Your camping buddy has an idea for a light to go inside your tent. He happens to have a powerful and heavy horseshoe magnet that he bought at a surplus store. This magnet creates a $0.20T$field between two pole tips $10cm$apart. His idea is to build the hand-cranked generator shown in FIGURE .He thinks you can make enough current to fully light a $1.0\Omega$lightbulb rated at $4.0W$. That’s not super bright, but it should be plenty of light for routine activities in the tent.

a. Find an expression for the induced current as a function of time if you turn the crank at frequency $f$. Assume that the semicircle is at its highest point at $t=0s$.

b. With what frequency will you have to turn the crank for the maximum current to fully light the bulb? Is this feasible?

The magnetic flux $\Phi$is the amount of magnetic field that passes through a loop of area $A$. The magnetic flux is provided by when the magnetic field is parallel to the plane's normal.

${\Phi }_m=BA$

The magnetic flux, however, will be supplied by when the magnetic field makes an angle with the plane.

${\Phi }_{\mathrm{m}}=BA\cos \omega t=BA\cos \left(2\pi ft\right)$

Where $f$is the rotation's frequency. The semicircle has a radius of $r=5\mathrm{cm}$. As a result, the semicircle's area is

$A=\left(\pi r^2\right)/2=\pi (0.05\mathrm{m}{)}^2/2=3.9\times {10}^{-3}{\mathrm{m}}^2$

The induced emf is the change in magnetic flux inside the loop, as defined by Faraday's law, and it is given by an equation in the form

$\epsilon =-\frac{d{\Phi }_{\mathrm{m}}}{dt}$

$=-\frac{dBA\cos \left(2\pi ft\right)}{dt}$

$=-BA\frac{d\cos \left(2\pi ft\right)}{dt}$

$=2\pi fBA\left(\sin 2\pi ft\right)$

Ohm's law is used to compute the induced current through the coil, as indicated in the following equation.

$I_{\text{induced}}=\frac{\epsilon }{R}=\frac{2\pi fBA\left(\sin 2\pi ft\right)}{R}$

To get $I$text induced in terms of $t$and $f$, we plug the values for$B,R,andA$into equation

$I_{\text{induced}}=\frac{2\pi fBA\left(\sin 2\pi ft\right)}{R}$

$=\frac{2\pi (0.20\mathrm{T})\left(3.9\times {10}^{-3}\mathrm{m}\right)f\sin \left(2\pi ft\right)}{1\Omega }$

$=\left(4.9\times {10}^{-3}\right)f\sin \left(2\pi ft\right)$

We have the bulb's power and resistance, so we can use this information to calculate the bulb's induced current or maximum current.

From the relation $\left(P=I^{2}R\right)$

$I_{\text{induced}}=I_{\max }=\sqrt{\frac{P}{R}}=\sqrt{\frac{4\mathrm{W}}{1\Omega }}=2\mathrm{A}$

The maximum current will be induced when the semicircle is perpendicular to the magnetic field, so the term localid="1648963189984" $\sin \left(2\pi ft\right)=\sin 90°=1$.

As a consequence of our calculations, the maximum current will be

$I_{\max }=\left(4.9\times {10}^{-3}\right)f$

Fill in the value of $Imax$in this equation for $f$

$f=\frac{I_{\max }}{4.9\times {10}^{-3}}=\frac{2\mathrm{A}}{4.9\times {10}^{-3}}=408\mathrm{Hz}$