Electromagnetic Induction

The chapter builds in a deliberate sequence, and it pays to follow the order. It starts with the three classic experiments of Faraday and Henry — magnet and coil, two coils, and two stationary coils with a tapping key — which together isolate the one variable that matters: changing flux, however that change is produced. From there it defines magnetic flux precisely, states Faraday's law of induction as a quantitative rule, and fixes the sign and direction of the induced effect using Lenz's law, tied explicitly to the conservation of energy. The chapter then specializes to a conductor physically sliding through a field, the motional emf, and shows it can be derived two independent ways — from the flux rule and from the Lorentz force on moving charges — which is where the deep link between electricity and magnetism becomes concrete. Next comes inductance: how a changing current in one coil induces emf in a neighbour (mutual inductance) and how a changing current induces a back emf in the very coil carrying it (self-inductance), along with the magnetic energy stored in that self-induction. The chapter ends with the AC generator, the practical machine built entirely on Faraday's law, converting mechanical rotation into the alternating emf that powers the electrical grid.

Magnetic flux — the quantity that must change

Before any law can be stated, we need a number describing "how much magnetic field passes through a surface." For a flat area $A$ sitting in a uniform field $\vec{B}$, the magnetic flux is defined exactly as electric flux was defined earlier, as the component of $\vec{B}$ along the area's normal, multiplied by the area:

where: $\Phi_B$ is the magnetic flux in weber (Wb), equivalently tesla-metre-squared (T m$^2$); $B$ is the magnetic field magnitude; $A$ is the area of the surface; and $\theta$ is the angle between $\vec{B}$ and the area vector $\vec{A}$ (the outward normal to the surface, scaled by its area). Flux is a scalar — it has no direction of its own, only a sign that depends on the chosen normal. When the surface is not flat or the field is non-uniform, the same idea is applied element by element and summed: $\Phi_B = \sum_i \vec{B}_i\cdot\Delta\vec{A}_i$, adding up the field through every small patch $\Delta A_i$ of the surface. Flux is maximum when the field is perpendicular to the surface ($\theta = 0$, flux $=BA$) and zero when the field grazes along the surface ($\theta = 90^\circ$). Every mechanism of induction in this chapter — moving a magnet, moving a coil, changing a current, rotating a loop — is really just a different way of changing one or more of $B$, $A$, or $\theta$ in this single formula.

Faraday's law of induction — the central rule

Faraday distilled his experiments — magnet and coil, two coils, tapping key and iron core — into one statement: the magnitude of the induced emf in a circuit equals the time rate of change of magnetic flux through the circuit. For a single loop,

and for a tightly wound coil of $N$ turns, where the same flux $\Phi_B$ threads every turn, the emfs of all turns add:

where: $\varepsilon$ is the induced emf (volt), $N$ is the number of turns, $\Phi_B$ is the flux through one turn, and the minus sign encodes Lenz's law — the direction of the induced emf opposes the change producing it (discussed fully below). This is the whole chapter compressed into one line: no matter how the flux is made to change — sliding a magnet in, sliding a coil out, tapping a key to ramp a neighbouring current, or spinning a loop — the induced emf depends only on $d\Phi_B/dt$, the rate of change, never on the flux itself. A large steady flux induces nothing; a small but rapidly changing flux induces a substantial emf. If the circuit is closed with resistance $R$, an induced current $I = \varepsilon/R$ flows; if the circuit is open, the same $\varepsilon$ appears across the open ends without any current.

Lenz's law and the conservation of energy

Faraday's law gives the magnitude; Lenz's law, proposed by Heinrich Lenz in 1834, fixes the direction hidden in the minus sign: 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. Picture the north pole of a bar magnet being pushed toward a coil. The flux through the coil is increasing, so the induced current must flow in the sense that makes the coil's own magnetic moment point toward the approaching pole — a repulsion, opposing the magnet's advance. Withdraw the magnet, and the flux decreases, so the induced current reverses to create an attractive pull, again opposing the change — this time opposing the decrease. Lenz's law is not an independent law of nature; it is the conservation of energy wearing a magnetic disguise. Suppose the induced current instead assisted the magnet's motion. The magnet would then accelerate on its own, gaining kinetic energy with no external work done — a perpetual-motion machine, forbidden outright. Because the induced current always opposes the motion that creates it, an external agent must do positive work to keep the magnet moving, and that work is exactly what appears as electrical energy delivered to the circuit (ultimately dissipated as heat in its resistance). Every induced effect in this chapter, without exception, respects this bookkeeping.

