From Flux to Induced EMF
From the experiments of Faraday and Henry, a single common thread emerged. The motion of a magnet towards or away from a coil, the motion of a current-carrying coil near another, and even the making or breaking of a steady current in a neighbouring coil all changed one quantity: the magnetic flux linked with the coil. As NCERT §6.4 states, Faraday concluded that an emf is induced in a coil when the magnetic flux through the coil changes with time.
Recall that magnetic flux through a plane of area $A$ in a uniform field $\mathbf{B}$ is $\Phi_B = \mathbf{B}\cdot\mathbf{A} = BA\cos\theta$, where $\theta$ is the angle between $\mathbf{B}$ and the area vector $\mathbf{A}$. Its SI unit is the weber (Wb), equal to $\text{T}\,\text{m}^2$. The NIOS lesson (§19.1.1) frames the same idea: the number of field lines threading a surface is proportional to the flux through it, and an emf appears across a loop precisely when that flux changes with time.
The crucial physical insight is that it is the time rate of change of flux — not the flux value — that drives induction. A loop sitting in an enormous but steady field carries no induced current; a loop in a weak field that changes rapidly can carry a large one.
As the N-pole approaches, the flux through the coil rises; the galvanometer deflects only while the magnet moves. Holding it still gives zero deflection — no change in flux, no emf. (NCERT Experiment 6.1.)
Statement and the Governing Equation
Faraday's law of induction is stated in NCERT §6.4 as follows: the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically,
$$\varepsilon = -\frac{d\Phi_B}{dt}$$
The negative sign indicates the direction of the induced emf (and hence the induced current in a closed loop); for magnitude problems it is dropped. The NIOS form is identical, $|\varepsilon| = \left|\dfrac{d\Phi_B}{dt}\right|$, and gives the useful unit relation $1\ \text{V} = 1\ \text{Wb s}^{-1}$.
| Quantity | Symbol | SI Unit | Defining relation |
|---|---|---|---|
| Magnetic flux | $\Phi_B$ | weber (Wb) = T m² | BA cos θ |
| Induced emf (single turn) | $\varepsilon$ | volt (V) | −dΦ/dt |
| Induced emf (N turns) | $\varepsilon$ | volt (V) | −N dΦ/dt |
| Induced current (closed loop) | $I$ | ampere (A) | ε / R |
If the circuit is closed and has resistance $R$, a current $I = \varepsilon/R$ flows. This is how almost every numerical proceeds: compute the rate of change of flux to get $\varepsilon$, then divide by $R$ for the current.
The N-Turn Factor
For a closely wound coil of $N$ turns, the same flux change links each individual turn. Because the turns are connected in series, the emfs induced in them add. NCERT (Eq. 6.4) and NIOS (Eq. 19.4) therefore give the total induced emf as
$$\varepsilon = -N\,\frac{d\Phi_B}{dt}$$
Here $\Phi_B$ is the flux linked with a single turn. Increasing the number of turns increases the induced emf proportionally — the design reason coils in generators and transformers carry hundreds or thousands of turns. The product $N\Phi_B$ is called the flux linkage.
EMF tracks the rate of change, not the flux value — and never drop the N.
Two errors recur. First, students reason "large field, therefore large emf." But $\varepsilon$ depends on $d\Phi_B/dt$; a steady flux of any magnitude induces zero emf. Second, when a problem says "coil of N turns," forgetting the factor $N$ scales the answer down by that factor — fatal when $N = 100$ or $1000$.
Checklist: (1) Is the flux actually changing? If $d\Phi_B/dt = 0$, then $\varepsilon = 0$. (2) Did you multiply by $N$? Use $\varepsilon = N\,\Delta\Phi_B/\Delta t$ for a coil.
Three Ways to Change the Flux
Since $\Phi_B = BA\cos\theta$, NCERT §6.4 notes that the flux can be varied by changing any one or more of $B$, $A$ and $\theta$. This classification tells you instantly which mechanism a question is testing.
| What changes | How it is done | Example |
|---|---|---|
| Field $B$ | Move a magnet, switch a neighbouring current on/off, ramp the field | Experiments 6.1–6.3; field decreased to zero |
| Area $A$ | Shrink, stretch, or move a loop into/out of a field region | Sliding rod (motional emf); deforming a loop |
| Angle $\theta$ | Rotate the coil so $\theta$ between $\mathbf{B}$ and $\mathbf{A}$ changes | AC generator; coil flipped through 180° |
A flux rising linearly (constant $d\Phi_B/dt$) yields a steady induced emf equal to the negative slope of the flux–time graph. The emf is the slope of $\Phi_B(t)$, not its height.
