoids \(S_1\) and \(S_2\) of length \(l\).

A. Mutual Induction

Definition: The phenomenon in which an electromotive force is induced in a coil (Secondary) due to a time-varying current in a neighboring coil (Primary).

Let current $I_2$ in outer coil $S_2$ produce magnetic flux $\Phi_1$ through inner coil $S_1$ (having $N_1$ turns):

$$N_1 \Phi_1 = M_{12} I_2 \implies \epsilon_1 = -M_{12} \frac{dI_2}{dt}$$

Where $M_{12}$ is the Coefficient of Mutual Induction (Mutual Inductance).

Board Derivation: Mutual Inductance of Two Long Co-axial Solenoids

Consider two long co-axial solenoids of length $l$ ($l \gg r_2 > r_1$):

Case 1: Current $I_2$ in Outer Solenoid $S_2$:
Magnetic field inside $S_2$ is $B_2 = \mu_0 n_2 I_2$.
This magnetic field links with the inner solenoid $S_1$ over area $A_1 = \pi r_1^2$.
Total flux linkage with $S_1$ is:

$$\begin{aligned} N_1 \Phi_1 &= (n_1 l) \cdot (\pi r_1^2) \cdot (\mu_0 n_2 I_2) \\ &= \mu_0 n_1 n_2 \pi r_1^2 l I_2 \end{aligned}$$

Since $N_1 \Phi_1 = M_{12} I_2$, we obtain:

$$M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l$$

Case 2: Current $I_1$ in Inner Solenoid $S_1$:
Magnetic field inside $S_1$ is $B_1 = \mu_0 n_1 I_1$. Since $B_1$ is confined solely inside area $A_1 = \pi r_1^2$, the flux linkage with outer solenoid $S_2$ is:

$$\begin{aligned} N_2 \Phi_2 &= (n_2 l) \cdot (\pi r_1^2) \cdot (\mu_0 n_1 I_1) \\ &= \mu_0 n_1 n_2 \pi r_1^2 l I_1 \end{aligned}$$

Since $N_2 \Phi_2 = M_{21} I_1$, we obtain:

$$M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$$

Reciprocity Theorem:

$$M_{12} = M_{21} = M = \mu_0 n_1 n_2 \pi r_1^2 l = \frac{\mu_0 N_1 N_2 A_1}{l}$$

If the solenoids are filled with a magnetic core of relative permeability $\mu_r$:

$$M = \mu_r \mu_0 n_1 n_2 \pi r_1^2 l$$
Example 6.8 Question: Two concentric circular coils, one of small radius $r_1$ and the other of large radius $r_2$ ($r_1 \ll r_2$), are placed co-axially with centers coinciding. Obtain the mutual inductance of the arrangement.
Solution:
Let current $I_2$ flow through the outer large coil $C_2$ (radius $r_2$).
Magnetic field at the common center: $$B_2 = \frac{\mu_0 I_2}{2 r_2}$$ Since the inner coil $C_1$ is very small ($r_1 \ll r_2$), field $B_2$ can be treated as completely uniform over its cross-sectional area $A_1 = \pi r_1^2$.
Flux linked with inner coil $C_1$: $$\Phi_1 = B_2 A_1 = \left( \frac{\mu_0 I_2}{2 r_2} \right) (\pi r_1^2) = \left( \frac{\mu_0 \pi r_1^2}{2 r_2} \right) I_2$$ Since $\Phi_1 = M_{12} I_2$, comparing gives: $$M_{12} = M_{21} = M = \mathbf{\frac{\mu_0 \pi r_1^2}{2 r_2}}$$
Factors Affecting Mutual Inductance & Coupling Coefficient
  1. Geometry & Size: Cross-sectional area $A$ and length $l$.
  2. Number of Turns: $M \propto N_1 N_2$.
  3. Relative Permeability ($\mu_r$): High permeability iron core increases $M$ significantly.
  4. Separation & Relative Orientation: Maximum when coils are co-axial; decreases to zero when coils are perpendicular ($\theta = 90^\circ$).

Coefficient of Coupling ($K$): Measures the fraction of flux linkage between two coils ($0 \le K \le 1$):

$$K = \frac{M}{\sqrt{L_1 L_2}}$$

7. Self-Inductance and Stored Magnetic Energy

Definition: Self-induction is the phenomenon by which an induced EMF is produced in a single isolated coil when the electric current passing through the same coil changes with time.

Total flux linkage $N\Phi_B \propto I \implies$ $N\Phi_B = L I$

The induced Back EMF is:

$$\epsilon = -\frac{d(N\Phi_B)}{dt} = -L \frac{dI}{dt}$$

Where $L$ is the Coefficient of Self-Inductance of the coil.

