Question 29: EM waves

Question

A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{j}$. Its associated electric field vector is $\_\_\_\_$ .

Accepted Answer

Given,
$$ \vec{B}=B_0 \sin \left(2 \pi 
u t-\frac{2 \pi x}{\lambda}ight)\hat{j} $$
We compare it with the standard wave form
$$ \sin(\omega t-kx) $$
This represents a wave travelling in the $+x$ direction.
In an electromagnetic wave:
$\vec{E}$, $\vec{B}$ and direction of propagation are mutually perpendicular.
Direction of propagation is given by $\vec{E} 	imes \vec{B}$.
Here,
propagation is along $+\hat{i}$
magnetic field is along $+\hat{j}$
So electric field must be along $\hat{k}$ or $-\hat{k}$.
Now use
$$ \vec{E} 	imes \vec{B} = +\hat{i} $$
If $\vec{B}$ is along $\hat{j}$, then for
$$ \hat{k} 	imes \hat{j} = -\hat{i} $$
which is wrong.
But
$$ (-\hat{k}) 	imes \hat{j} = \hat{i} $$
So $\vec{E}$ must be along $-\hat{k}$.
Also, for electromagnetic waves,
$$ E_0 = c B_0 $$
and since
$$ c=
u \lambda $$
we get
$$ E_0=
u \lambda B_0 $$
Hence,
$$ \vec{E}=-
u \lambda B_0 \sin \left(2 \pi 
u t-\frac{2 \pi x}{\lambda}ight)\hat{k} $$
So the correct answer is:
$$ \boxed{\vec{E}=-
u \lambda B_0 \sin \left(2 \pi 
u t-\frac{2 \pi x}{\lambda}ight)\hat{k}} $$
Option A.

Answer

$$ \vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{k} $$

Answer

$$ \vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{i} $$

Answer

$$ \vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{k} $$

Answer

$$ \vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{i} $$

EM waves

A magnetic field vector in an electromagnetic wave is represented by \vec{B}=B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{j}. Its associated electric field vector is \_\_\_\_ .

Choose the correct answer: