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# In a plane electromagnetic wave, the electric field of amplitude $1\,V{m^{ - 1}}$ varies with time in free space. What is the average energy density of a magnetic field? ( in$J{m^{ - 3}}$ )A. $8.86 \times {10^{ - 12}}$B. $4.43 \times {10^{ - 12}}$C. $17.72 \times {10^{ - 12}}$D. $8.86 \times {10^{ - 14}}$E. $2.21 \times {10^{ - 12}}$

Last updated date: 11th Jun 2024
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Hint: In an electromagnetic wave, the total energy density of the wave is the sum of density electric field energy and density magnetic field energy. Also the electric field and magnetic field are proportional to each other with the constant of proportionality being speed of light(c).

Formula Used:
Average energy density of magnetic field is given by the equation
${U_B} = \dfrac{1}{2}\dfrac{{{B^2}}}{{{\mu _0}}}$
$B$ is the RMS value and ${\mu _0}$ is the magnetic permeability.
The relation between RMS value and peak value of magnetic field is given as
$B = \dfrac{{{B_0}}}{{\sqrt 2 }}$
Where ${B_0}$ is the peak value.

The relationship between peak value of magnetic field ${B_0}$ ,and peak value of electric field, ${E_0}$ is
${B_0} = \dfrac{{{E_0}}}{c}$
Where value of c is given by the equation
$c = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$
${\varepsilon _0}$ is the electrical permittivity. Its value is given as ${\varepsilon _0} = 8.854 \times {10^{ - 12}}\,{C^{ 2}{N^{-1}}{m^{-2}}}$
Step by step solution:
Average energy density of magnetic field is given by the equation
${U_B} = \dfrac{1}{2}\dfrac{{{B^2}}}{{{\mu _0}}}$ …….(1)
$B$ is the RMS value and ${\mu _0}$ is the magnetic permeability.
The relation between RMS value and peak value of magnetic field is given as
$B = \dfrac{{{B_0}}}{{\sqrt 2 }}$
Where ${B_0}$ is the peak value.
Substituting this in equation (1). We get,
${U_B} = \dfrac{1}{2}\dfrac{{{{\left( {\dfrac{{{B_0}}}{{\sqrt 2 }}} \right)}^2}}}{{{\mu _0}}} \\ = \dfrac{1}{4}\dfrac{{B_0^2}}{{{\mu _0}}} \\$
We know that the relationship between peak value of magnetic field, ${B_0}$ and peak value of electric field, ${E_0}$ is
${B_0} = \dfrac{{{E_0}}}{c}$ ……(2)
Where value of c is given by the equation
$c = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ ……(3)
${\varepsilon _0}$ is the electrical permittivity. Its value is given as ${\varepsilon _0} = 8.854 \times {10^{ - 12}}\,{C^{ 2}{N^{-1}}{m^{-2}}}$
Substitute equation (2) and (3) in equation (1). We get,
${U_B} = \dfrac{1}{4}{\dfrac{{\left( {\dfrac{{{E_0}}}{c}} \right)}}{{{\mu _0}}}^2}$
${U_B} = \dfrac{1}{4}\dfrac{{E_0^2}}{{{c^2}{\mu _0}}} \\ = \dfrac{1}{4}\dfrac{{E_0^2{\mu _0}{\varepsilon _0}}}{{{\mu _0}}} \\ = \dfrac{1}{4}E_0^2{\varepsilon _0} \\$
The amplitude of electric field which is the peak value of electric field ${E_0}$ is given as $1\,V{m^{ - 1}}$.
Now substitute the given values.
${U_B} = \dfrac{1}{4}E_0^2{\varepsilon _0} \\ = \dfrac{{{{\left( {1\,} \right)}^2} \times 8.854 \times {{10}^{ - 12}}}}{4} \\ = 2.213 \times {10^{ - 12}}\,J{m^{ - 3}} \\$
Hence the correct answer is option E.

Note:
This could be solved also by using the property that the average electric energy density is equal to the average magnetic energy density. Here the amplitude of the electric field is given. That is, the peak value of the electric field is given. By using this we can find the peak value of the magnetic field. But while calculating the average energy density we need RMS value. So always remember to convert the peak value of the magnetic field into RMS value.