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Electromagnetic Induction Class 12 Physics Notes

Introduction to Electromagnetic Induction

For a long time, electricity and magnetism were thought to be separate forces. In the early 19th century, scientists like Oersted and Ampere discovered that they are interconnected, showing that moving electric charges (currents) create magnetic fields. This led to a logical question: can the reverse happen? Can a moving magnet create an electric current?

The answer is a definite yes. Around 1830, experiments by Michael Faraday and Joseph Henry proved that a changing magnetic field could indeed induce an electric current in a closed coil. This phenomenon is called electromagnetic induction.

This discovery was revolutionary. It is not just a fascinating scientific concept; it's the foundation of our modern world. Without electromagnetic induction, we wouldn't have electric generators, transformers, and therefore, no widespread electricity for lights, trains, computers, or phones.

The Experiments of Faraday and Henry

The principles of electromagnetic induction were uncovered through a series of key experiments.

Experiment 1: Magnet and Coil

Setup: A conducting coil is connected to a galvanometer (a device that detects small electric currents).
Observation 1: When the North pole of a bar magnet is pushed towards the coil, the galvanometer needle deflects, showing a current is produced. This deflection only lasts as long as the magnet is moving. If the magnet is held still, the current stops.
Observation 2: When the magnet is pulled away from the coil, the galvanometer deflects again, but in the opposite direction. This means the current is flowing the other way.
Observation 3: Using the South pole produces the same effects, but the current directions are reversed compared to using the North pole.
Observation 4: The faster the magnet is moved, the larger the deflection, meaning a stronger current is induced.
Conclusion: It is the relative motion between the magnet and the coil that generates, or induces, an electric current.

Experiment 2: Two Coils

Setup: The bar magnet is replaced by a second coil ( $C_2$ ) connected to a battery. This coil produces a steady magnetic field due to the steady current flowing through it.
Observation: When coil $C_2$ is moved towards the first coil ( $C_1$ ), the galvanometer connected to $C_1$ deflects. When $C_2$ is moved away, the deflection is in the opposite direction. The current only exists while there is relative motion.
Conclusion: The magnetic field produced by a current-carrying coil behaves just like a bar magnet. Relative motion between two coils induces a current in one of them.

Experiment 3: Changing Current, No Motion

Setup: Two coils, $C_1$ (with galvanometer) and $C_2$ (with battery and a switch), are held stationary near each other.
Observation 1: When the switch in the circuit for $C_2$ is pressed, the current in it goes from zero to its maximum value. During this brief moment, the galvanometer in $C_1$ shows a momentary deflection and then returns to zero.
Observation 2: As long as the current in $C_2$ is steady (switch held down), there is no deflection in the galvanometer.
Observation 3: When the switch is released, the current in $C_2$ drops back to zero. During this brief moment, the galvanometer shows a momentary deflection in the opposite direction.
Conclusion: Relative motion is not strictly necessary. An electric current can be induced by changing the magnetic field. A changing current in one coil creates a changing magnetic field, which in turn induces a current in a nearby coil.

Magnetic Flux

To understand and quantify electromagnetic induction, we first need the concept of magnetic flux ( $\Phi_B$ ). It is a measure of the total number of magnetic field lines passing through a given surface area.

For a flat surface of area $A$ placed in a uniform magnetic field $\mathbf{B}$ , the magnetic flux is calculated as the dot product of the magnetic field vector and the area vector.

$\Phi_{\mathrm{B}}=\mathbf{B} \cdot \mathbf{A}=B A \cos \theta$

$\mathbf{B}$ is the magnetic field strength.
$\mathbf{A}$ is the area vector (its magnitude is the area, and its direction is perpendicular to the surface).
$\theta$ is the angle between the magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$ .

If the magnetic field is not uniform, the flux is calculated by summing up the flux through many small area elements:

$\boldsymbol{\Phi}_{B}=\sum_{\text{all}} \mathbf{B}_{i} \cdot \mathrm{d} \mathbf{A}_{i}$

Magnetic flux is a scalar quantity. Its SI unit is the weber (Wb), which is equivalent to tesla meter squared ( $\text{T m}^2$ ).

Faraday's Law of Induction

Based on his experiments, Michael Faraday concluded that an electromotive force (emf), and therefore a current if the circuit is closed, is induced in a coil whenever the magnetic flux passing through it changes with time.

Faraday's Law of Electromagnetic Induction states: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Mathematically, this is expressed as:

$\varepsilon=-\frac{\mathrm{d} \Phi_{B}}{\mathrm{~d} t}$

$\varepsilon$ is the induced electromotive force (emf), measured in volts (V).
$\frac{\mathrm{d} \Phi_{B}}{\mathrm{~d} t}$ is the rate of change of magnetic flux.
The negative sign relates to the direction of the induced emf, which is explained by Lenz's Law.

For a coil with N turns, the total induced emf is N times the emf in a single turn:

$\varepsilon=-N \frac{\mathrm{d} \Phi_{B}}{\mathrm{~d} t}$

This shows that you can increase the induced emf by increasing the number of turns in the coil.

Note

Magnetic flux (

\Phi_B = B A \cos \theta

) can be changed in three ways:

Changing the magnetic field strength ( $B$ ).
Changing the area of the coil ( $A$ ).
Changing the angle ( $\theta$ ) between the magnetic field and the area vector (e.g., by rotating the coil).

Example

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 get a large deflection, we need to maximize the rate of change of magnetic flux. This can be done by:

(i) Using a soft iron rod inside the current-carrying coil ( $C_2$ ) to strengthen its magnetic field.
(ii) Connecting coil $C_2$ to a more powerful battery to increase the current and thus the magnetic field.
(iii) Moving the coils relative to each other more rapidly.

(b) To show the presence of an induced current without a galvanometer, you could replace it with a small light bulb. The relative motion between the coils would induce a current, causing the bulb to glow.

Example

A square loop of side 10 cm and resistance

0.5 \Omega

is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Given

Side of square loop, $s = 10 \text{ cm} = 0.1 \text{ m}$
Area of loop, $A = (0.1 \text{ m})^2 = 10^{-2} \text{ m}^2$
Resistance, $R = 0.5 \Omega$
Initial magnetic field, $B_{initial} = 0.10 \text{ T}$
Final magnetic field, $B_{final} = 0 \text{ T}$
Time interval, $\Delta t = 0.70 \text{ s}$
Angle $\theta$ between area vector (east-west plane) and B (north-east) is $45^{\circ}$ .

To Find

Magnitude of induced emf, $\varepsilon$
Magnitude of current, $I$

Formula

$\Phi = B A \cos \theta$ $\varepsilon = \frac{|\Delta \Phi_{B}|}{\Delta t}$ $I = \frac{\varepsilon}{R}$

Solution

First, calculate the initial magnetic flux. The area vector is normal to the east-west plane, so it points either north or south. The magnetic field is north-east. The angle $\theta$ is $45^{\circ}$ .

$\Phi_{initial} = B_{initial} A \cos \theta = (0.10 \text{ T})(10^{-2} \text{ m}^2) \cos 45^{\circ} = \frac{0.1 \times 10^{-2}}{\sqrt{2}} \text{ Wb}$

The final flux is zero since the magnetic field is decreased to zero. $\Phi_{final} = 0$

Now, calculate the magnitude of the change in flux, $|\Delta \Phi_B|$ . $|\Delta \Phi_B| = |\Phi_{final} - \Phi_{initial}| = |0 - \frac{10^{-3}}{\sqrt{2}}| = \frac{10^{-3}}{\sqrt{2}} \text{ Wb}$

Next, calculate the magnitude of the induced emf. $\varepsilon = \frac{|\Delta \Phi_{B}|}{\Delta t} = \frac{10^{-3} / \sqrt{2}}{0.7} \approx \frac{10^{-3}}{1.414 \times 0.7} \approx 1.0 \times 10^{-3} \text{ V} = 1.0 \text{ mV}$

Finally, calculate the current. $I = \frac{\varepsilon}{R} = \frac{1.0 \times 10^{-3} \text{ V}}{0.5 \Omega} = 2.0 \times 10^{-3} \text{ A} = 2 \text{ mA}$

Final Answer The induced emf is $1.0$ mV and the induced current is $2$ mA.