Comfortable with $\Phi_B = BA\cos\theta$ and the area-vector convention? Revise it in Magnetic Flux before pushing into rate-of-change problems.
Worked NCERT Examples
The following examples are taken from NCERT §6.4 and the NIOS lesson. Each isolates one of the three flux-change mechanisms.
A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east–west plane. A uniform field of 0.10 T is set up across the plane in the north-east direction. The field is decreased to zero in 0.70 s at a steady rate. Find the induced emf and current.
The area vector makes $\theta = 45^\circ$ with the field, so the initial flux is $\Phi = BA\cos\theta = 0.10 \times (0.1)^2 \times \cos 45^\circ = 0.70\times10^{-3}\ \text{Wb}$. Final flux is zero. Magnitude of emf: $\varepsilon = \dfrac{|\Delta\Phi|}{\Delta t} = \dfrac{0.70\times10^{-3}}{0.70} = 1.0\ \text{mV}$. Current: $I = \dfrac{\varepsilon}{R} = \dfrac{1.0\times10^{-3}}{0.5} = 2\ \text{mA}$. The earth's steady field threads the loop but, being constant in time, induces no emf.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's field ($3.0\times10^{-5}$ T). It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the emf and current.
Initial flux per turn: $\Phi_i = BA\cos 0^\circ = 3.0\times10^{-5} \times \pi(0.1)^2 = 3\pi\times10^{-7}\ \text{Wb}$. After a 180° flip, $\Phi_f = -3\pi\times10^{-7}\ \text{Wb}$, so $|\Delta\Phi| = 6\pi\times10^{-7}\ \text{Wb}$. With $N = 500$: $\varepsilon = N\dfrac{|\Delta\Phi|}{\Delta t} = \dfrac{500 \times 6\pi\times10^{-7}}{0.25} \approx 3.8\times10^{-3}\ \text{V}$, and $I = \varepsilon/R = 1.9\times10^{-3}\ \text{A}$. These are average (estimated) values.
A 75-turn circular coil of radius 35 mm has its axis parallel to a uniform field. The field changes at a constant rate from 25 mT to 50 mT in 250 ms. Find the magnitude of the induced emf.
Flux per turn is $\Phi_B = B\pi R^2$, so $\varepsilon = N\pi R^2 \dfrac{\Delta B}{\Delta t}$. Here $\Delta B/\Delta t = \dfrac{(50-25)\times10^{-3}}{0.250} = 0.10\ \text{T s}^{-1}$. Hence $|\varepsilon| = 75\pi (0.035)^2 (0.10) = 0.030\ \text{V} = 30\ \text{mV}$.
The Negative Sign and Direction
The negative sign in $\varepsilon = -d\Phi_B/dt$ is not optional decoration — it encodes the direction of the induced emf. The polarity is always such that the induced current opposes the change in flux that produced it, a statement formalised as Lenz's law and rooted in conservation of energy. For the magnitude of emf or current asked in most NEET items, the sign is dropped; for direction questions, it is everything.
Faraday's law fixes how large the induced emf is; Lenz's law fixes which way it drives current. The two are a single statement split into magnitude and direction.
To assign the direction of every induced current confidently, work through Lenz's Law, the companion that supplies the sign.
Faraday's Law in one screen
- Induced emf equals the time rate of change of flux: $\varepsilon = -\dfrac{d\Phi_B}{dt}$ (single turn).
- For a closely wound coil of $N$ turns, $\varepsilon = -N\dfrac{d\Phi_B}{dt}$; never omit the $N$.
- Flux $\Phi_B = BA\cos\theta$ can change three ways: vary $B$, vary $A$, or vary $\theta$.
- It is the rate of flux change that matters — a steady flux of any size induces zero emf.
- In a closed loop of resistance $R$, the induced current is $I = \varepsilon/R$.
- The negative sign gives direction (Lenz's law); drop it for magnitude calculations.