Self-Induction — Question 8

Question 8

A zero-resistance isolated loop preserves linkage while its inductance changes

An isolated ideal loop has zero resistance and no changing external flux. Its inductance changes quasistatically from L0 to 2L0. The initial current is I0. Use the zero-emf condition $d(Li)/dt=0$.

1. Find current as a function of instantaneous L.

2. For $L=L_0(1+\alpha t)$, find di/dt and verify that the two linkage-derivative terms cancel.

Question 8 - Solution

A zero-resistance isolated loop preserves linkage while its inductance changes

(a) Method and calculation

$$Li=L_0I_0\Rightarrow\boxed{i=L_0I_0/L}.$$$$L=2L_0\Rightarrow\boxed{i=I_0/2}.$$

(b) Interpretation and further calculation

$$i=\frac{I_0}{1+\alpha t}\Rightarrow\dot i=-\frac{\alpha I_0}{(1+\alpha t)^2}.$$$$\boxed{L\dot i+i\dot L=0}.$$

Physical check

Current itself is not conserved when the inductance varies; total linkage is the conserved quantity in the stated ideal loop.