Example

A circular coil of radius

10 \text{ cm}

, 500 turns and resistance

2 \Omega

is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. 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. Horizontal component of the earth's magnetic field at the place is

3.0 \times 10^{-5} \text{ T}

.

Given

Radius, $r = 10 \text{ cm} = 0.1 \text{ m}$
Area, $A = \pi r^2 = \pi (0.1)^2 = \pi \times 10^{-2} \text{ m}^2$
Number of turns, $N = 500$
Resistance, $R = 2 \Omega$
Time interval, $\Delta t = 0.25 \text{ s}$
Magnetic field, $B = 3.0 \times 10^{-5} \text{ T}$

To Find

Magnitude of induced emf, $\varepsilon$
Magnitude of induced current, $I$

Formula

$\Phi_B = B A \cos \theta$ $\varepsilon = N \frac{|\Delta \Phi_B|}{\Delta t}$ $I = \frac{\varepsilon}{R}$

Solution

Initially, the plane is perpendicular to the field, so the area vector is parallel to the field. Thus, the initial angle $\theta_{initial} = 0^{\circ}$ . $\Phi_{B(\text{initial})} = B A \cos 0^{\circ} = (3.0 \times 10^{-5})(\pi \times 10^{-2})(1) = 3\pi \times 10^{-7} \text{ Wb}$

After rotating $180^{\circ}$ , the area vector points in the opposite direction to the field. Thus, the final angle $\theta_{final} = 180^{\circ}$ . $\Phi_{B(\text{final})} = B A \cos 180^{\circ} = (3.0 \times 10^{-5})(\pi \times 10^{-2})(-1) = -3\pi \times 10^{-7} \text{ Wb}$

The change in flux is: $\Delta \Phi_B = \Phi_{B(\text{final})} - \Phi_{B(\text{initial})} = (-3\pi \times 10^{-7}) - (3\pi \times 10^{-7}) = -6\pi \times 10^{-7} \text{ Wb}$

The magnitude of the induced emf is: $\varepsilon = N \frac{|\Delta \Phi_B|}{\Delta t} = 500 \times \frac{|-6\pi \times 10^{-7}|}{0.25} = 500 \times \frac{6 \times 3.14 \times 10^{-7}}{0.25} \approx 3.8 \times 10^{-3} \text{ V}$

The induced current is: $I = \frac{\varepsilon}{R} = \frac{3.8 \times 10^{-3} \text{ V}}{2 \Omega} = 1.9 \times 10^{-3} \text{ A}$

Final Answer The estimated induced emf is $3.8 \times 10^{-3}$ V and the induced current is $1.9 \times 10^{-3}$ A.

Lenz's Law and Conservation of Energy

Faraday's Law tells us the magnitude of the induced emf, but what about its direction? This is given by Lenz's Law.

Lenz's Law states: The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

The negative sign in Faraday's Law, $\varepsilon = -N \frac{d\Phi_B}{dt}$ , is the mathematical representation of Lenz's Law.

Example

Let's reconsider Experiment 1. When the North pole of a magnet is pushed towards a coil, the magnetic flux through the coil increases. According to Lenz's law, the induced current will flow in a direction that creates its own magnetic field to oppose this increase. To oppose an approaching North pole, the coil must create its own North pole on the side facing the magnet. This requires a counter-clockwise current (as seen from the magnet's side), which creates a repulsive force.

If the North pole is pulled away, the flux decreases. The induced current will now flow in a direction to oppose this decrease, meaning it tries to pull the magnet back. It does this by creating a South pole on the side facing the magnet, which requires a clockwise current and creates an attractive force.

Lenz's Law and Conservation of Energy

Lenz's Law is a direct consequence of the law of conservation of energy.

