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Question: The electric field of a plane electromagnetic wave varies with time of amplitude \(2Vm^{- 1}\)propag...

The electric field of a plane electromagnetic wave varies with time of amplitude 2Vm−12Vm^{- 1}propagating along z-axis. The average energy density of the magnetic field (in J m−3- 3):

A

13.29×10−1213.29 \times 10^{- 12}

B

8.86×10−128.86 \times 10^{- 12}

C

17.72×10−1217.72 \times 10^{- 12}

D

4.43×10−124.43 \times 10^{- 12}

Answer

8.86×10−128.86 \times 10^{- 12}

Explanation

Solution

: Amplitude of electric field and magnetic field are related by the relation

E0B0=c\frac{E_{0}}{B_{0}} = c

Average energy density of the magnetic field is

μB=14B02μ0\mu_{B} = \frac{1}{4}\frac{B_{0}^{2}}{\mu_{0}}

=E02μ0c2= \frac{E_{0}^{2}}{\mu_{0}c^{2}} (∵B0=E0c)\left( \because B_{0} = \frac{E_{0}}{c} \right)

=14ε0E02= \frac{1}{4}\varepsilon_{0}E_{0}^{2} (∵c=1μ0ε0)\left( \because c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}} \right)

=14×8.854×10−12×(2)2= \frac{1}{4} \times 8.854 \times 10^{- 12} \times (2)^{2}

8.854×10−12Jm−38.854 \times 10^{- 12}Jm^{- 3}

≈8.86×10−12Jm−3\approx 8.86 \times 10^{- 12}Jm^{- 3}