The electric field of a plane electromagnetic wave of amplitude 2 ~V ~m^ -1 varies with time and propagates along z -axis. The average energy density of the magnetic field (in J m^ -3 ) is

Correct answer: B. 8.86 × 10^ -12 Average energy density of the magnetic field is aligned & u_B=(1)/(4) (B_0^2)/(μ_0) ( B_0=(E_0)/(c)) & =(1)/(4) _0 E_0^2 ( c= 1 μ_0 _0 ) & =(1)/(4) × 8.854 × 10^ -12 ×(2)^2 & =8.854 × 10^ -12 ~J ~m^ -3 8.86 × 10^ -12 ~J ~m^ -3 aligned

AIIMS 201926 May

Exam: AIIMS
Year: 2019
Date: 26 May
Subject: Physics
Chapter: Magnetic Properties of Matter

AIIMS 2019 · 26 May · Physics · Magnetic Properties of Matter

The electric field of a plane electromagnetic wave of amplitude 2...

The electric field of a plane electromagnetic wave of amplitude

2 \mathrm{~V} \mathrm{~m}^{-1}

varies with time and propagates along

z

-axis. The average energy density of the magnetic field (in

\mathrm{J} \mathrm{m}^{-3}

) is

Options

A $13.29 \times 10^{-12}$
B $8.86 \times 10^{-12}$
C $17.72 \times 10^{-12}$
D $4.43 \times 10^{-12}$

Correct answer

B. $8.86 \times 10^{-12}$

Step-by-step solution

Average energy density of the magnetic field is

\begin{aligned} & u_B=\frac{1}{4} \frac{B_0^2}{\mu_0} \quad\left(\because \quad B_0=\frac{E_0}{c}\right) \\ & =\frac{1}{4} \varepsilon_0 E_0^2 \quad\left(\because \quad c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right) \\ & =\frac{1}{4} \times 8.854 \times 10^{-12} \times(2)^2 \\ & =8.854 \times 10^{-12} \mathrm{~J} \mathrm{~m}^{-3} \approx 8.86 \times 10^{-12} \mathrm{~J} \mathrm{~m}^{-3}\end{aligned}

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