ELECTROMAGNETISM & RELATIVITY - Chapter 6, Exercise 1 Solution
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Verifying the Speed of Light from mu0 and epsilon0

PROBLEM
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Using the same formula c = 1/sqrt(mu0 epsilon0), verify the
calculation yourself by computing mu0 x epsilon0 first, then taking
the square root and its reciprocal, using mu0 = 4 x pi x 10^-7 and
epsilon0 = 8.854 x 10^-12. Show each intermediate step.

SOLUTION
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Step 1: Compute mu0 numerically.

    mu0 = 4 x pi x 10^-7
    mu0 = 4 x 3.14159 x 10^-7
    mu0 = 12.566 x 10^-7
    mu0 = 1.2566 x 10^-6 T.m/A

Step 2: Multiply mu0 by epsilon0.

    mu0 x epsilon0 = (1.2566 x 10^-6) x (8.854 x 10^-12)
    mu0 x epsilon0 = 1.1127 x 10^-17

Step 3: Take the square root.

    sqrt(1.1127 x 10^-17) = sqrt(11.127 x 10^-18)
    sqrt(11.127 x 10^-18) = 3.336 x 10^-9

Step 4: Take the reciprocal to find c.

    c = 1 / (3.336 x 10^-9)
    c = 2.998 x 10^8 m/s

ANSWER: c is approximately 2.998 x 10^8 m/s - matching the real,
independently measured speed of light exactly.

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WHY THIS WORKS AS AN ANSWER

Working through each step by hand shows exactly why this result was
so genuinely striking to Maxwell and his contemporaries: mu0 and
epsilon0 are both measured entirely through laboratory experiments
involving electric charges, currents, and magnetic forces - nothing
in either constant's own definition has anything to do with light or
optics. That combining these two purely electrical/magnetic
quantities, via this exact formula, reproduces the speed of light to
within the measurement precision of the era is not a coincidence
built into the formula - it is real, direct evidence that light
itself is fundamentally an electromagnetic phenomenon, exactly as
Maxwell's own 1865 paper concluded.