Electromagnetic Induction

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Home Concept Boosters CBSE Class 12 Physics Electromagnetic Induction

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page covers Electromagnetic Induction (EMI) completely for CBSE Class 12 Physics.

Key Concepts

Class 12 · Physics · Electromagnetic Induction

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Core concepts you must know

Class 12 · Ch 6

Magnetic Flux Definition

The total number of magnetic field lines crossing normally through a given surface area. It is the dot product of the magnetic field vector and the area vector. SI Unit: Weber (Wb).

$$\phi = \oint \vec{B} \cdot d\vec{S} = BA \cos\theta$$

Faraday’s Law of Induction Formula

The magnitude of the induced EMF in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit.

$$e = -N \frac{d\phi}{dt}$$

Lenz’s Law Definition

The direction of the induced current (and hence the induced EMF) is always such that it creates a magnetic field that opposes the change in the magnetic flux that produced it. This justifies the negative sign in Faraday’s Law and is a consequence of the Conservation of Energy.

Motional EMF (Translational & Rotational) Formula

EMF induced across the ends of a conductor moving in a magnetic field. For a straight rod of length $l$ moving with velocity $v$, it’s $Blv$. For a rod rotating with angular velocity $\omega$ pivoted at one end, the formula adapts.

$$\text{Translational: } e = Blv \quad | \quad \text{Rotational: } e = \frac{1}{2} B \omega l^2$$

Self-Inductance ($L$) & Solenoids Formula

The property of a coil by virtue of which it opposes any change in the strength of current flowing through it by inducing an opposing EMF. SI Unit: Henry (H).

$$\phi = Li \quad \Rightarrow \quad e = -L \frac{di}{dt}$$

$$\text{Solenoid } L = \mu_0 n^2 (\pi r^2 l) = \mu_0 n^2 A l = \frac{\mu_0 N^2 A}{l}$$

Mutual Inductance ($M$) & Coupling Formula

The phenomenon of inducing EMF in a secondary coil due to a change in current in a primary coil. The coefficient of coupling ($k$) determines how perfectly they are linked magnetically.

$$\phi_2 = M i_1 \quad \Rightarrow \quad e_2 = -M \frac{di_1}{dt}$$

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

LR Circuit: Transients & Time Constant Formula

Current does not reach its maximum ($i_0 = e/R$) instantly due to the opposing induced EMF. It grows and decays exponentially. The time constant $\tau$ is the time taken for the current to reach $63.2\%$ of its max value.

$$\text{Growth: } i = \frac{e}{R}\left[1 – e^{-\frac{t}{L/R}}\right] \quad | \quad \text{Decay: } i = i_0 e^{-\frac{t}{L/R}}$$

$$\text{Time Constant: } \tau = \frac{L}{R}$$

Energy & Energy Density in an Inductor Formula

Work must be done against the back EMF to establish a current in an inductor. This work is stored as magnetic potential energy ($U$). The energy stored per unit volume is the energy density ($u$).

$$U = \frac{1}{2} L i^2 \quad | \quad u = \frac{U}{V} = \frac{B^2}{2 \mu_0}$$

AC Generator (Rotating Coil) Formula

When a coil of $N$ turns and area $A$ rotates with angular velocity $\omega$ in a uniform magnetic field $B$, the magnetic flux changes continuously, producing an alternating EMF.

$$e = NAB\omega \sin(\omega t) \quad \text{where } e_{max} = NAB\omega$$

Concept Deep Dive

Eddy Currents (Foucault Currents)

The invisible whirlpools of energy

Core Concept

When bulk pieces of conductors are subjected to changing magnetic flux, induced currents circulate throughout the volume of the metal, looking like swirling eddies in water. Because the resistance of a bulk piece of metal is very low, these Eddy Currents can be massive, leading to immense heat dissipation ($P = i^2 R$).

While often a nuisance that wastes energy in transformer cores, we utilize eddy currents practically in magnetic braking of trains, induction furnaces for melting metals, and electromagnetic damping in galvanometers.

How to minimize them?

We slice the solid metal core into thin sheets (laminations) and paint them with an insulating varnish. This breaks the large continuous paths, severely restricting the whirlpools and minimizing heat loss.

Lenz’s Law and Energy Conservation

Why the negative sign exists

Crucial Concept

Imagine pushing the North pole of a magnet into a coil. The coil must induce a current. If that current created a South pole, it would attract your magnet, pulling it in faster and faster, creating infinite free energy!

Instead, Lenz’s law dictates it creates a North pole to repel your magnet. You have to physically push against this repulsion (doing mechanical work). This mechanical work is exactly what gets converted into electrical energy in the coil. Lenz’s law is simply the Law of Conservation of Energy applied to electromagnetism.

