A long, straight solenoid with a cross-sectional area of 8.00 cm^2 is wound with 90 turns of wi

1. First, we need to find the initial magnetic field inside the solenoid when the current is on.

A long, straight solenoid with a cross-sectional area of...

A long, straight solenoid with a cross-sectional area of $8.00 \mathrm{~cm}^{2}$ is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in $0.0400 \mathrm{~s}$. What is the average induced emf in the second winding?

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Key Concepts

Magnetic Field in a Solenoid

In a long solenoid, the magnetic field inside is uniform and is calculated using the formula B = ??nI, where ?? is the permeability of free space, n is the number of turns per unit length, and I is the current. This concept is critical because it allows one to determine the magnetic field produced by the solenoid based on its physical windings and current.

Magnetic Flux

Magnetic flux is the measure of the quantity of magnetic field passing through a given area and is defined as ? = B·A when the field is perpendicular to the surface. This concept connects the magnetic field produced by the solenoid with the area of the loop that encircles it, quantifying how much of the field 'links' with that loop.

Faraday’s Law of Electromagnetic Induction

Faraday's law states that a changing magnetic flux through a circuit induces an electromotive force (emf) in that circuit, given by emf = -N (??/?t), where N is the number of turns in the circuit and ??/?t is the rate of change of magnetic flux. This principle is central to understanding how turning off the solenoid current produces an induced emf in an adjacent winding.

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