Respuesta :
Answer:
(a). The magnitude of the electric field in the wire is 0.498 V/m.
(b). The displacement current in the wire at that time is [tex]4.120\times10^{-16}\ A[/tex]
(c). The ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire is [tex]2.28\times10^{-18}[/tex]
Explanation:
Given that,
Resistivity [tex]\rho=1.94\times10^{-8}\ \Omega m[/tex]
Area of cross section area A =7.00 mm²
Rate = 2400 A/s
Current = 180 A
(a). We need to calculate the magnitude of the electric field in the wire
Using for of electric field
[tex]E=\rho\times J[/tex]
[tex]E=\dfrac{\rho\times i}{A}[/tex]
Put the value into the formula
[tex]E=\dfrac{1.94\times10^{-8}\times180}{7.00\times10^{-6}}[/tex]
[tex]E= 0.498\ V/m[/tex]
(b). We need to calculate the displacement current in the wire
Using formula of displacement current
[tex]I_{d}=\epsilon_{0}\dfrac{d\phi}{dt}[/tex]
[tex]I_{d}=\epsilon_{0}(\dfrac{d(EA)}{dt})[/tex]
[tex]I_{d}=\epsilon_{0}d(\rho I)dt[/tex]
[tex]I_{d}=\epsilon_{0}\times\rho\dfrac{dI}{dt}[/tex]
Put the value into the formula
[tex]I_{d}=8.85\times10^{-12}\times1.94\times10^{-8}\times2400[/tex]
[tex]I_{d}=4.120\times10^{-16}\ A[/tex]
(c). We need to calculate the ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire
Using formula of magnetic field
[tex]\dfrac{B_{d}}{B}=\dfrac{\dfrac{\mu_{0}I_{d}}{2\pi r}}{\dfrac{\mu_{0}I}{2\pi r}}[/tex]
[tex]\dfrac{B_{d}}{B}=\dfrac{I_{d}}{I}[/tex]
Put the value into the formula
[tex]\dfrac{B_{d}}{B}=\dfrac{4.120\times10^{-16}}{180}[/tex]
[tex]\dfrac{B_{d}}{B}=2.28\times10^{-18}[/tex]
Hence, (a). The magnitude of the electric field in the wire is 0.498 V/m.
(b). The displacement current in the wire at that time is [tex]4.120\times10^{-16}\ A[/tex]
(c). The ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire is [tex]2.28\times10^{-18}[/tex]
