The magnetic field component of an electromagnetic wave is 25 $\mu \mathrm { T }$ . The electric energy density of the wave is
A) $2.5 \times 10 ^ { - 4 } \mathrm {~J} / \mathrm { m } ^ { 3 }$
B) $2.5 \times 10 ^ { - 4 } \mathrm {~J} / \mathrm { m } ^ { 2 }$
C) $2.5 \times 10 ^ { - 2 } \mathrm {~J} / \mathrm { m } ^ { 3 }$
D) $2.5 \times 10 ^ { - 2 } \mathrm {~J} / \mathrm { m } ^ { 2 }$

Question

Quizplus · Accepted Answer

The electric energy density ($u_E$) in an electromagnetic wave can be found using the relation between the magnetic field component ($B$) and the electric field component ($E$) in the wave, given by $E = cB$, where $c$ is the speed of light in vacuum ($3 \times 10^8 \, \text{m/s}$). The energy density of the electric field is given by $u_E = \frac{1}{2} \epsilon_0 E^2$, where $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$). Substituting $E = cB$ into the formula for $u_E$, we get $u_E = \frac{1}{2} \epsilon_0 (cB)^2$. Substituting the given value of $B = 25 \mu T = 25 \times 10^{-6} T$ and the constants, we find the electric energy density to be $2.5 \times 10^{-4} \, \text{J/m}^3$, making option A correct.

The Magnetic Field Component of an Electromagnetic Wave Is 25 μT\mu \mathrm { T }μT

The Magnetic Field Component of an Electromagnetic Wave Is 25 $\mu \mathrm { T }$