Respuesta :
Answer:
[tex]P_{avg} = 2.513\times 10^{- 9}\ W[/tex]
Solution:
As per the question:
Current, I(t) = 5.55 A
Frequency, f = 359 Hz
B = 72.2 cm
c = 32.5 cm
a = 80.2 cm
Resistance, R = 54.3 [tex]\Omega [/tex]
Now,
Th magnetic flux of the loop is given by:
[tex]\phi = \frac{\mu_{o}cI}{2\pi }ln(\frac{a + b}{a})[/tex]
Now,
I(t) = [tex]I_{o}sin\omega t[/tex]
Now, the magnitude of the induced emf is given by:
[tex]e = \frac{d\phi}{dt}[/tex]
[tex]e = \frac{d}{dt}(\frac{\mu_{o}c\times I_{o}sin\omega t}{2\pi }ln(\frac{a + b}{a}))[/tex]
[tex]e = \frac{\mu_{o}c\times I_{o}.ln(\frac{a + b}{a}))\frac{d}{dt}(sin\omega t}{2\pi }[/tex]
[tex]e = \frac{\omega \mu_{o}c\times I_{o}cos\omega t.ln(\frac{a + b}{a}))}{2\pi }[/tex]
[tex]\omega = 2\pi f[/tex]
The rms value of the induced emf is given by:
[tex]e_{rms} = \frac{\frac{2\pi f\mu_{o}I_{o}c}{2\pi}.ln(\frac{a + b}{a})}{\sqrt{2}}[/tex]
Substituting appropriate values in the above eqn:
[tex]e_{rms} = \frac{\frac{2\pi \times 359\times 4\pi \times 10^{- 7}\times 0.325\times 5.55}{2\pi}.ln(\frac{0.802 + 0.722}{0.802})}{\sqrt{2}}[/tex]
[tex]e_{rms} = 3.69\times 10^{- 4}\ V[/tex]
Now,
To calculate average power:
[tex]P_{avg} = \frac{e_{rms}^{2}}{R}[/tex]
[tex]P_{avg} = \frac{3.69\times 10^{- 4}^{2}}{54.3} = 2.513\times 10^{- 9}\ W[/tex]
