Magnetic Flux & Faraday's Law of Induction

Magnetic Flux: Counting Field Through a Surface

Faraday's genius was finding the one quantity whose change explains every observation: magnetic flux.

For a plane of area $A$ in a uniform field $\vec{B}$:

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

where $\theta$ is the angle between $\vec{B}$ and the area vector $\vec{A}$ (normal to the surface - recall Chapter 1).

For non-uniform fields or curved surfaces, sum over area elements:

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

Essentials:

• Flux is a scalar.
• SI unit: weber (Wb) = T m$^2$. (And from Faraday's law below, 1 Wb = 1 V s.)
• $\theta = 0$: maximum flux $BA$ (field perpendicular to the surface, i.e. along the normal). $\theta = 90^{\circ}$: zero flux (field in the plane of the surface).

Key Point: The classic trap - when the coil's plane is parallel to $\vec{B}$, the area vector is perpendicular to $\vec{B}$ and the flux is zero. Always work with the angle to the normal.

Faraday's Law of Induction

The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

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

The negative sign indicates the direction of the emf and current - that's Lenz's law, coming in Section 3.

For a closely wound coil of N turns, the flux change through each turn is the same, so the emfs add:

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

This is why practical coils have many turns - the induced emf is multiplied by N.

The law instantly explains all three experiments of Section 1: the moving magnet (Experiment 1) and moving coil (Experiment 2) change $\Phi_B$ through C$_1$ by changing B at the coil; the tapping key (Experiment 3) changes B from zero to maximum and back. In every case, emf appears only while $\Phi_B$ is changing.

Three Handles on the Flux

Since $\Phi_B = BA\cos\theta$, you can induce an emf by changing any one (or more) of the three factors:

1. Change B — move a magnet, vary a neighbouring current (Experiments 1-3), or collapse a field with time.
2. Change A — shrink or stretch a loop in a field, or slide a rod that changes the enclosed area (motional emf, Section 4).
3. Change $\theta$ — rotate a coil in a field, so $\theta$ varies with time. This is exactly how the AC generator works (Section 7).

[JEE Tip] The induced charge that flows when the flux changes is independent of how fast you change it:

$$q = \frac{N\,\Delta\Phi_B}{R}$$

(From $I = \varepsilon/R = -\frac{N}{R}\frac{d\Phi_B}{dt}$, integrate over time.) Fast flip or slow flip, the same charge flows - only the current differs. Ballistic galvanometers exploit exactly this.

[NEET Important] A steady field through a stationary loop - however strong - induces nothing. Only $\frac{d\Phi_B}{dt}$ matters.

Solved Examples

Example 1: The collapsing north-east field

A square loop of side 10 cm and resistance 0.5 $\Omega$ 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, then decreased to zero in 0.70 s at a steady rate. Find the induced emf and current.

Solution:

1. Geometry: the loop's area vector points north (normal to the east-west vertical plane); the field points north-east, so $\theta = 45^{\circ}$.
2. Initial flux: $\Phi_0 = BA\cos 45^{\circ} = \frac{0.1 \times 10^{-2}}{\sqrt{2}} = 7.07 \times 10^{-4}$ Wb.
3. EMF: $\varepsilon = \frac{\Phi_0 - 0}{0.7} \approx 1.0 \times 10^{-3}$ V = 1.0 mV.
4. Current: $I = \varepsilon/R = 10^{-3}/0.5 = 2$ mA.
5. Note: the earth's field also threads the loop, but it is steady — it induces nothing.

Example 2: Flipping a coil in the earth's field

A circular coil of radius 10 cm, 500 turns and resistance 2 $\Omega$ has its plane perpendicular to the horizontal component of the earth's field, $B_H = 3.0 \times 10^{-5}$ T. It is rotated about its vertical diameter through 180 degrees in 0.25 s. Estimate the induced emf and current.

Solution:

1. Initial flux (per turn): $\Phi_i = B_H A = 3.0 \times 10^{-5} \times \pi \times 10^{-2} = 3\pi \times 10^{-7}$ Wb.
2. Final flux: after a 180-degree flip, $\Phi_f = -3\pi \times 10^{-7}$ Wb.
3. EMF: $\varepsilon = N\frac{\Delta\Phi}{\Delta t} = \frac{500 \times 6\pi \times 10^{-7}}{0.25} = 3.8 \times 10^{-3}$ V.
4. Current: $I = \varepsilon/R = 1.9 \times 10^{-3}$ A.
5. These are estimated (average) values — instantaneous values depend on the rotation speed at each instant.