Motional emf — a moving conductor generates its own emf

A different-looking but equivalent route to induction is to keep the field steady and move the conductor. Consider a rod PQ of length $l$, part of a closed rectangular circuit PQRS, sliding with constant velocity $v$ inside a uniform field $B$ perpendicular to the plane of the loop. As PQ slides, the enclosed area — and therefore the flux $\Phi_B = Blx$ — changes purely because the geometry changes, with $B$ itself never varying in time. Applying Faraday's law to this changing area gives the motional emf:

where: $B$ is the field strength, $l$ is the length of the rod between the rails, and $v$ is the rod's speed perpendicular to both the rod and the field. This single formula recurs throughout the chapter and its applications: a rotating rod, a falling wire, and the AC generator's coil are all disguised versions of Equation (4). The full derivation, along with the independent Lorentz-force route to the same result, is worked out below.

Inductance — flux linked to current, not to an external source

Where Faraday's law connects flux change to emf, inductance connects flux to current itself, independent of how that current arose. If a coil's geometry is fixed, its flux linkage $N\Phi_B$ is directly proportional to the current producing it, and the constant of proportionality is called inductance — the electromagnetic analogue of capacitance, which depended on geometry and the medium, not on the charge or voltage applied.

Mutual inductance describes one coil's flux linkage due to a neighbouring coil's current: $N_1\Phi_1 = M_{12} I_2$, where $M_{12}$ depends only on the geometry and relative placement of the two coils. Remarkably, it does not matter which coil is called the source: $M_{12} = M_{21} = M$, a reciprocity proven in the second derivation below. When $I_2$ varies with time, the induced emf in coil 1 is $\varepsilon_1 = -M\,dI_2/dt$ — one coil's changing current speaks to its neighbour purely through the shared flux.

Self-inductance is the same idea applied to a single, isolated coil and its own current: $N\Phi_B = LI$, so that when $I$ itself changes, the coil induces an emf in itself, $\varepsilon = -L\,dI/dt$ — this is the back emf, always opposing the change (growth or decay) of its own current. Self-inductance is the electrical inertia of a circuit, exactly as mass is mechanical inertia; a coil resists any change in current, not the current itself. The SI unit of both mutual and self inductance is the henry (H), with dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$, obtained from flux divided by current.

The AC generator — Faraday's law turned into a machine

The most consequential engineering application of this chapter is the AC generator (alternator), which converts mechanical rotation into electrical energy. A coil (the armature) of $N$ turns and area $A$ is mechanically rotated at constant angular speed $\omega$ inside a uniform magnetic field $B$, with its axis of rotation perpendicular to $B$. At time $t$, the angle between the field and the coil's area vector is $\theta = \omega t$ (taking $\theta = 0$ at $t=0$), so the flux is $\Phi_B = BA\cos\omega t$. Differentiating and applying Faraday's law for $N$ turns gives the instantaneous emf:

where: $\varepsilon_0$ is the peak (maximum) emf, reached whenever the coil plane lies along the field ($\sin\omega t = \pm 1$); $N$, $B$, $A$, $\omega$ are as before. The emf is zero when the coil plane is perpendicular to the field (flux at its extreme, but momentarily not changing) and maximum when the coil plane is parallel to the field (flux passing through zero, but changing fastest) — flux and emf are exactly a quarter-cycle out of step. Since $\omega = 2\pi\nu$, the same result is often written $\varepsilon = \varepsilon_0\sin(2\pi\nu t)$, with $\nu$ the rotation frequency ($50\ \text{Hz}$ in India). Real generators keep the coils fixed and rotate the electromagnets instead, but the physics — and Equation (5) — is identical.

Board level - square loop losing flux in a uniform field (NCERT Example 6.2 pattern)

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, making an angle of $45^\circ$ with the plane of the loop's area vector. The field is decreased steadily to zero in $0.70\ \text{s}$. Determine the induced emf and induced current during this interval.

