Respuesta :
Answer:
(a) The maximum conduction current is [tex]40 \times 10^{-6}[/tex] A
(b) Value of [tex]\epsilon _{o} = 4.41 \times 10^{3}[/tex] V
(c) Maximum value of [tex]\frac{d\phi}{dt} = 4.5 \times 10^{6}[/tex] [tex]\frac{V}{m}[/tex]
Explanation:
Given:
Radius of circular plates [tex]R = 3.2 \times 10^{-2}[/tex] m
Separation between plates [tex]d = 1.8 \times 10^{-3}[/tex] m
Frequency [tex]f = 91[/tex] Hz
Maximum displacement current [tex]I_{d} = 40 \times 10^{-6}[/tex] A
(a)
Displacement current is equal to the conduction current so we write,
[tex]I _{max} = 40 \times 10^{-6}[/tex] A
(b)
From the formula of displacement current,
[tex]I _{d} = \frac{\omega \epsilon _{o} \mu_{o} \pi r^{2} }{d}[/tex]
Where [tex]\mu _{o} = 8.85 \times 10^{-12}[/tex], [tex]\epsilon _{o} =[/tex] peak value of emf, [tex]\omega = 2\pi f[/tex]
[tex]\epsilon _{o} = \frac{I_{d} d}{2\pi f \times \mu_{o} \times \pi r^{2} }[/tex]
[tex]\epsilon _{o} = \frac{40 \times 10^{-6} \times 1.8 \times 10^{-3} }{6.28 \times 91 \times 8.85 \times 10^{-12} \times 3.14 (3.2 \times 10^{-2} )^{2} }[/tex]
[tex]\epsilon _{o} = 4.41 \times 10^{3}[/tex] V
(c)
From another formula of displacement current,
[tex]I_{d} = \mu_{o} \frac{d\phi}{dt}[/tex]
Where [tex]\frac{d\phi}{dt} =[/tex] change in flux
[tex]\frac{d\phi}{dt} = \frac{40\times 10^{-6} }{8.85 \times 10^{-12} }[/tex]
[tex]\frac{d\phi}{dt} = 4.5 \times 10^{6}[/tex] [tex]\frac{V}{m}[/tex]
Therefore, the maximum conduction current is [tex]40 \times 10^{-6}[/tex] A and value of
[tex]\epsilon _{o} = 4.41 \times 10^{3}[/tex] V and maximum value of [tex]\frac{d\phi}{dt} = 4.5 \times 10^{6}[/tex] [tex]\frac{V}{m}[/tex]
