Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 11 - 2006

The relationship between frequency (f) and tension (T) for a stretched string is given by the formula:

$f ext{ } ext{proportional to} ext{ } \sqrt{T}$

This indicates that frequency increases as tension increases, assuming other factors remain constant.

To find the speed of the sound wave (v) in the stretched string, we use the formula:

$v = f \times \lambda$

Since the fundamental frequency corresponds to a wavelength (\lambda) equal to twice the length (L) of the string,

the wavelength is ( \lambda = 2L = 2 \times 0.8 ext{ m} = 1.6 ext{ m} ). Thus,

$v = 400 \text{ Hz} \times 1.6 ext{ m} = 640 ext{ m/s}$.

The strings of an electric guitar must be made of steel to ensure they can interact effectively with the magnetic field produced by the pickups. Steel is a ferromagnetic material, which allows the strings to induce a varying current in the coil when they vibrate, thus generating the electric signal.

Magnetic flux (Φ) is defined as the product of the magnetic field (B) and the area (A) through which the field lines pass, represented mathematically as:

$\Phi = B \cdot A$

Where the unit of magnetic flux is the Weber (Wb).

The current produced in the coil of the electric guitar varies due to changes in the magnetic flux cutting through the coil as the strings vibrate. This variation in magnetic field induces an electromotive force (emf) according to Faraday's law of electromagnetic induction.

When the number of turns in a coil is increased, the induced emf becomes greater, resulting in a louder sound produced. This can be represented as:

Louder sound / greater (sound intensity) greater amplitude.

The emf (E) induced in a coil can be calculated using Faraday's law:

$E = -N \frac{\Delta \Phi}{\Delta t}$

Where N is the number of turns, and (\Delta \Phi) is the change in magnetic flux. Substituting the values:

$E = -5000 \times \frac{8 \times 10^{-4} ext{ Wb}}{0.1 ext{ s}} = 40 ext{ V}$.