Question

In a plane electromagnetic wave, the electric field of amplitude $1V{m}^{-1}$ varies with time in free space. The average energy density of magnetic field is (in $J{m}^{-2}$)

• A. $8.86\times {10}^{-12}$
• B. $4.43\times {10}^{-12}$
• C. $17.72\times {10}^{-12}$
• D. $1.11\times {10}^{-12}$
• E. $2.21\times {10}^{-12}$

Answer 1

The correct option is E $2.21\times {10}^{-12}$

We know that,
The energy density (energy per unit volume) in an electric field $E$ in vacuum is given by
$U=\frac{1}{2}{\epsilon }_0{E}^2$
where ${\epsilon }_0=8.86\times {10}^{-12}{C}^2/N-m^2$
Given : $E=1V/m$
$U=\frac{1}{2}\times 8.86\times {10}^{-12}\times (1{)}^2$
$U=4.43\times {10}^{-12}$
So, the average energy density of magnetic field
$U=\frac{4.43\times {10}^{-12}}{2}$
$U=2.21\times {10}^{-12}J/m^2$