Compare & Contrast

✗ Mass (Mechanics)

Inertia of linear motion.
Opposes any change in the object’s velocity.
Cannot instantly jump from 0 to $v$ (requires force over time).
Kinetic Energy: $K = \frac{1}{2} m v^2$

✓ Self-Inductance (Electricity)

“Inertia” of electrical circuits.
Opposes any change in the circuit’s current.
Current cannot instantly jump from 0 to max (requires EMF over time).
Magnetic Energy: $U = \frac{1}{2} L i^2$

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Remember

Inductors do NOT oppose current itself; they only oppose the change in current. A steady DC current flows through an ideal inductor with zero resistance as if it’s just a normal wire.

Common Mistakes to Avoid

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Mistake 1

Misunderstanding the Angle $\theta$ in Flux: In $\phi = BA \cos \theta$, the angle $\theta$ is between the Magnetic Field ($\vec{B}$) and the Area Vector ($d\vec{S}$, which is the normal/perpendicular to the surface). If a question says “the plane of the coil is at $30^\circ$ to the field”, then $\theta$ is actually $90^\circ – 30^\circ = 60^\circ$!

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Mistake 2

Ignoring the minus sign in Lenz’s Law: When writing Faraday’s law, writing $e = N \frac{d\phi}{dt}$ instead of $e = -N \frac{d\phi}{dt}$ loses marks in board exams. Even if you only need the magnitude, state $|e| = N \frac{d\phi}{dt}$ explicitly.

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Mistake 3

Confusing Induced Charge with Time: The induced EMF depends on how fast the flux changes ($dt$). However, the total Induced Charge ($q = \frac{\Delta \phi}{R}$) is completely independent of time. Whether you pull a magnet out in 1 second or 1 hour, the total charge that flowed through the circuit is exactly the same!

Exam Tips

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Tip 1

Use Fleming’s Right-Hand Rule to find the direction of Induced Current (Motional EMF). (Thumb = Motion of conductor, Forefinger = Magnetic Field, Middle Finger = Induced Current). Don’t confuse it with the Left-Hand rule used for force on a motor!

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Tip 2

For questions involving an airplane flying horizontally: The metallic wings of the airplane cut the Vertical component ($B_V$) of the Earth’s magnetic field. Therefore, the induced EMF across the wingtips is $e = B_V l v$. (Do not use the horizontal component $B_H$).

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Did You Know

Modern induction cooktops use high-frequency changing magnetic fields. These fields pass harmlessly through the glass top and induce massive eddy currents directly inside the iron or steel cooking pan. The pan itself becomes the heater, while the glass top stays relatively cool!

Expected Exam Questions

Board Pattern Questions

Class 12 · EMI · CBSE Exam

Class 12 · Physics

A metallic rod falls freely under gravity, keeping its length horizontal and parallel to the East-West direction. Will an EMF be induced across its ends? [1 mark]

Answer Yes 📝

Explanation

As the rod falls downwards (vertically), it cuts the horizontal component ($B_H$) of the Earth’s magnetic field, which points from South to North. Since the rod’s length (E-W), velocity (downward), and magnetic field ($B_H$ S-N) are all mutually perpendicular, a motional EMF ($e = B_H l v$) is induced across its ends.

Two identical loops, one of copper and the other of aluminum, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced EMF and (ii) the induced current produced in the two coils. [2 marks]

Answer (i) EMF is same; (ii) Current is higher in copper 📝

Explanation

(i) The induced EMF ($e = NAB\omega \sin \omega t$) depends only on the number of turns, area, magnetic field, and angular speed. Since these are identical for both loops, the induced EMF is the same.
(ii) Induced current is given by $i = e/R$. Since copper has a lower resistivity (and thus lower resistance $R$) than aluminum, the induced current in the copper loop will be greater.

A long solenoid with 15 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 A to 4.0 A in 0.1 s, what is the induced EMF in the loop while the current is changing? [3 marks]

Answer $$7.54 \times 10^{-6} \text{ V}$$ 📝

Explanation

Given: $n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$. Area $A = 2.0 \times 10^{-4} \text{ m}^2$. $\Delta I = 4.0 – 2.0 = 2.0 \text{ A}$. $dt = 0.1 \text{ s}$.
Magnetic field of solenoid $B = \mu_0 n I$.
Flux through loop $\phi = BA = (\mu_0 n I) A$.
Induced EMF $e = \frac{d\phi}{dt} = \mu_0 n A \left(\frac{di}{dt}\right)$.
$e = (4\pi \times 10^{-7}) \times 1500 \times (2.0 \times 10^{-4}) \times \left(\frac{2.0}{0.1}\right)$
$e = 4\pi \times 10^{-7} \times 1500 \times 2.0 \times 10^{-4} \times 20$
$e = 24000\pi \times 10^{-11} = 2.4\pi \times 10^{-6} \approx 7.54 \times 10^{-6} \text{ V}$.

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