Faraday's Law of Induction

Statement of the law

• Faraday's law of electromagnetic induction: the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

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

• The negative sign indicates the direction (polarity) of the induced emf, and hence of the induced current in a closed loop; that is the content of Lenz's law. Always write it.
• For a closely wound coil of $N$ turns, in which the same flux change is linked with each turn,

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

• $\Phi_B$ here is the flux through ONE turn. The induced emf can therefore be increased by increasing the number of turns.

How Faraday's law explains the Faraday-Henry experiments

• Experiment 6.1: moving a magnet towards or away from coil $C_1$ changes $B$ at the coil, so $\Phi_B$ changes, so an emf appears.
• Experiment 6.2: moving current-carrying coil $C_2$ towards or away from $C_1$ changes the field of $C_2$ at $C_1$, so $\Phi_B$ changes.
• Experiment 6.3: pressing the key makes the current in $C_2$ — and so its field, and so the flux through $C_1$ — rise from zero to a maximum in a short time, giving a momentary emf. While the key is held down the current is steady, the flux is constant, and the emf drops to zero. Releasing the key makes the current fall to zero in a short time, giving another momentary emf, of the opposite sign.

The three ways to change the flux

• Since $\Phi_B = BA\cos\theta$, the flux can be changed by varying any of $B$, $A$ or $\theta$.

Induced current and induced charge

• If the circuit is closed, the induced current is

$$I = \dfrac{\varepsilon}{R} = -\dfrac{N}{R}\dfrac{d\Phi_B}{dt}$$

• The charge that flows in a finite flux change is obtained by integrating:

$$q = \int I\,dt = \dfrac{N\,\Delta\Phi_B}{R}$$

• This is worth memorising: the induced CHARGE depends only on the TOTAL change of flux and on the resistance, never on how quickly the change happens. The induced emf and current, by contrast, do depend on the time taken.
• On an $I$ against $t$ graph, $q$ is the area under the curve. CBSE has used exactly this: a current falling linearly from $0.4\ \text{A}$ at $t=0$ to zero at $t=1.0\ \text{s}$ gives $q = \dfrac{1}{2}\times 0.4\times 1.0 = 0.2\ \text{C}$, and then $\Delta\Phi = qR$.

Worked patterns CBSE sets on this section

• Flux given as a function of time. If $\phi = 5t^2+3t+16$ Wb, then

$$\varepsilon = -\dfrac{d\phi}{dt} = -(10t+3)$$

so at $t = 4\ \text{s}$, $\varepsilon = -(40+3) = -43\ \text{V}$.

• Field changing at a steady rate. A square coil of side $10\ \text{cm}$ with $100$ turns is placed perpendicular to a field increasing at $1\ \text{T s}^{-1}$. Then $A = (0.1)^2 = 0.01\ \text{m}^2$ and

$$\varepsilon = NA\dfrac{dB}{dt} = 100\times 0.01\times 1 = 1.0\ \text{V}$$

• Field switched off. A coil of area $100\ \text{cm}^2$ at $30^{\circ}$ to a field of $10^{-1}\ \text{T}$ which is reduced to zero in $10^{-4}\ \text{s}$:

$$\lvert\varepsilon\rvert = A\cos 30^{\circ}\left\lvert\dfrac{\Delta B}{\Delta t}\right\rvert = 100\times 10^{-4}\times\dfrac{\sqrt{3}}{2}\times\dfrac{10^{-1}}{10^{-4}} = 5\sqrt{3}\ \text{V}$$

• Field varying in a piecewise-linear way (a $B$-$t$ graph). Compute $\varepsilon = -A\dfrac{\Delta B}{\Delta t}$ segment by segment: a rising ramp gives a constant negative emf, a flat section gives zero, a falling ramp gives a constant positive emf of the same size. For a loop of radius $12\ \text{cm}$ and $R = 8.5\ \Omega$ with $B$ going $0\to1\ \text{T}$ in $2\ \text{s}$: $A = 4.5\times10^{-2}\ \text{m}^2$, $\varepsilon = 2.25\times10^{-2}\ \text{V}$, $I = 2.6\ \text{mA}$, and the $I$-$t$ graph is three flat steps: $-2.6\ \text{mA}$, $0$, $+2.6\ \text{mA}$.
• Rotating a coil through $180^{\circ}$ (NCERT Example 6.3). The flux reverses, so $\Delta\Phi_B = 2BA$ and $\varepsilon = N\dfrac{2BA}{\Delta t}$.

Standard results

• $\varepsilon = -N\dfrac{d\Phi_B}{dt}$, in volt; $[\varepsilon] = [\text{M L}^2\text{T}^{-3}\text{A}^{-1}]$.
• $I = \varepsilon/R$; $q = N\Delta\Phi_B/R$, independent of time.
• Reversing a coil through $180^{\circ}$: $\Delta\Phi_B = 2NBA$.
• Coil perpendicular to a changing field: $\varepsilon = NA\dfrac{dB}{dt}$.
• No flux change means no emf, whatever the strength of the field.