The energy density associated with electric field E⃗ and magnetic field B⃗ of an electromagnetic wave in free space is given by (ε₀- permittivity of free space, μ₀- permeability of free space)

Solution

Physics sir · 4 steps

Recall the capacitor and solenoid energy densities, then check that E = cB makes the two equal.

1. Electric part. From a parallel-plate capacitor, the energy stored per unit volume in a field E is U_E = 1/2ε₀ E². Here ε₀ multiplies E² in the numerator.
2. Magnetic part. From a solenoid, the energy stored per unit volume in a field B is U_B = B²/2μ₀. Here μ₀ sits in the denominator.
3. Cross-check. In an EM wave E = cB and c² = 1/(μ₀ ε₀). Putting these into U_E should give U_B, because the two fields carry equal energy.

   U_E = 1/2ε₀ E² = 1/2ε₀ c² B² = B²/2μ₀ = U_B

4. So. The pair U_E = (ε₀ E²)/2 and U_B = B²/2μ₀ is correct, which is option (a).

The idea to keep. Energy density: ε₀ goes up with E², μ₀ goes down under B². If you forget, use E = cB: the correct pair always gives U_E = U_B in an EM wave.

Answer from the NTA final answer key. Solution by PrepWiser.