Question

The electric field of a plane electromagnetic wave varies with the time of amplitude $2V{m}^{-1}$propagating along the z-axis. The average energy density of the magnetic field is $(inJ{m}^{-3})$

• A. $13.29\times {10}^{-12}$
• B. $8.85\times {10}^{-12}$
• C. $17.72\times {10}^{-12}$
• D. $4.43\times {10}^{-12}$
• E. $2.22\times {10}^{-12}$

Answer 1

The correct option is B

$8.85\times {10}^{-12}$

Step 1: Given data

Electric field is $E=2V{m}^{-1}$

Step 2: Formula used

The following formula is used to compute the $U_B=\frac{1}{2}\frac{B_{\circ }^2}{{\mu }_{\circ }}\dots \dots \dots \dots \dots \left(1\right),$

Where $U_B=$average energy density of the magnetic field, $B_0=$magnetic field, ${\mu }_0=$vacuum permeability

Speed of light $c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}\dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Step 3: Compute the average energy density of the magnetic field

We know that $Electricfield\left(E\right)=Magneticfield\left(B_{\circ }\right)\times Speedoflight\left(c\right)\dots \dots \dots \dots \dots \left(3\right)$

Substitute equation $\left(3\right)$ in equation $\left(1\right)$.

$\begin{array}{rcl}{U}_B& =& \frac{1}{4}\frac{E^2}{c^2{\mu }_{\circ }}\cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(4\right)\end{array}$

Substitute equation $\left(2\right)$ in equation $\left(4\right)$,

$\begin{array}{rcl}{U}_B& =& \frac{1}{4}\frac{E^2}{{\mu }_{\circ }\times \left({\displaystyle \frac{1}{{\mu }_{\circ }{\epsilon }_{\circ }}}\right)}\\ & =& \frac{1}{4}{\epsilon }_{\circ }{E}^2\\ & =& \frac{1}{4}\times \left(8·85\times {10}^{-12}\right)\times {\left(2\right)}^2\\ & =& 8.85\times {10}^{-12}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.\end{array}$

The average energy density of the magnetic field is $8.85\times {10}^{-12}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$

Hence, option B is the correct answer.