Question 44: EM waves

Question

The electric field of an electromagnetic wave in free space is $\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 	imes 10^6 \mathrm{t}-5 	imes 10^{-3}(3 x+4 y)ight](4 \hat{i}-3 \hat{j}) N / C$.
The associated magnetic field in Tesla is

Accepted Answer

We start with the given electric field:
$$ \vec{E} = 57 \cos\Bigl[7.5	imes10^6\, t - 5	imes10^{-3}(3x+4y)\Bigr](4\,\hat{i} - 3\,\hat{j}) \quad 	ext{(N/C)} $$
Since we are dealing with an electromagnetic wave in free space, the electric field, magnetic field, and the propagation direction are all mutually perpendicular. In free space the magnetic field is related to the electric field by:
$$ \vec{B} = \frac{1}{c}\,\hat{n} 	imes \vec{E}, $$
where:
Ã¢ÂÂ¢Ã¢ÂÂ$$c = 3	imes10^8 \, 	ext{m/s}$$ is the speed of light, and
Ã¢ÂÂ¢Ã¢ÂÂ$$\hat{n}$$ is the unit vector in the direction of wave propagation.
Step 1. Identify the propagation direction:
The phase of the wave is written as
$$ 7.5	imes10^6\,t - 5	imes10^{-3}(3x+4y). $$
This tells us that the wave vector, up to a multiplicative constant, is
$$ \vec{k} = 5	imes10^{-3}(3\,\hat{i} + 4\,\hat{j}). $$
The unit vector in the direction of propagation is then
$$ \hat{n} = \frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}+4\hat{j}}{5}. $$
Step 2. Compute the cross product:
We have
$$ \vec{E} \propto (4\,\hat{i}-3\,\hat{j}), $$
so we need to calculate
$$ \hat{n} 	imes \vec{E} = \frac{1}{5}(3\hat{i}+4\hat{j}) 	imes \Bigl[57 \cos(\ldots)(4\hat{i}-3\hat{j})\Bigr]. $$
Taking the constant factors out of the cross product gives
$$ \hat{n} 	imes \vec{E} = \frac{57 \cos(\ldots)}{5}\,\Bigl[(3\hat{i}+4\hat{j}) 	imes (4\hat{i}-3\hat{j})\Bigr]. $$
Recall that the cross product in the plane (where both vectors have zero $$\hat{k}$$ components) will point in the $$\hat{k}$$ (or $$-\hat{k}$$) direction. To compute the cross product of vectors in the $$xy$$-plane, we use the formula
$$ \vec{a} 	imes \vec{b} = (a_xb_y - a_yb_x)\,\hat{k}. $$
Let
Ã¢ÂÂ¢Ã¢ÂÂ$$\vec{a} = 3\hat{i}+4\hat{j},$$
Ã¢ÂÂ¢Ã¢ÂÂ$$\vec{b} = 4\hat{i}-3\hat{j}.$$
Then
$$ \vec{a} 	imes \vec{b} = \Bigl(3\cdot (-3) - 4\cdot 4\Bigr)\,\hat{k} = (-9 - 16)\,\hat{k} = -25\,\hat{k}. $$
Thus,
$$ \hat{n} 	imes \vec{E} = \frac{57 \cos(\ldots)}{5}\,(-25\,\hat{k}) = -\frac{57 	imes 25}{5}\, \cos(\ldots) \,\hat{k}. $$
Simplify the constant:
$$ \frac{25}{5} = 5, $$
so we get
$$ \hat{n} 	imes \vec{E} = -\frac{57 	imes 5}{1} \cos(\ldots)\,\hat{k} = -285 \cos(\ldots) \,\hat{k}. $$
Step 3. Apply the relation for the magnetic field:
Now, substituting back into the formula for $$\vec{B}$$,
$$ \vec{B} = \frac{1}{c}\,\hat{n} 	imes \vec{E} = -\frac{285}{3	imes10^8} \cos(\ldots)\,\hat{k}. $$
Notice that
$$ \frac{285}{3	imes10^8} = \frac{57 	imes 5}{3	imes10^8}, $$
so we have
$$ \vec{B} = -\frac{57}{3	imes10^8} \cos\Bigl[7.5	imes10^6\,t - 5	imes10^{-3}(3x+4y)\Bigr](5\,\hat{k})\quad 	ext{(T)}. $$
This expression exactly matches Option D.
Final Answer: Option D

Answer

$\overrightarrow{\mathrm{B}}=\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 \mathrm{t}-5 	imes 10^{-3}(3 x+4 y)ight](\hat{k})$

Answer

$\overrightarrow{\mathrm{B}}=\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 \mathrm{t}-5 	imes 10^{-3}(3 x+4 y)ight](5 \hat{k})$

Answer

$\overrightarrow{\mathrm{B}}=-\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 \mathrm{t}-5 	imes 10^{-3}(3 x+4 y)ight](\hat{k})$

Answer

$\overrightarrow{\mathrm{B}}=-\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 \mathrm{t}-5 	imes 10^{-3}(3 x+4 y)ight](5 \hat{k})$

EM waves

The electric field of an electromagnetic wave in free space is \overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C. The associated magnetic field in Tesla is

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