Answer:
[tex]25.39Hz[/tex]
Explanation:
#First, we need to determine the actual emf required. The generator's internal resistance will cause a voltage drop inside the generator/
Internal resistance is defined using the formula:
[tex]R=\rho L/A\\\rho=1.68\times10^-^8 \Omega m\\L=0.050m\times4\times 60=12.0m\\A=\pi r^2=\pi d^2/4=\pi(0.00059m)^2/4=2.734\times10^-^7m^2\\\\R=(1.68\times10^-^8 \Omega \ m\times 12.0m)/2.734\times10^-^7m^2\\=0.7374\ \Omega[/tex]
#The bulb is rated 12.0V,25.0W
Current, [tex]I=25.0W/12.0V=2.083A[/tex]
Therefore, the voltage drop in the generator is calculated as:
[tex]2.083A\times0.7374\Omega=1.5360V[/tex]
Actual EMF required is thus 1.536V+12.0V=13.536V
#peak voltage is [tex]13.536V\sqrt 2=19.143V[/tex]
#For a generator, by Faraday's Law
[tex]E_m_a_x=NBA\ \omega\\19.143=60\times 0.800T\times (0.05m)^2\ \omega\\\\\omega=159.525rad/s[/tex]
[tex]f=\omega/2\pi\\=159.525/2\pi=25.39Hz[/tex]
#The rate of the generator is 25.39Hz
