Ranking the rate of change of flux

Problem

Four identical loops undergo flux changes: (A) $\Phi$ goes from $0$ to $0{,}5 \ \mathrm{Wb}$ in $0{,}1 \ \mathrm{s}$; (B) $\Phi$ goes from $0{,}3$ to $0 \ \mathrm{Wb}$ in $0{,}05 \ \mathrm{s}$; (C) $\Phi$ stays at $0{,}4 \ \mathrm{Wb}$ for $2 \ \mathrm{s}$; (D) $\Phi$ goes from $-0{,}2$ to $+0{,}2 \ \mathrm{Wb}$ in $0{,}4 \ \mathrm{s}$. Rank the magnitudes of the EMF from smallest to largest.

Solution

The EMF depends only on the rate of change of the flux, not on its value: $\abs{\mathcal{E}} = \abs{\frac{\Delta \Phi}{\Delta t}}$

Case-by-case calculation: $\begin{aligned} \text{A:} \ \abs{\mathcal{E}} = \frac{0{,}5}{0{,}1} = 5 \ \mathrm{V} \qquad \text{B:} \ \abs{\mathcal{E}} = \frac{0{,}3}{0{,}05} = 6 \ \mathrm{V} \\ \text{C:} \ \abs{\mathcal{E}} = \frac{0}{2} = 0 \ \mathrm{V} \qquad \text{D:} \ \abs{\mathcal{E}} = \frac{0{,}4}{0{,}4} = 1 \ \mathrm{V} \end{aligned}$

In C the flux is large but constant: zero derivative, zero EMF. In D what matters is the total amplitude of the change, $\abs{+0{,}2 - (-0{,}2)} = 0{,}4 \ \mathrm{Wb}$.

Order from smallest to largest: $\ev{\text{C} < \text{D} < \text{A} < \text{B}}$

Collegamenti

Argomenti: Electromagnetic induction Concetti: Faraday-Neumann-Lenz law Competenze: Ranking reasoning · Flux balance