Question

A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivity σ. The current I is radially outward from the inner surface to the outer surface.
Find an expression for the electric field strength inside the metal as a function of the radius r from the cylinder's axis.Express your answer in terms of some or all of the variables a, b, L, σ, I, r, and the constant π.
Evaluate the electric field strength at the inner surface of an iron cylinder if a = 0.50 cm , b = 2.3 cm , L = 10 cm, and I = 27 A .
Evaluate the electric field strength at the outer surface of an iron cylinder if a = 0.50 cm , b = 2.3 cm , L =10 cm, and I = 27 A .

Answer 1

Answer

current density is given by

$I = \int J.da$

where J = σ E

$I = \int J.da\\ =\int \sigma E .da\\ = \sigma E \int da \\ =\sigma E \int_0^{2\pi} rL$

now,

$E=\dfrac{I}{2\pi r L \sigma}$

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

$E=\dfrac{I}{2\pi a L \sigma}$

$E=\dfrac{27}{2\pi\times 0.005\times 0.1\times 10^7}$

      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

$E=\dfrac{I}{2\pi b L \sigma}$

$E=\dfrac{27}{2\pi\times 0.023\times 0.1\times 10^7}$

      E =1.87 x 10⁻⁴ V/m

Answer 2

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,

$I = \int J\cdot da$

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,

$I = \int_0^{2\pi} (\sigma E)\cdot RL\\\\I = {2\pi} (\sigma E)\cdot RL\\\\E = \dfrac{I }{ {2\pi} \sigma\cdot RL}$

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

$\overrightarrow E = \dfrac{I }{ {2\pi} \sigma\cdot RaL}\\\\\overrightarrow E = \dfrac{27 }{ {2\pi} \times 10^7 \times 0.1 \times 0.1}\\\\\overrightarrow E = 8.59 \times 10^{-4}\rm \ V/m$

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

$\overrightarrow E = \dfrac{I }{ {2\pi} \sigma\cdot bL}\\\\\overrightarrow E = \dfrac{27 }{ {2\pi} \times 10^7 \times 0.023 \times 0.1}\\\\\overrightarrow E = 1.87 \times 10^{-4}\rm \ V/m$

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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