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Question: An electromagnetic wave travelling in the x-direction has frequency of \( 2\times {{10}^{14}}Hz \) a...

An electromagnetic wave travelling in the x-direction has frequency of 2×1014Hz2\times {{10}^{14}}Hz and electric field amplitude of 27Vm−127V{{m}^{-1}} , from the options below, which one describes the magnetic field for this
(A) B→(x,t)=(3×10−8 T)j^ sin⁡[2π(1⋅5×10−8x−2×1014t)]\overrightarrow{B}\left( x,t \right)=\left( 3\times {{10}^{-8}}\text{ T} \right)\hat{j}\text{ }\sin \left[ 2\pi \left( 1\cdot 5\times {{10}^{-8}}x-2\times {{10}^{14}}t \right) \right]
(B) B→(x,t)=(9×10−8T)j^ sin⁡[(1⋅5×10−6x−2×1014t)]\overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{j}\text{ }\sin \left[ \left( 1\cdot 5\times {{10}^{-6}}x-2\times {{10}^{14}}t \right) \right]
(C) B→(x,t)=(9×10−8T)k^ sin⁡[2π(0⋅66×106x−2×1014t)]\overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{k}\text{ }\sin \left[ 2\pi \left( 0\cdot 66\times {{10}^{6}}x-2\times {{10}^{14}}t \right) \right]
(D) B→(x,t)=(9×10−8T)i^ sin⁡[2π(1⋅5×10−8x−2x×1014t)]\overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{i}\text{ }\sin \left[ 2\pi \left( 1\cdot 5\times {{10}^{-8}}x-2x\times {{10}^{14}}t \right) \right]

Explanation

Solution

Wave function for a given plane electromagnetic wave is written as
B=Bosin⁡(kx−wt)\text{B=Bo}\sin \left( kx-wt \right)
In case of plane electromagnetic waves, direction of propagation is along the cross product E→×B→\overrightarrow{E}\times \overrightarrow{B} . Hence, if direction of propagation is along x axis then E→\overrightarrow{E} is along y-axis and B→\overrightarrow{B} is along Z axis. From this, the direction of the magnetic field is found.

Complete step by step solution
We know that
Bo=Eoc\text{Bo}=\dfrac{\text{Eo}}{\text{c}}
Now,
Eo=27\text{Eo=27} …. Given
c=3×108\text{c}=3\times {{10}^{8}}
Hence
Bo=273×108=9×10−8\text{Bo}=\dfrac{27}{3\times {{10}^{8}}}=9\times {{10}^{-8}}
As the direction of propagation is along the x-axis, therefore Bo\text{Bo} is along the z axis.
So Bo→=(9×10−8)k^\overrightarrow{\text{Bo}}=\left( 9\times {{10}^{-8}} \right)\hat{k}
Now k=2πλk=\dfrac{2\pi }{\lambda } and w=2πTw=\dfrac{2\pi }{T}
So putting these values in equation
B=Bo sin⁡(kx−wt)\text{B=Bo }\sin \left( kx-wt \right)
Now
λ=cv =3×1082×1014=1⋅5×10−6m \begin{aligned} & \lambda =\dfrac{c}{v} \\\ & =\dfrac{3\times {{10}^{8}}}{2\times {{10}^{14}}}=1\cdot 5\times {{10}^{-6}}m \\\ \end{aligned}
And
T=1v=12×1014=0⋅5×10−14 secT=\dfrac{1}{v}=\dfrac{1}{2\times {{10}^{14}}}=0\cdot 5\times {{10}^{-14}}\text{ sec}
Putting these values in equation (1), we get
B=(9×10−8)k^ sin[2π(x1⋅5×10−6−t0⋅5×1014)] B=(9×10−8)k^ sin⁡[2π(0⋅666×106x−2×1014t)] \begin{aligned} & \text{B}=\left( 9\times {{10}^{-8}} \right)\hat{k}\text{ sin}\left[ 2\pi \left( \dfrac{x}{1\cdot 5\times {{10}^{-6}}}-\dfrac{t}{0\cdot 5\times {{10}^{14}}} \right) \right] \\\ & \text{B=}\left( 9\times {{10}^{-8}} \right)\hat{k}\text{ }\sin \left[ 2\pi \left( 0\cdot 666\times {{10}^{6}}x-2\times {{10}^{14}}t \right) \right] \\\ \end{aligned} .

Note
EoE_o and BoB_o are the maximum magnitude of electric and magnetic fields in case of an electromagnetic wave.
Number of cycles of electric field or magnetic field completed in one second is called frequency of electromagnetic wave.
For mathematical representation, we define angular frequency by multiplying frequency with 2π2\pi
w=2πvw=2\pi v
Time period of electromagnetic oscillation can be written as:
T=1vT=\dfrac{1}{v}
Distance travelled by wave in one time period is called wavelength and is represented by λ\lambda
λ=CT λ=CV \begin{aligned} & \lambda =CT \\\ & \lambda =\dfrac{C}{V} \\\ \end{aligned} .