Example 3: Flux at an angle

A plane loop of area 0.05 m$^2$ sits in a uniform field of 0.3 T, with its area vector at 60 degrees to the field. Find the flux through it.

Solution:

1. $\Phi_B = BA\cos\theta = 0.3 \times 0.05 \times \cos 60^{\circ}$.
2. $= 0.3 \times 0.05 \times 0.5 = 7.5 \times 10^{-3}$ Wb.

Example 4: The parallel-plane trap

A coil's plane is parallel to a 2 T magnetic field. What is the flux through the coil?

Solution:

1. Plane parallel to $\vec{B}$ means the area vector is perpendicular to $\vec{B}$: $\theta = 90^{\circ}$.
2. $\Phi_B = BA\cos 90^{\circ} = 0$ — zero flux, no matter how strong the field.
3. Maximum flux requires the field along the normal ($\theta = 0$), i.e. perpendicular to the coil's plane.

Example 5: EMF from a steadily dying field

A 100-turn coil of area $10^{-2}$ m$^2$ lies with its plane perpendicular to a field that falls steadily from 0.5 T to zero in 0.1 s. Find the induced emf.

Solution:

1. $\frac{dB}{dt} = \frac{0.5}{0.1} = 5$ T/s.
2. $\varepsilon = N A \frac{dB}{dt} = 100 \times 10^{-2} \times 5$.
3. Answer: $\varepsilon = 5$ V (while the field is collapsing).

Example 6: Time-dependent flux (JEE pattern)

The flux through a loop varies as $\Phi_B = (3t^2 + 4t + 5)$ mWb. Find the magnitude of the induced emf at $t = 2$ s.

Solution:

1. $\varepsilon = \left|\frac{d\Phi_B}{dt}\right| = (6t + 4)$ mV.
2. At $t = 2$ s: $\varepsilon = 6(2) + 4 = 16$ mV.
3. Note: emf depends on the rate of change — the constant 5 mWb contributes nothing.

Example 7: Why N turns help

A single loop develops 2 mV when a flux change occurs. What emf develops in a closely wound 250-turn coil for the same flux change per turn?

Solution:

1. Each turn links the same changing flux, and the turn emfs add in series.
2. $\varepsilon = N \times 2$ mV $= 250 \times 2 = 500$ mV $= 0.5$ V.
3. This is why Faraday wound coils with hundreds of turns — and why $\varepsilon = -N\,d\Phi_B/dt$.

Example 8: Weber, volt-second and tesla

Show that 1 weber = 1 volt-second, and express the weber in terms of tesla.

Solution:

1. From Faraday's law $\varepsilon = d\Phi_B/dt$: V = Wb/s, so 1 Wb = 1 V s.
2. From $\Phi_B = BA$: 1 Wb = 1 T m$^2$.
3. Both forms appear in exams; they are the same unit seen through two laws.

Example 9: Average vs instantaneous emf

In the flipped-coil example above, why are the computed emf and current called 'estimated' values?

Solution:

1. We used $\varepsilon = N\Delta\Phi/\Delta t$ — the average emf over the quarter second.
2. The instantaneous emf $-N\,d\Phi_B/dt$ depends on the rotation speed at each instant, which varies unless the rotation is uniform.
3. Average values answer 'how much overall'; instantaneous values answer 'how much right now'. Know which one a question wants!

Example 10: Induced charge is rate-independent (JEE pattern)

A 100-turn coil of area $10^{-2}$ m$^2$ and total circuit resistance 2 $\Omega$ is flipped through 180 degrees in a field B = 0.01 T (initially along the normal). How much charge flows, and does flipping faster change it?

Solution:

1. Flux change per turn: $\Delta\Phi = 2BA = 2 \times 0.01 \times 10^{-2} = 2 \times 10^{-4}$ Wb.
2. Charge: $q = \frac{N\Delta\Phi}{R} = \frac{100 \times 2 \times 10^{-4}}{2} = 10^{-2}$ C.
3. Flipping faster raises the emf and current but shortens the time — the charge is unchanged: $q$ depends only on the total flux change, not the rate.