Respuesta :
Answer:
[tex]P_{avg} = 6.283*10^{-9} \ W[/tex]
Explanation:
Given that;
I₀ = 9.55 A
f = 359 cycles/s
b = 72.2 cm
c = 32.5 cm
a = 80.2 cm
Using the formula;
[tex]\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})[/tex]
where;
[tex]E= \frac{d \phi}{dt}[/tex]
[tex]E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t[/tex]
[tex]E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}[/tex]
Replacing our values into above equation; we have:
[tex]E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}[/tex]
[tex]E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}[/tex]
[tex]E_{rms} = 6.356*10^{-4} \ V[/tex]
Then the [tex]P_{avg[/tex] is calculated as:
[tex]P_{avg} = \frac{E^2}{R}[/tex]
[tex]P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}[/tex]
[tex]P_{avg} = 6.283*10^{-9} \ W[/tex]
The average power dissipated through the loop is 6.283 × 10⁻⁹ W
Given information:
Current I = 9.55A
frequency, f = 359 cycles/s
so, ω = 2πf
ω = 2π×359
ω = 2254.52 rad/s
dimensions of the loop:
b = 72.2 cm
c = 32.5 cm
distance from wire
a = 80.2 cm
Average Power:
The magnetic flux through the coil:
[tex]\phi=\frac{\mu_oIc}{2\pi}\ln(\frac{b+a}{b})\\\\\phi=\frac{\mu_oc}{2\pi}\ln(\frac{b+a}{b})I_osin\omega t[/tex]
The EMF generated :
[tex]E = \frac{d\phi}{dt} \\\\E = \frac{\mu_oc}{2\pi}\ln(\frac{b+a}{b})I_o\omega cos\omega t\\\\E = \frac{\mu_oIc}{2\pi}\ln(\frac{b+a}{b})\omega\\\\E=\frac{4\pi\times10^{-7}\ln(\frac{72.2+80.2}{80.2})\times9.55\times2254.52 }{2\pi}\\\\[/tex]
E ≈ 9 × 10⁻⁴ V
The RMS value will be:
[tex]E_{rms}=E/\sqrt{2}\\\\E_{rms}=6.36\times10^{-4}V[/tex]
Now the average power dissipated is given by:
[tex]P = \frac{E_{rms}^2}{R}\\\\P=\frac{(6.36\times10^{-4})^2}{64.3}[/tex]
P = 6.283 × 10⁻⁹ W
Learn more about magnetic flux:
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