Respuesta :
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux
[tex]\phi=NBA[/tex]
Put the value into the formula
[tex]\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2[/tex]
[tex]\phi=7.85\times10^{-7}\ Tm^2[/tex]
We need to calculate the induced emf
Using formula of induced emf
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{7.85\times10^{-7}}{dt}[/tex]
Put the value of emf from ohm's law
[tex]\epsilon =IR[/tex]
[tex]IR=\dfrac{7.85\times10^{-7}}{dt}[/tex]
[tex]Idt=\dfrac{7.85\times10^{-7}}{R}[/tex]
[tex]Idt=\dfrac{7.85\times10^{-7}}{0.50}[/tex]
[tex]Idt=0.00000157=1.57\times10^{-6}\ C[/tex]
We know that,
[tex]Idt=dq[/tex]
[tex]dq=1.57\times10^{-6}\ C[/tex]
We need to calculate the voltage across the capacitor
Using formula of charge
[tex]dq=C dV[/tex]
[tex]dV=\dfrac{dq}{C}[/tex]
Put the value into the formula
[tex]dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}[/tex]
[tex]dV=1.57\ V[/tex]
Hence, The voltage across the capacitor is 1.57 V.