Step 1. Find the initial flux. The area is $A = (0.10\ \text{m})^2 = 10^{-2}\ \text{m}^2$, and the angle between $B$ and the area vector is $\theta = 45^\circ$. By Equation (1), $\Phi_{\text{initial}} = BA\cos\theta = (0.10)(10^{-2})(\cos 45^\circ) = 7.07\times 10^{-4}\ \text{Wb}$.

Step 2. Find the change in flux and the induced emf. The final flux is zero, so $\Delta\Phi_B = 7.07\times 10^{-4}\ \text{Wb}$ over $\Delta t = 0.70\ \text{s}$. By Equation (2), $\varepsilon = \Delta\Phi_B/\Delta t = 1.0\times10^{-3}\ \text{V} = 1.0\ \text{mV}$.

Step 3. Find the induced current: $I = \varepsilon/R = (1.0\times10^{-3}\ \text{V})/(0.5\ \Omega) = 2\ \text{mA}$. Answer: induced emf $\varepsilon = 1.0\ \text{mV}$, induced current $I = 2\ \text{mA}$, both present only while the field is changing; the steady Earth's field also threads the loop but, being unchanging, induces nothing.

Competitive level (JEE / NEET) - the rotating rod generator (NCERT Example 6.6 pattern)

A metallic rod of length $1.0\ \text{m}$ is rotated with a frequency of $50\ \text{rev/s}$, one end hinged at the centre and the other sliding on a circular metallic ring of radius $1.0\ \text{m}$, the axis of rotation being perpendicular to the plane of the ring. A uniform magnetic field of $1.0\ \text{T}$, parallel to the axis, fills the region. Find the emf induced between the centre and the ring.

Step 1. Treat the rod as infinitesimal segments $dr$ at radius $r$, each moving with its own local speed $v = \omega r$ (unlike a straight sliding rod, this is not constant along the length). Each element contributes $d\varepsilon = Bv\,dr = B\omega r\,dr$, the motional-emf law from Equation (4) applied locally.

Step 2. Integrate from the hinge ($r=0$) to the ring ($r=R$): $\varepsilon = \displaystyle\int_0^R B\omega r\,dr = \dfrac{B\omega R^2}{2}$. This is the key extension of the sliding-rod formula: every element has its own $v=\omega r$, and integrating $Bv\,dr$ replaces simply multiplying $Blv$.

Step 3. Substitute: $\omega = 2\pi\nu = 2\pi(50) = 314\ \text{rad/s}$, $B = 1.0\ \text{T}$, $R = 1.0\ \text{m}$, giving $\varepsilon = \dfrac{(1.0)(314)(1.0)^2}{2} = 157\ \text{V}$.

Step 4. Cross-check with the area-sweep method: the rod sweeps a sector of area $\tfrac12 R^2\theta$, so $d(\text{area})/dt = \tfrac12 R^2\omega$ and $\varepsilon = B\,d(\text{area})/dt = \tfrac12 B\omega R^2$ — identical to Step 2. Answer: the induced emf between the centre and the ring is $157\ \text{V}$. Timing note: this pattern tempts a direct (wrong) substitution of $v=\omega R$ into $\varepsilon = Blv$ with the full length $l=R$, overcounting by a factor of $2$ — that shortcut assumes every point moves at the rim speed, which only the tip does. Recognise "rotating rod" as an integration problem, and the factor $\tfrac12$ falls out automatically. Under two minutes once $\int_0^R \omega r\,dr = R^2/2$ is memorised.

Competitive level (JEE / NEET) - self-inductance from a measured emf, then energy stored (NCERT-style, extending Exercises 6.6/6.9)

Current in a coil falls steadily from $5.0\ \text{A}$ to $0.0\ \text{A}$ in $0.10\ \text{s}$, and the average induced emf measured across the coil is $200\ \text{V}$. (a) Estimate the self-inductance of the coil. (b) Find the magnetic energy stored in the coil when it was carrying the original $5.0\ \text{A}$.

Step 1. Apply the self-induction law with average rates: $\varepsilon = -L\,dI/dt$ over a finite interval with roughly constant $dI/dt$ becomes $|\varepsilon| \approx L\,|\Delta I|/\Delta t$.

