Faraday's Law of Induction — Question 10

Question 10

Time-dependent resistance removes the endpoint-only rule for transported charge

A loop has linkage $\Lambda(t)=\Lambda_0+ct$ and resistance $R(t)=R_0(1+t/T)$ during $0<t<T$, with c positive. Ignore self-inductance.

1. Find emf, current and total signed charge.

2. Compare with a constant R0 loop undergoing the same linkage change and explain why the charges differ.

Question 10 - Solution

Time-dependent resistance removes the endpoint-only rule for transported charge

(a) Method and calculation

$$\mathcal E=-c,\quad i(t)=-\frac c{R_0(1+t/T)}.$$$$Q=-\frac{cT}{R_0}\int_1^2\frac{du}{u}=\boxed{-\frac{cT}{R_0}\ln2}.$$

(b) Interpretation and further calculation

$$Q_{R_0}=-cT/R_0.$$$$\boxed{Q/Q_{R_0}=\ln2<1}.$$

Physical check

Endpoint linkage alone fixes the voltage-time integral. It fixes charge only when resistance can be treated as constant.