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