Chapter 21, Problem 4Q

To determine

To Explain: The direction of the induced current in each loop.

Answer to Problem 4Q

The direction of an induced current will be clockwise.

Explanation of Solution

Introduction:Faraday’s law of induction is the fundamental law of electromagnetism. According to this law, the circuit contains $N$ closely wrapped loops, emfs induced in every particular loop add organized. So, $EMF=-N\frac{\Delta {\varphi }_B}{\Delta t}$ In this equation the minus sign represents direction the induced EMF acts.

According to the Faraday’s, rate of change magnetic flux will produce an induced current. As per Lenz’s law, the induced B field in a loop of wire will oppose the change in the magnetic flux through the loop.

For a ring on left-side of current-carrying wire, there will be no (zero) induced current. Reason for this: as ringmoves parallel to wire, magnetic flux through a ring does not vary. This means that there is no (zero) induced emf and also, no (zero) induced current.

For a ring on right side of current-carryingwire, direction of an induced current will be clockwise. Reason for this: as ring moves away from wire, magnetic fluxthrough ring remains decreasing (here are less magnetic field lines pointing into loop). In anattempt to oppose certaindecrement (due to Lenz’s law), EMF and current will be induced in a ring inclockwise direction (by using Right-Hand Rule).

Conclusion: There will bezero induced current for a ring on left-side of current-carrying wire.