The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :

Electromagnetic Waves PYQ — JEE Main 2019 · 9 Jan · Shift 2 · Q69 with Solution

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Correct answer: D

For an electromagnetic wave in free space, the energy density due to the electric field is

$u_E = \frac{1}{2}\varepsilon_0 E^2$

and the energy density due to the magnetic field is

$u_B = \frac{1}{2\mu_0} B^2.$

Here, $U_E$ and $U_B$ denote the energy associated with electric and magnetic fields respectively, so we compare these two expressions.

For an electromagnetic wave in free space,

$E = cB$

where

$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}.$

So,

$B = \frac{E}{c}.$

$u_B = \frac{1}{2\mu_0} B^2 = \frac{1}{2\mu_0}\left(\frac{E}{c}\right)^2.$

Using

$c^2 = \frac{1}{\mu_0\varepsilon_0},$

we get

$\frac{1}{c^2} = \mu_0\varepsilon_0.$

Therefore,

= \frac{1}{2\mu_0} E^2 (\mu_0\varepsilon_0) = \frac{1}{2}\varepsilon_0 E^2.$$ Thus, $$u_B = u_E.$$ Hence, $$U_E = U_B.$$ 4. **Option check** - A: $U_E = \dfrac{U_B}{2}$ — Incorrect - B: $U_E > U_B$ — Incorrect - C: $U_E < U_B$ — Incorrect - D: $U_E = U_B$ — Correct Therefore, the correct answer is **D**.