Imagine if the induced current helped the change instead of opposing it. If you pushed a magnet towards a coil, and the coil produced a South pole, it would attract the magnet, pulling it in faster. This would induce an even stronger current, creating an even stronger attraction, and so on. The magnet would accelerate, gaining kinetic energy seemingly from nowhere. This would be a perpetual motion machine, which violates the law of conservation of energy.

In reality, because the induced current opposes the motion, you must do mechanical work to push the magnet against the repulsive force. This work you do is converted into electrical energy in the coil, which is then dissipated as heat (Joule heating). Energy is conserved.

Example

Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz's law.

Solution

(i) Rectangular loop abcd moving into the field: The magnetic flux (out of the page) through the loop is increasing. To oppose this, the induced current must create a magnetic field directed into the page. Using the right-hand rule, the current must flow in the clockwise direction, along the path bcdab.

(ii) Triangular loop abc moving out of the field: The magnetic flux (out of the page) through the loop is decreasing. To oppose this, the induced current must create a magnetic field directed out of the page. This requires a counter-clockwise current, flowing along the path bacb.

(iii) Irregular loop abcd moving out of the field: The magnetic flux (out of the page) is decreasing. Similar to the triangular loop, the induced current must flow in a counter-clockwise direction to create a magnetic field out of the page to oppose the change. The current flows along the path cdabc.

Note

No current is induced as long as a loop is completely inside or completely outside the uniform magnetic field, because in those cases, the magnetic flux through the loop is not changing.

Motional Electromotive Force

An emf can be induced by moving a conductor through a magnetic field. This is called motional emf.

Consider a rectangular conductor PQRS where the arm PQ is free to move. The loop is in a uniform magnetic field $\mathbf{B}$ directed perpendicular to its plane. If the arm PQ is moved to the left with a constant velocity $\mathbf{v}$ , the area of the loop decreases.

Let the length RS be $l$ and the length RQ be $x$ . The magnetic flux through the loop is: $\Phi_B = B l x$

As the arm moves, $x$ changes with time. The rate of change of flux induces an emf: $\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(B l x) = -B l \frac{dx}{dt}$

Since the arm moves left, $\frac{dx}{dt} = -v$ . Substituting this in, we get: $\varepsilon = -B l (-v) = B l v$

This induced emf, $\varepsilon = B l v$ , is the motional emf.

Explanation using Lorentz Force

We can also understand motional emf by considering the forces on the charge carriers (electrons) inside the moving conductor PQ.

When the rod PQ moves with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ , each charge $q$ inside the rod also moves with this velocity.
These moving charges experience a magnetic Lorentz force, $F = qvB$ .
The direction of this force is along the rod from P to Q, pushing the free electrons towards one end.
This separation of charge creates an electric field and thus a potential difference (emf) across the ends of the rod.
The work done to move a charge $q$ from P to Q is $W = F \times l = (qvB)l$ .
Since emf is work done per unit charge, $\varepsilon = W/q$ .
Therefore, $\varepsilon = \frac{qvBl}{q} = B l v$ , which is the same result.

Example

A wheel with 10 metallic spokes each 0.5 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

H_E

at a place. If

H_E = 0.4 \text{ G}

at the place, what is the induced emf between the axle and the rim of the wheel? Note that

1 \text{ G} = 10^{-4} \text{ T}

.

Given

Number of spokes = 10
Length of each spoke (radius), $R = 0.5 \text{ m}$
Rotational speed, $\nu = 120 \text{ rev/min} = \frac{120}{60} \text{ rev/s} = 2 \text{ rev/s}$
Angular speed, $\omega = 2\pi\nu = 2\pi(2) = 4\pi \text{ rad/s}$
Magnetic field, $B = 0.4 \text{ G} = 0.4 \times 10^{-4} \text{ T}$

To Find

Induced emf, $\varepsilon$

Formula

The emf induced in a rod of length $R$ rotating with angular velocity $\omega$ in a magnetic field $B$ is: $\varepsilon = \frac{1}{2} B \omega R^2$

Solution

Substitute the given values into the formula: $\varepsilon = \frac{1}{2} \times (0.4 \times 10^{-4} \text{ T}) \times (4\pi \text{ rad/s}) \times (0.5 \text{ m})^2$ $\varepsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4 \times 3.14) \times (0.25)$ $\varepsilon \approx 6.28 \times 10^{-5} \text{ V}$

Note

The number of spokes is irrelevant. Each spoke is a conductor moving through the magnetic field, and an emf is induced across each one. Since all spokes are connected between the same two points (the axle and the rim), they are in parallel. The total emf across a parallel combination is the same as the emf across a single component.

