Source: High school physics (Chinese)
Problem Sets:
Problem
Wires $AB$ and $CD$ are parallel to each other, as shown. Wire $AB$ contains a battery and a switch $S$; wire $CD$ contains a galvanometer $G$. Initially the switch is open.
- Determine the direction of the induced current in wire $CD$ at the instant the switch $S$ is closed.
- Determine the direction of the induced current in wire $CD$ at the instant the switch $S$ is opened.
Closing $S$: induced current flows $D \to C$ (opposite to the current in $AB$). Opening $S$: induced current flows $C \to D$ (same direction as the current in $AB$).
The current in $AB$ changes only while the switch is operated, so an induced current appears in $CD$ only at those instants (mutual induction, Lenz's law). Take the battery to drive the current in $AB$ from $A$ to $B$.
Closing $S$: the current in $AB$ rises from zero, so the magnetic flux through the $CD$ circuit increases. By Lenz's law the induced current opposes this increase, flowing opposite to the current in $AB$, i.e. from $D$ to $C$.
Opening $S$: the current in $AB$ falls to zero, so the flux through $CD$ decreases. The induced current opposes the decrease and flows in the same direction as the current in $AB$, i.e. from $C$ to $D$.