The aluminum foil used in cooking has an electrical conductivity sigma = 3.5 tim

Question

The aluminum foil used in cooking has an electrical conductivity sigma = 3.5 times 10^7 (ohm m)^-1, and a typical thickness delta = 2 times 10^-4 m. Show that such foil can be used to shield a region from electromagnetic waves of a given frequency, provided that the skin-depth of the waves in the foil is less than about a third of its thickness. Because skin-depth increases as frequency decreases, it follows that the foil can only shield waves whose frequency exceeds a critical value. Estimate this critical frequency (in Hertz). What is the corresponding wavelength?

Explanation / Answer

given conductivity sigma = 3.5*10^7 ohm^-1 m^-1

   thickness delta = 2*10^-4 m

we know that the skindepth is Ds = sqrt(1/(pi*f*mue*sigma))

               (1/3)(2*10^-4) = sqrt(1/(pi*4pi*10^-7*3.5*10^7*f))

               f = 1628376.1 Hz

wavelength is L = c/f = 3*10^8/1628376.1 = 184.2326229180102 m

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