Respuesta :
Answer
current density is given by
[tex]I = \int J.da[/tex]
where J = σ E
[tex]I = \int J.da\\ =\int \sigma E .da\\ = \sigma E \int da \\ =\sigma E \int_0^{2\pi} rL [/tex]
now,
[tex]E=\dfrac{I}{2\pi r L \sigma}[/tex]
a) Electric field strength at the inner surface of an iron cylinder
a = 0.5 cm = 0.005 m b = 2.3 = 0.023
L = 10 cm = 0.1 m I = 27 A
[tex]E=\dfrac{I}{2\pi a L \sigma}[/tex]
[tex]E=\dfrac{27}{2\pi\times 0.005\times 0.1\times 10^7}[/tex]
E = 8.59 x 10⁻⁴ V/m
b) Electric field strength at the outer surface of an iron cylinder
a = 0.5 cm = 0.005 m b = 2.3 = 0.023
L = 10 cm = 0.1 m I = 27 A
[tex]E=\dfrac{I}{2\pi b L \sigma}[/tex]
[tex]E=\dfrac{27}{2\pi\times 0.023\times 0.1\times 10^7}[/tex]
E =1.87 x 10⁻⁴ V/m
The electric field strength at the inner surface and the outer surface of an iron cylinder is 8.59 x 10⁻⁴ V/m and 1.87 x 10⁻⁴ V/m respectively.
What is electric field strength?
Electric field strength is the strength or intensity of an electric field at a particular location.
We know that the current density is given as,
[tex]I = \int J\cdot da[/tex]
Also, the value of J = σ E, and the since the metal is cylindrical the area will be the product of Radius and Lenght which will be covered from 0 to 2π, therefore,
[tex]I = \int_0^{2\pi} (\sigma E)\cdot RL\\\\I = {2\pi} (\sigma E)\cdot RL\\\\E = \dfrac{I }{ {2\pi} \sigma\cdot RL}[/tex]
Now, substitute the inner and the outer radius to calculate the electric field around,
Electric field strength at the inner surface
For R = a = 10 cm = 0.1 m
Given to us,
Length, L = 10 cm = 0.10 m
I = 27 A
[tex]\overrightarrow E = \dfrac{I }{ {2\pi} \sigma\cdot RaL}\\\\\overrightarrow E = \dfrac{27 }{ {2\pi} \times 10^7 \times 0.1 \times 0.1}\\\\\overrightarrow E = 8.59 \times 10^{-4}\rm \ V/m[/tex]
Thus, the electric field strength at the inner surface of an iron cylinder is 8.59 x 10⁻⁴ V/m.
Electric field strength at the Outer surface
For R = b = 2.3 cm = 0.023 m
Given to us,
Length, L = 10 cm = 0.10 m
I = 27 A
[tex]\overrightarrow E = \dfrac{I }{ {2\pi} \sigma\cdot bL}\\\\\overrightarrow E = \dfrac{27 }{ {2\pi} \times 10^7 \times 0.023 \times 0.1}\\\\\overrightarrow E = 1.87 \times 10^{-4}\rm \ V/m[/tex]
Thus, the electric field strength at the outer surface of an iron cylinder is 1.87 x 10⁻⁴ V/m.
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