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 expressed as:

1. Frequency is proportional to the square root of tension:
   $f \propto \sqrt{T}$
2. The relationship can also be written as:
   $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
   where L is the length of the string and $\, \mu \,$ is the linear mass density.

The speed of sound (v) in the stretched string can be calculated using the formula:
$v = f \cdot \, \lambda$
For the fundamental frequency:
$v = 400 \, Hz \times 2L \quad (L=0.8 \, m)$
Thus,
$v = 400 \, Hz \times 1.6 \, m = 640 \, ms^{-1}$.

The strings in an electric guitar must be made of steel because steel is ferromagnetic. This allows the strings to interact with the magnetic field produced by the magnets in the pickups, facilitating the generation of an induced emf when the strings vibrate.

Magnetic flux (Φ) is defined as the product of the magnetic field (B) and the area (A) through which the field lines pass, given by:
$\Phi = B \cdot A$
It is measured in Weber (Wb).

The current produced in a coil of the electric guitar varies due to the amplitude of the vibrations of the string. As the string vibrates more vigorously, it changes the magnetic flux through the coil, inducing a varying emf that leads to changes in the current.

Increasing the number of turns in a coil results in a louder sound, as the induced emf becomes greater due to a higher magnetic flux linkage. The sound intensity will thus increase, enhancing the overall output of the electric guitar.

The emf (E) induced in the coil can be calculated using Faraday's law of electromagnetic induction, given by:
$E = -N \frac{d\Phi}{dt}$
For 5000 turns and a change in magnetic flux of $8 \times 10^{-4} \,Wb$, over a time of 0.1 s:
$E = -5000 \cdot \frac{8 \times 10^{-4}}{0.1} = -40 \, V$
Hence, the induced emf is 40 V.