Board Derivation: Self-Inductance of a Long Solenoid

Consider a long solenoid of length $l$, area $A$, with $n$ turns per unit length (total turns $N = nl$). Current is $I$.

1. Magnetic field inside solenoid: $B = \mu_0 n I = \mu_0 \left(\frac{N}{l}\right) I$.

2. Magnetic flux linked per single turn: $\phi = BA = \left(\mu_0 \frac{N}{l} I\right) A$.

3. Total flux linkage for all $N$ turns:

$$N\Phi_B = N \phi = N \left( \mu_0 \frac{N}{l} I A \right) = \frac{\mu_0 N^2 A}{l} I = \mu_0 n^2 A l I$$

4. Since $N\Phi_B = LI$, we obtain:

$$L = \mu_0 n^2 A l = \frac{\mu_0 N^2 A}{l}$$

With an iron core of relative permeability $\mu_r$:

$$L = \mu_r \mu_0 n^2 A l = \frac{\mu_r \mu_0 N^2 A}{l}$$

Magnetic Energy Stored in an Inductor:

To establish a current $I$ against the opposing back-EMF, the power supply must perform work:

$$\frac{dW}{dt} = |\epsilon| I = \left( L \frac{dI}{dt} \right) I \implies dW = L I dI$$

Total work done (stored as magnetic potential energy $U_B$):

$$U_B = \int_0^I L I dI = \mathbf{\frac{1}{2} L I^2}$$
Magnetic Energy Density ($u_B$)

For a solenoid of volume $V = Al$, using $L = \mu_0 n^2 Al$ and $I = \frac{B}{\mu_0 n}$:

$$U_B = \frac{1}{2} (\mu_0 n^2 Al) \left( \frac{B}{\mu_0 n} \right)^2 = \frac{B^2}{2\mu_0} (Al)$$

Energy per unit volume (Magnetic Energy Density):

$$u_B = \frac{U_B}{\text{Volume}} = \mathbf{\frac{B^2}{2\mu_0}}$$

Comparison with Electrostatics: Energy density in electric field of capacitor is $u_E = \frac{1}{2}\epsilon_0 E^2$. In both fields, energy density is proportional to the square of field strength!

Example 6.9 Question: (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $B$, area $A$ and length $l$. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?
Solution:
(a) Energy $U_B = \frac{1}{2} L I^2$. Substituting $L = \mu_0 n^2 Al$ and $I = \frac{B}{\mu_0 n}$ gives: $$U_B = \frac{1}{2} (\mu_0 n^2 Al) \left(\frac{B}{\mu_0 n}\right)^2 = \frac{1}{2\mu_0} B^2 Al$$ (b) Magnetic energy density is $u_B = \frac{B^2}{2\mu_0}$. Electrostatic energy density in a capacitor is $u_E = \frac{1}{2}\epsilon_0 E^2$. Both expressions show that electromagnetic field energy is stored directly in the space containing the field and scales quadratically with field strength ($B^2$ and E^2).

8. AC Generator

Principle: An AC Generator (Alternator) converts mechanical energy into electrical energy based on the principle of Electromagnetic Induction. When a closed armature coil is rotated rapidly in a uniform magnetic field, the magnetic flux changes periodically, generating an alternating EMF.

Figure 6.13: AC Generator schematic
Figure 6.13: Schematic diagram of an AC Generator mechanism.
Figure 6.14: Alternating EMF generation stages
Figure 6.14: Stages of armature coil rotation and generated alternating EMF.
Board Derivation: Mathematical Theory of AC Generator

Consider a coil of $N$ turns, cross-sectional area $A$, rotating with constant angular speed $\omega$ about an axis perpendicular to uniform magnetic field $\vec{B}$.

Let $\theta$ be the angle between area vector $\vec{A}$ and field $\vec{B}$ at any time $t$. If $\theta = 0$ at $t = 0$, then $\theta = \omega t$.

Magnetic flux at time $t$:

$$\Phi_B = B A \cos(\omega t)$$

Total flux linkage for $N$ turns:

$$\Phi_{\text{total}} = N B A \cos(\omega t)$$

According to Faraday's Law, instantaneous induced EMF is:

$$\begin{aligned} \epsilon &= -\frac{d\Phi_{\text{total}}}{dt} = -NBA \frac{d}{dt}[\cos(\omega t)] \\ &= -NBA [-\omega \sin(\omega t)] \\ &= \mathbf{NBA\omega \sin(\omega t)} \end{aligned}$$

