A voice coil in an earbud speaker is a simple electromagnet constructed by winding a length of wire around a circular piece of metal that is then connected to a small battery. The current's direction is reversed continuously causing the voice coil to move up and down between 20 times per second and 20,000 times per second. The earbud speaker has a pottential difference of 0.8V and a current of 61.5mA. The wire has a resistance of 5 Ohms/m. Calculate the length of the wire. Calculate the wavelength of the sound in the air if the coil moves up and down 6000 times per second.

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Assuming that this wire is made of ohmic material then we use Ohm's law. `V=IR` .

Rearranged gives us that `R=V/I` .

From the problem, `V =0.8text(V)` and `I = 61.5times10^-3text(A)`

so, `R=V/IrArr(0.8text(V))/(61.5times10^-3text(A))~~13Omega`

Since the wire has a resitance per length,` rho=5Omega/text(m)` , then the length would be

`L=R/rhorArrL=(13Omega)/(5Omega/text(m))=2.6text(m)`

For the second part, you need to recognize that the vibration is the frequency of the sound. 6000 times per second is the same thing as 6000 Hz.

Given that `v=lambdaf` , then ` lambda=v/f` where `v` is the speed of sound in air, `lambda` is the wavelength, and `f` is the frequency of the sound.

Let `v=340 text(m)/text(s)`

So `lambda=(340 text(m)/text(s))/(6000text(Hz))~~5.7times10^-1m`

**Answer: L=2.6m, wavelength=.57m**

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