Peak EMF Of The G string

Consider a small element $dx$ at a distance of $x$ from the left end of the string.
Given that,

$y = a\cos(2\pi ft)\sin(\frac{\pi x}{L})$

Let $2\pi f = w$ and $\frac{\pi}{L} = k$

$y = a\cos(wt)\sin(kx)$

Differentiating wrt $t$ ,

$v_x = -aw\sin(wt)\sin(kx)$

Now, consider a charge $q$ that is situated at the position $x$ . It is experiencing $2$ forces : Electric force and Magnetic force.
For the charge to be in equilibrium, the two forces must balance each other.

$F_e = F_M$

$\Rightarrow qE = Bv_xq$

$\Rightarrow E = Bv_x$

$\Rightarrow E = -Baw\sin(wt)\sin(kx)$

Thus, the total EMF induced across the ends of the string is,

$V = \int_0^LE\cdot dx$

$V = \int_0^L -Baw\sin(wt)\sin(kx)dx$

$\Rightarrow V = -Baw\sin(wt)\int_0^L\sin(kx)dx = -Baw\sin(wt)\left(\frac{-2L}{\pi}\right)$

$\Rightarrow |V| = \frac{2BawL}{\pi}\sin(wt)$

Thus,

$V_{max} = \frac{2BawL}{\pi} = \boxed{1.02V}$

This can be modeled as a "rail problem", that is, we've to find the emf across a rod moving in a magnetic field.

Divide up the string into lots of tiny elements of lengths $dx$ . This element vibrates in a plane perpendicular to the direction of the magnetic field. If it's moving with a speed $v$ , then the emf across it is $dV = Bvdx$ .

The speed of an element at a distance $x$ from the left end is given by: $\displaystyle v = \dfrac{ \partial y } { \partial t } = -2\pi f a \sin(2\pi f t) \sin( \dfrac{\pi x}{L} )$

Thus, the "net" emf across the string can be obtained by integrating the emf's across the elements.

$\begin{aligned} \displaystyle V &= \int_0^L dV \\ &= \int_0^L B(-2\pi f a \sin(2\pi f t) \sin( \dfrac{\pi x}{L} )) dx \\ &= -4BLfa\sin(2\pi f t) \end{aligned}$

This obviously varies with time, and the peak occurs only at certain instants. The peak value being: $4BLfa = 1.0192 V$ .

Using Faraday's Law of Electromagnetic Induction, we can calculate the induced emf by knowing how magnetic flux changes with respect to time. Magnetic flux is defined as $\Phi_{B}=\oint{\vec{B} \bullet d\vec{A}}$ Since the area vector is always parallel to the magnetic field, we may rewrite this as $\Phi_{B}=BA$ . The magnetic field is always constant, but the area is variable: $A = \int^{L}_{x=0} y(t, x) \, dx = \int^{L}_{x=0} a\, \cos{(2\pi f t)} \sin{\frac{\pi x}{L}}\, dx = \frac{2aL}{\pi}\cos{2\pi f t}$ emf is related to magnetic flux in the following way $\varepsilon = \frac{-d\Phi_{B}}{dt}$ Using the area previously calculated area we obtain: $\varepsilon = \frac{-d}{dt}\left( \frac{2aBL}{\pi}\cos{2\pi f t}\right) = 4aBLf\sin{2\pi f t}$ $\varepsilon$ is maximized when $\sin{2\pi f t}$ is equal to 1. Therefore $\varepsilon_{max} = 4aBLf = 1.02 V$