Peak (Maximum) EMF occurs when $\sin(\omega t) = \pm 1$:

$$\mathbf{\epsilon_0 = NBA\omega = 2\pi \nu NBA}$$

Therefore, the instantaneous alternating EMF is:

$$\mathbf{\epsilon = \epsilon_0 \sin(\omega t) = \epsilon_0 \sin(2\pi \nu t)}$$

If the circuit has resistance $R$, the instantaneous alternating current is:

$$I = \frac{\epsilon}{R} = \frac{\epsilon_0}{R}\sin(\omega t) = \mathbf{I_0 \sin(\omega t)}$$
Example 6.10 Question: Kamla peddles a stationary bicycle. The pedals are attached to a $100\text{ turn}$ coil of area $0.10\text{ m}^2$. The coil rotates at half a revolution per second ($\nu = 0.5\text{ rev/s}$) in a uniform magnetic field of $0.01\text{ T}$ perpendicular to the axis of rotation. What is the maximum voltage generated in the coil?
Solution:
Given: $N = 100$, $A = 0.10\text{ m}^2$, $\nu = 0.5\text{ Hz}$, $B = 0.01\text{ T}$.
Angular speed: $$\omega = 2\pi\nu = 2 \times 3.1416 \times 0.5 = \pi\text{ rad/s} = 3.1416\text{ rad/s}$$ Maximum (Peak) Voltage: $$\begin{aligned} \epsilon_0 &= NBA\omega \\ &= 100 \times 0.01 \times 0.10 \times 3.1416 \\ &= \mathbf{0.314\text{ V}} \end{aligned}$$

9. Comprehensive Formula Revision & Physical Quantities

Physical Quantity Symbol Defining Formula SI Unit Dimensions
Magnetic Flux $\Phi_B$ $\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$ Weber ($\text{Wb}$) or $\text{T}\cdot\text{m}^2$ $[\text{M}\text{L}^2\text{T}^{-2}\text{A}^{-1}]$
Induced EMF $\epsilon$ $\epsilon = -N \frac{d\Phi_B}{dt}$ Volt ($\text{V}$) $[\text{M}\text{L}^2\text{T}^{-3}\text{A}^{-1}]$
Induced Charge $\Delta q$ $\Delta q = \frac{|\Delta\Phi|}{R}$ Coulomb ($\text{C}$) $[\text{A}\text{T}]$
Translational Motional EMF $\epsilon$ $\epsilon = Blv$ Volt ($\text{V}$) $[\text{M}\text{L}^2\text{T}^{-3}\text{A}^{-1}]$
Rotational Motional EMF $\epsilon$ $\epsilon = \frac{1}{2}B\omega l^2 = BA\nu$ Volt ($\text{V}$) $[\text{M}\text{L}^2\text{T}^{-3}\text{A}^{-1}]$
Self-Inductance (Solenoid) $L$ $L = \frac{\mu_0 N^2 A}{l} = \mu_0 n^2 Al$ Henry ($\text{H}$) $[\text{M}\text{L}^2\text{T}^{-2}\text{A}^{-2}]$
Mutual Inductance (Co-axial) $M$ $M = \mu_0 n_1 n_2 \pi r_1^2 l$ Henry ($\text{H}$) $[\text{M}\text{L}^2\text{T}^{-2}\text{A}^{-2}]$
Magnetic Energy Stored $U_B$ $U_B = \frac{1}{2} L I^2$ Joule ($\text{J}$) $[\text{M}\text{L}^2\text{T}^{-2}]$
Magnetic Energy Density $u_B$ $u_B = \frac{B^2}{2\mu_0}$ $\text{J}/\text{m}^3$ $[\text{M}\text{L}^{-1}\text{T}^{-2}]$
AC Generator Peak EMF $\epsilon_0$ $\epsilon_0 = NBA\omega = 2\pi\nu NBA$ Volt ($\text{V}$) $[\text{M}\text{L}^2\text{T}^{-3}\text{A}^{-1}]$

10. Key Exercise Problems with Step-by-Step Solutions

Figure 6.15: Exercise 6.1 situations
Figure 6.15: Configurations (a) to (f) for predicting the direction of induced current using Lenz's law.
Exercise 6.1 Question: Predict the direction of induced current in the situations described by Figures 6.15(a) to (f):
Solution:
  1. Fig. 6.15(a): As South pole approaches coil from right, end facing magnet becomes South pole (clockwise). Current flows along $p \to r \to q \to p$.
  2. Fig. 6.15(b): Moving magnet causes South pole to recede and North pole to approach. Current flows along $p \to r \to q \to p$ in left coil and $y \to z \to x \to y$ in right coil.
  3. Fig. 6.15(c): Tapping key is just closed (make). Current grows $\implies$ flux increases. Induced current opposes growth, flowing along $y \to z \to x \to y$.
  4. Fig. 6.15(d): Rheostat resistance is decreased $\implies$ primary current increases. Induced current opposes growth, flowing along $z \to y \to x \to z$.
  5. Fig. 6.15(e): Tapping key is just released (break) $\implies$ current drops to zero. Induced current supports decaying current, flowing along $x \to r \to y \to x$.
  6. Fig. 6.15(f): Straight current-carrying wire lies in the plane of the loop. Magnetic field lines are parallel to the plane of the loop ($\theta = 90^\circ$). Flux linked $\Phi_B = 0$ at all times $\implies$ No induced current.
Figure 6.16: Exercise 6.2 loop deformations
Figure 6.16: (a) Irregular loop becoming circular, (b) Circular loop turning into narrow straight wire.
Exercise 6.2 Question: Use Lenz's law to determine the direction of induced current in the situations described by Figure 6.16:
  1. A wire of irregular shape turning into a circular shape;
  2. A circular loop being deformed into a narrow straight wire.
Solution:
(a) Irregular loop $\to$ Circular loop (inward field $\otimes$):
For a given perimeter, a circle encloses the maximum area. As the area increases, magnetic flux pointing into the page increases. By Lenz's law, induced current must oppose this increase by generating outward field lines ($\odot$). Hence, induced current flows in the anti-clockwise direction ($a \to d \to c \to b \to a$).

