Question

The magnetic field of an electromagnetic wave is given by $\vec{B}=1.6\times {{10}^{-6}}\cos (2\times {{10}^{-7}}z+6\times {{10}^{15}}t)(2\overset{\scriptscriptstyle\frown}{i}+\overset{\scriptscriptstyle\frown}{j})\dfrac{Wb}{{{m}^{2}}}$
​The associated electric field will be:-
A. $\vec{E}=4.8\times {{10}^{2}}\cos (2\times {{10}^{-7}}z+6\times {{10}^{15}}t)(\overset{\scriptscriptstyle\frown}{i}-2\overset{\scriptscriptstyle\frown}{j})\dfrac{V}{{{m}^{{}}}}$
B. $\vec{E}=4.8\times {{10}^{2}}\cos (2\times {{10}^{-7}}z-6\times {{10}^{15}}t)(2\overset{\scriptscriptstyle\frown}{i}+\overset{\scriptscriptstyle\frown}{j})\dfrac{V}{{{m}^{{}}}}$
C. $\vec{E}=4.8\times {{10}^{2}}\cos (2\times {{10}^{-7}}z-6\times {{10}^{15}}t)(-2\overset{\scriptscriptstyle\frown}{j}+\overset{\scriptscriptstyle\frown}{i})\dfrac{V}{{{m}^{{}}}}$
D. $\vec{E}=4.8\times {{10}^{2}}\cos (2\times {{10}^{-7}}z+6\times {{10}^{15}}t)(-\overset{\scriptscriptstyle\frown}{i}+2\overset{\scriptscriptstyle\frown}{j})\dfrac{V}{{{m}^{{}}}}$

Answer 1

Hint: An electromagnetic wave is generated from a charged particle that produces an electric field. When it accelerates, it generates ripples and a magnetic field is also formed. This leads to the development of an electromagnetic wave.

Formula used: $c=\dfrac{E}{B}$, where $c$ is the velocity of light, $E$ is the electric field and $B$is the magnetic field.

Complete answer:
The standard form for magnetic field of an electromagnetic wave is given by
\[B\left( x,t \right)\text{ }=~{{B}_{max}}cos\left( kx\text{ }-\text{ }\omega t\text{ }+\text{ }\varphi \right).\]
We are given the vector form of B that is
$\vec{B}=1.6\times {{10}^{-6}}\cos (2\times {{10}^{-7}}z+6\times {{10}^{15}}t)(2\overset{\scriptscriptstyle\frown}{i}+\overset{\scriptscriptstyle\frown}{j})\dfrac{Wb}{{{m}^{2}}}$
We know that $c=\dfrac{E}{B}$
Hence, magnitude of $E$ is given by $E=cB=3\times {{10}^{8}}\times 1.6\times {{10}^{-6}}$
$=4.8\times {{10}^{2}}$
We know that the electromagnetic waves is transverse wave means amplitude of waves is perpendicular to the direction of propagation and both these vectors are perpendicular to the propagation vector.
$\Rightarrow \vec{E}.\vec{B}=0$
⇒ either direction of E is $\overset{\scriptscriptstyle\frown}{i}-2\overset{\scriptscriptstyle\frown}{j}$ or $-\overset{\scriptscriptstyle\frown}{i}+2\overset{\scriptscriptstyle\frown}{j}$
from given option
Also wave propagation direction is parallel to $\vec{E}\times \vec{B}$ which is $(\overset{\scriptscriptstyle\frown}{k})$
Hence we can say that the electric field is along $\overset{\scriptscriptstyle\frown}{i}-2\overset{\scriptscriptstyle\frown}{j}$.

Combining the results that we obtained from the magnitude calculation and the direction analysis,
The value of $\vec{E}=4.8\times {{10}^{2}}\cos (2\times {{10}^{-7}}z+6\times {{10}^{15}}t)(\overset{\scriptscriptstyle\frown}{i}-2\overset{\scriptscriptstyle\frown}{j})\dfrac{V}{{{m}^{{}}}}$

So, the correct answer is Option A .

Note:
Both magnetic and electric fields in an electromagnetic wave will vary with time, one leading the other to change.
Electromagnetic waves are considered to be ubiquitous like light and widely utilized in modern technology that includes AM and FM radio, phones including cordless and cellular, door openers for garages , wireless networks, microwave ovens, RADAR etc.