Step 2. Solve for $L$: $L = \dfrac{|\varepsilon|\,\Delta t}{|\Delta I|} = \dfrac{(200\ \text{V})(0.10\ \text{s})}{5.0\ \text{A}} = 4.0\ \text{H}$.

Step 3. Use $U_B = \tfrac{1}{2}LI^2$ with the initial current $I = 5.0\ \text{A}$, the current the coil carried just before it began to fall: $U_B = \tfrac{1}{2}(4.0)(5.0)^2 = 50\ \text{J}$. Answer: self-inductance $L = 4.0\ \text{H}$; magnetic energy stored at $5.0\ \text{A}$ is $50\ \text{J}$. Timing note: extract $L$ from an emf–di/dt measurement, then reuse it for stored energy — a standard JEE/NEET two-part pattern. A common slip is plugging the $200\ \text{V}$ emf into the energy formula instead of the $5.0\ \text{A}$ current: $U_B=\tfrac12 LI^2$ always needs current, not emf.

Confusing flux with rate of change of flux. A large, constant flux induces nothing at all; only $d\Phi_B/dt$ matters. A tiny coil in a rapidly switching field can out-induce a huge coil sitting in a strong but steady field.

Forgetting the area vector's angle. $\Phi_B = BA\cos\theta$ is routinely miscalculated by students who use $BA$ directly, ignoring that $\theta$ is measured between $B$ and the area's normal, not between $B$ and the plane of the coil (the two are $90^\circ$ apart).

Getting Lenz's law backwards on withdrawal. Students correctly reason that an approaching magnet induces a repelling pole, but then forget that a receding magnet induces an attracting pole (opposing the decrease in flux, which means pulling the magnet back, not pushing it further away).

Using $\varepsilon = Blv$ for a rotating rod without integrating. As Worked Example 2 shows, every point of a rotating rod has a different speed $v=\omega r$; only the correctly integrated $\varepsilon = \tfrac12 B\omega R^2$ is right, never $Blv$ with $v$ taken at the tip and $l$ the full length.

Mixing up mutual and self inductance in circuits with two coils. When two coils each carry their own changing current, the total induced emf in one coil has two terms — a self term $-L\,dI_1/dt$ and a mutual term $-M\,dI_2/dt$ — and dropping either term is a common exam error.

Treating the induced emf's sign as arbitrary. The minus sign in Faraday's law is not decorative; it is Lenz's law itself, and any answer that predicts an induced current assisting the change that caused it is guaranteed wrong by conservation of energy, without needing to check the geometry at all.

• State Faraday's law of induction, $\varepsilon = -N\,d\Phi_B/dt$, and explain in words why only a changing flux (not flux itself) induces an emf.
• Compute magnetic flux $\Phi_B = BA\cos\theta$ for a plane surface, correctly identifying the angle between the field and the area vector.
• Predict the direction of an induced current in any given scenario using Lenz's law, and justify that direction by appeal to conservation of energy.
• Derive the motional emf $\varepsilon = Blv$ for a sliding rod both from the flux rule and from the Lorentz force on free charges, and state which of the two extends more naturally to a stationary conductor in a changing field.
• Set up and evaluate the integral for the emf of a rotating rod, $\varepsilon = \tfrac12 B\omega R^2$, without falling back on the flat $Blv$ formula.
• Define mutual inductance and self-inductance, state their SI unit (henry), and write the reciprocity relation $M_{12}=M_{21}$.
• Calculate the self-inductance of a long solenoid, $L = \mu_0 n^2 Al$, and the magnetic energy it stores, $U_B = \tfrac12 LI^2$.
• Derive the instantaneous emf of an AC generator, $\varepsilon = \varepsilon_0\sin\omega t$ with $\varepsilon_0 = NBA\omega$, from first principles using Faraday's law applied to a rotating coil.
• Distinguish clearly between a changing magnetic flux (which induces emf) and a changing electric flux through the same loop (which, in this chapter's scope, does not).

Ask the tutor: Want to work through a rotating-loop or LR-circuit numerical step by step, or see how this chapter's motional emf sets up the alternating current studied next?