(b) Circular loop $\to$ Narrow straight wire (outward field $\odot$):
As the circle is deformed, its enclosed area decreases, causing the outward magnetic flux to decrease. By Lenz's law, induced current must oppose this decrease by producing outward magnetic field lines ($\odot$). Hence, induced current flows in the anti-clockwise direction ($a \to d \to c \to b \to a$).
Exercise 6.3 Question: A long solenoid with $15\text{ turns per cm}$ has a small loop of area $2.0\text{ cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0\text{ A}$ to $4.0\text{ A}$ in $0.1\text{ s}$, what is the induced emf in the loop while the current is changing?
Solution:
Number of turns per unit length: $$n = 15\text{ turns/cm} = 1500\text{ turns/m}$$ Area of loop: $$A = 2.0\text{ cm}^2 = 2.0 \times 10^{-4}\text{ m}^2$$ Rate of change of current: $$\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20\text{ A/s}$$ Magnetic field inside solenoid: $$B = \mu_0 n I \implies \frac{dB}{dt} = \mu_0 n \frac{dI}{dt}$$ Induced EMF in the loop: $$\begin{aligned} \epsilon &= A \frac{dB}{dt} = A \mu_0 n \frac{dI}{dt} \\ &= (2.0 \times 10^{-4}) \times (4\pi \times 10^{-7}) \times 1500 \times 20 \\ &= \mathbf{7.54 \times 10^{-6}\text{ V}} = 7.54\,\mu\text{V} \end{aligned}$$
Exercise 6.6 Question: A horizontal straight wire $10\text{ m}$ long extending from east to west is falling with a speed of $5.0\text{ m/s}$ at right angles to the horizontal component of Earth's magnetic field ($B_H = 0.30 \times 10^{-4}\text{ Wb/m}^2$).
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Solution:
Given: $l = 10\text{ m}$, $v = 5.0\text{ m/s}$, $B_H = 0.30 \times 10^{-4}\text{ T}$.
(a) Instantaneous induced EMF: $$\begin{aligned} \epsilon &= B_H l v \\ &= (0.30 \times 10^{-4}\text{ T}) \times (10\text{ m}) \times (5.0\text{ m/s}) \\ &= \mathbf{1.5 \times 10^{-3}\text{ V}} = 1.5\text{ mV} \end{aligned}$$ (b) Direction of EMF: By Fleming's Right-Hand Rule (Forefinger pointing North along $B_H$, Thumb pointing downwards along motion $v$), the induced current is directed from West to East.
(c) Higher Potential End: Since the wire acts as a battery/source of EMF with current driven from West to East internally, the Western end is at the higher electrical potential (+).
Exercise 6.7 & 6.8 Question 6.7: Current in a circuit falls from $5.0\text{ A}$ to $0.0\text{ A}$ in $0.1\text{ s}$. If an average emf of $200\text{ V}$ is induced, give an estimate of the self-inductance of the circuit.

Question 6.8: A pair of adjacent coils has a mutual inductance of $1.5\text{ H}$. If the current in one coil changes from $0$ to $20\text{ A}$ in $0.5\text{ s}$, what is the change of flux linkage with the other coil?
Solution 6.7:
$$\begin{aligned} |\epsilon| &= L \left|\frac{dI}{dt}\right| \implies 200 = L \left(\frac{5.0 - 0}{0.1}\right) = L (50) \\ &\implies \mathbf{L = \frac{200}{50} = 4.0\text{ H}} \end{aligned}$$
Solution 6.8:
$$\begin{aligned} \Delta(N_2\Phi_2) &= M \Delta I_1 = 1.5\text{ H} \times (20\text{ A} - 0) \\ &= \mathbf{30\text{ Wb}}\quad (\text{or }\text{Wb-turns}) \end{aligned}$$
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