Question

The amplitude of a monochromatic 1000 Hz EM wave reduces by a factor of \(1/e\) after penetrating to a depth of 100 m in a homogeneous medium. Given the magnetic permeability of free space \(\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}\), the electrical conductivity of the medium in S/m is _________ (rounded off to three decimal places).

Hint:

The skin depth formula \(\delta = \sqrt{\tfrac{2}{\mu \sigma \omega}}\) is essential in EM induction problems. Always check units: frequency in Hz, \(\mu\) in H/m, and depth in meters.

Answer 1

Step 1: Identify the skin depth relation.
The skin depth \(\delta\) is the depth at which the wave amplitude decreases to \(1/e\). \[ \delta = \sqrt{\frac{2}{\mu \sigma \omega}} \] 

Step 2: Substitute given values.
\(\delta = 100\,\text{m},\ f = 1000\,\text{Hz},\ \omega = 2\pi f = 2\pi \times 1000 = 6283.19\,\text{rad/s},\ \mu = \mu_0 = 4\pi \times 10^{-7}\,\text{H/m}.\) 

Step 3: Solve for conductivity \(\sigma\).
\[ \sigma = \frac{2}{\mu \omega \delta^2} \] \[ \sigma = \frac{2}{\,(4\pi \times 10^{-7})(6283.19)(100^2)} \] 

Step 4: Simplify denominator.
\((100^2) = 10000.\)
\(\mu \omega \delta^2 = (4\pi \times 10^{-7})(6283.19)(10000).\) \[ = 1.2566 \times 10^{-6} \times 6283.19 \times 10000. \] First multiply: \(1.2566 \times 10^{-6} \times 6283.19 \approx 0.007896.\)
Then multiply by \(10000 \Rightarrow 78.96.\) 

Step 5: Final calculation.
\[ \sigma = \frac{2}{78.96} \approx 0.0253\ \text{S/m}. \] \[ \boxed{0.025} \]