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# The magnetic field of a plane electromagnetic wave is given by:$\overrightarrow B = {B_0}\hat i\left[ {\cos \left( {kz - wt} \right)} \right] + {B_1}\hat j\left[ {\cos \left( {kz + wt} \right)} \right]$ where ${B_0} = 3 \times {10^{ - 5}}T$;${B_1} = 2 \times {10^{ - 6}}T$. The rms value of the force experienced by a stationary charge$Q = {10^{ - 4}}C$ at $z = 0$ is closest to:A. $0.9\,N$B. $0.1\,N$C. $3 \times {10^{ - 2}}\,N$D. $0.6\,N$

Last updated date: 06th Sep 2024
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Hint-The relation between peak value of electric field and magnetic field in an electromagnetic wave is given as
$E = Bc$
Where, $E$ is the peak value of the electric field and $B$ is the peak value of the magnetic field and $c$ is the speed of light.
The net peak value of magnetic field is given as
$B = \sqrt {B_0^2 + B_1^2}$
Force due to electric field on a charge $q$is given as
$F = qE$
Where, $q$ is the charge and $E$ is the electric field.
The relation connecting peak value and RMS value of force is
${F_{rms}} = \dfrac{F}{{\sqrt 2 }}$

Step by step solution:
Electromagnetic waves are created as a result of vibrations of electric field and magnetic field. So, we can say that they are composed of oscillating electric and magnetic fields
The relation between peak value of electric field and magnetic field in an electromagnetic wave is given as
$E = Bc$ ….. (1)
Where, $E$ is the peak value of the electric field and $B$ is the peak value of the magnetic field and $c$ is the speed of light.
Given,
$Q = {10^{ - 4}}C$
The value of magnetic field of plane electromagnetic wave as
$\overrightarrow B = {B_0}\hat i\left[ {\cos \left( {kz - wt} \right)} \right] + {B_1}\hat j\left[ {\cos \left( {kz + wt} \right)} \right]$
From this we can see that the peak value of magnetic field in the x direction is ${B_0}$ and the peak value of magnetic field in the y direction is ${B_1}$.
So, the net peak value of magnetic field is given as
$B = \sqrt {B_0^2 + B_1^2}$
Substitute the given values,${B_0} = 3 \times {10^{ - 5}}T$ and ${B_1} = 2 \times {10^{ - 6}}T$.
$B = \sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} \\ = 3.006 \times {10^{ - 5}}\,T \\$
Now let us substitute this value in equation (1)
$E = Bc \\ = 3.006 \times {10^{ - 5}} \times 3 \times {10^8} \\ = 9.018 \times {10^3}V{m^{ - 1}} \\$
This is the peak value of the electric field.
Force due to electric field on a charge $q$is given as
$F = qE$
Where, $q$ is the charge and $E$ is the electric field.
Substitute the given values. Then, we get
$F = {10^{ - 4}} \times 9.018 \times {10^3} \\ = 0.9018\,N \\$
This is the peak value of force. We need to find the RMS value of force.
The relation connecting peak value and RMS value of force is
${F_{rms}} = \dfrac{F}{{\sqrt 2 }}$
On substituting the value of force in this equation, we get
${F_{rms}} = \dfrac{{0.9018\,}}{{\sqrt 2 }} \\ = 0.637\,N \\$

So, the correct answer is option D.

Note: Here, it is given that the charge is stationary. Thus, only force due to the electric field is present. Effect of force due to the magnetic field will be felt only by moving charges. So, if the charge is moving then we should also consider the force due to the magnetic field.