QJEEM25Ph9815: JEE Main 2025 Physics - Class 12 JEEM25Ph9815 Solution: The electric field of an electromagnetic wave in free space is…

We start with the given electric field:

$\vec{E} = 57 \cos\Bigl[7.5\times10^6\, t - 5\times10^{-3}(3x+4y)\Bigr](4\,\hat{i} - 3\,\hat{j}) \quad \text{(N/C)}$

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:

$\vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E},$

where:

• $c = 3\times10^8 \, \text{m/s}$ is the speed of light, and

• $\hat{n}$ is the unit vector in the direction of wave propagation.

Step 1. Identify the propagation direction:

The phase of the wave is written as

$7.5\times10^6\,t - 5\times10^{-3}(3x+4y).$

This tells us that the wave vector, up to a multiplicative constant, is

$\vec{k} = 5\times10^{-3}(3\,\hat{i} + 4\,\hat{j}).$

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

$\hat{n} \times \vec{E} = \frac{1}{5}(3\hat{i}+4\hat{j}) \times \Bigl[57 \cos(\ldots)(4\hat{i}-3\hat{j})\Bigr].$

Taking the constant factors out of the cross product gives

$\hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,\Bigl[(3\hat{i}+4\hat{j}) \times (4\hat{i}-3\hat{j})\Bigr].$

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

$\vec{a} \times \vec{b} = (a_xb_y - a_yb_x)\,\hat{k}.$

Let

• $\vec{a} = 3\hat{i}+4\hat{j},$

• $\vec{b} = 4\hat{i}-3\hat{j}.$

Then

$\vec{a} \times \vec{b} = \Bigl(3\cdot (-3) - 4\cdot 4\Bigr)\,\hat{k} = (-9 - 16)\,\hat{k} = -25\,\hat{k}.$

Thus,

$\hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,(-25\,\hat{k}) = -\frac{57 \times 25}{5}\, \cos(\ldots) \,\hat{k}.$

Simplify the constant:

$\frac{25}{5} = 5,$

so we get

$\hat{n} \times \vec{E} = -\frac{57 \times 5}{1} \cos(\ldots)\,\hat{k} = -285 \cos(\ldots) \,\hat{k}.$

Step 3. Apply the relation for the magnetic field:

Now, substituting back into the formula for $\vec{B}$ ,

$\vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E} = -\frac{285}{3\times10^8} \cos(\ldots)\,\hat{k}.$

Notice that

$\frac{285}{3\times10^8} = \frac{57 \times 5}{3\times10^8},$

so we have

$\vec{B} = -\frac{57}{3\times10^8} \cos\Bigl[7.5\times10^6\,t - 5\times10^{-3}(3x+4y)\Bigr](5\,\hat{k})\quad \text{(T)}.$

This expression exactly matches Option D.

Final Answer: Option D