Final Answer The induced emf between the axle and the rim is $6.28 \times 10^{-5}$ V.

Inductance

Inductance is the property of an electrical conductor by which a change in current through it induces an electromotive force in both the conductor itself (self-inductance) and in any nearby conductors (mutual inductance).

In any coil, the magnetic flux ( $\Phi_B$ ) it produces is proportional to the current ( $I$ ) flowing through it. The total flux, called flux linkage, for a coil of $N$ turns is $N\Phi_B$ . $N\Phi_B \propto I$ The constant of proportionality is the inductance. Inductance depends on the geometry of the coil (size, shape, number of turns) and the material inside it. It is a scalar quantity, and its SI unit is the henry (H).

Mutual Inductance

When two coils are placed near each other, a changing current in one coil will induce an emf in the second coil. This is called mutual induction.

Let the flux linkage in coil 1 ( $N_1\Phi_1$ ) due to a current $I_2$ in coil 2 be given by $N_1\Phi_1 = M_{12}I_2$ .
$M_{12}$ is the mutual inductance of coil 1 with respect to coil 2.

According to Faraday's Law, the emf induced in coil 1 ( $\varepsilon_1$ ) is: $\varepsilon_1 = - \frac{d(N_1\Phi_1)}{dt} = -M_{12} \frac{dI_2}{dt}$

An important principle is that the mutual inductance is reciprocal: $M_{12} = M_{21} = M$ .

Example

For two long co-axial solenoids of length

l

and radius

r_1

(inner) and

r_2

(outer), with

n_1

and

n_2

turns per unit length respectively, the mutual inductance is given by:

M = \mu_0 n_1 n_2 \pi r_1^2 l

Self-Inductance

A changing current in a coil will induce an emf in that same coil. This phenomenon is called self-induction. This induced emf always opposes the change in current and is often called back emf.

The flux linkage in a coil of $N$ turns due to its own current $I$ is given by $N\Phi_B = LI$ .
$L$ is the self-inductance of the coil.

According to Faraday's Law, the self-induced emf ( $\varepsilon$ ) is: $\varepsilon = - \frac{d(N\Phi_B)}{dt} = -L \frac{dI}{dt}$

Self-inductance acts like inertia for electric circuits; it opposes any change (increase or decrease) in the current.

Example

For a long solenoid of length

l

, cross-sectional area

A

, and

n

turns per unit length, the self-inductance is:

L = \mu_0 n^2 A l

Energy Stored in an Inductor

To establish a current in an inductor, work must be done against the back emf. This work is stored as magnetic potential energy in the magnetic field of the inductor. The energy ( $W$ ) stored in an inductor with self-inductance $L$ carrying a current $I$ is:

$W = \frac{1}{2} L I^2$

Example

(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field

B

, area

A

and length

l

of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?

Solution

(a) Magnetic Energy in a Solenoid

The magnetic energy stored is $U_B = \frac{1}{2} L I^2$ . For a long solenoid, the magnetic field is $B = \mu_0 n I$ , so $I = \frac{B}{\mu_0 n}$ . The self-inductance of the solenoid is $L = \mu_0 n^2 A l$ .

Substitute $L$ and $I$ into the energy equation: $U_B = \frac{1}{2} (\mu_0 n^2 A l) \left( \frac{B}{\mu_0 n} \right)^2$ $U_B = \frac{1}{2} (\mu_0 n^2 A l) \frac{B^2}{\mu_0^2 n^2}$ $U_B = \frac{1}{2\mu_0} B^2 A l$

This is the total magnetic energy stored in the solenoid.

(b) Comparison with Electrostatic Energy

The magnetic energy per unit volume ( $u_B$ ) is the total energy divided by the volume ( $V=Al$ ): $u_B = \frac{U_B}{V} = \frac{\frac{1}{2\mu_0} B^2 A l}{Al} = \frac{B^2}{2\mu_0}$

The electrostatic energy stored per unit volume ( $u_E$ ) in a parallel plate capacitor is: $u_E = \frac{1}{2} \varepsilon_0 E^2$

Comparison: In both cases, the energy density (energy per unit volume) is proportional to the square of the field strength ( $B^2$ for magnetic fields and $E^2$ for electric fields).

AC Generator

One of the most important applications of electromagnetic induction is the AC generator, a device that converts mechanical energy into electrical energy.

Principle and Operation

An AC generator works by rotating a coil of wire (called an armature) in a uniform magnetic field. This rotation continuously changes the angle $\theta$ between the coil's area vector $\mathbf{A}$ and the magnetic field vector $\mathbf{B}$ , which causes a change in magnetic flux.

The magnetic flux through a coil of $N$ turns and area $A$ , rotating with constant angular speed $\omega$ in a magnetic field $B$ , is: $\Phi_B = B A \cos \theta = B A \cos(\omega t)$ (assuming $\theta = 0$ at $t = 0$ ).
According to Faraday's Law, the induced emf is: $\varepsilon = -N \frac{d\Phi_B}{dt} = -N \frac{d}{dt}(B A \cos(\omega t))$ $\varepsilon = -N B A (-\omega \sin(\omega t))$ $\varepsilon = N B A \omega \sin(\omega t)$

The value of the emf varies sinusoidally with time. The current produced also varies in this way and is called an alternating current (ac).

The maximum or peak value of the emf ( $\varepsilon_0$ ) occurs when $\sin(\omega t) = 1$ : $\varepsilon_0 = N B A \omega$

So, the instantaneous emf can be written as: $\varepsilon = \varepsilon_0 \sin(\omega t)$

In commercial generators, the mechanical energy to rotate the armature is provided by sources like falling water (hydro-electric), high-pressure steam from burning coal or nuclear fuel (thermal or nuclear generators). The frequency of rotation in India is 50 Hz, while in countries like the USA, it is 60 Hz.

Example

Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area

0.10 \text{ m}^2

. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

Given

Number of turns, $N = 100$
Area, $A = 0.10 \text{ m}^2$
Frequency of rotation, $\nu = 0.5 \text{ rev/s}$ (Hz)
Magnetic field, $B = 0.01 \text{ T}$

To Find

Maximum voltage (peak emf), $\varepsilon_0$

Formula

The angular speed is $\omega = 2\pi\nu$ . The maximum voltage is $\varepsilon_0 = N B A \omega = N B A (2\pi\nu)$ .

Solution

Substitute the given values into the formula: $\varepsilon_0 = 100 \times (0.01 \text{ T}) \times (0.10 \text{ m}^2) \times (2 \times 3.14 \times 0.5 \text{ Hz})$ $\varepsilon_0 = 100 \times 0.01 \times 0.1 \times 3.14$ $\varepsilon_0 = 0.314 \text{ V}$

Final Answer The maximum voltage generated in the coil is $0.314$ V.

Electromagnetic InductionClass 12 Physics Notes

Introduction to Electromagnetic Induction

The Experiments of Faraday and Henry

Experiment 1: Magnet and Coil

Experiment 2: Two Coils

Experiment 3: Changing Current, No Motion

Magnetic Flux

Faraday's Law of Induction

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Lenz's Law and Conservation of Energy

Lenz's Law and Conservation of Energy

Solution

Motional Electromotive Force

Explanation using Lorentz Force

Given

To Find

Formula

Solution

Inductance

Mutual Inductance

Self-Inductance

Energy Stored in an Inductor

Solution

AC Generator

Principle and Operation

Given

To Find

Formula

Solution