The electric field in a plane electromagnetic wave is given by E_z = 60 cos (5x + 1.5 ×10⁹ t)V/m. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

The correct option is: B_y = 2 × 10^–7 cos (5x + 1.5 × 10⁹t)T

Step-by-step Derivation:

1. Understanding the Wave Propagation and Direction:
The given electric field expression is:
$E_z=60\cos (5x+1.5\times {10}^{9}t)\text{V/m}$
This tells us two important things about the directions of the fields:
• The electric field $\overrightarrow{E}$ is directed along the z-axis (indicated by $E_z$). Therefore, we can write:
$\overrightarrow{E}=E_z\hat{k}$
• The argument of the cosine function is $(5x+1.5\times {10}^{9}t)$. Since the spatial coordinate in the argument is $x$ and the sign between the spatial and temporal terms is positive (+), the electromagnetic wave propagates along the negative x-direction. Thus, the unit vector of wave propagation is:
$\hat{n}=-\hat{i}$

2. Finding the Direction of the Magnetic Field:
In a plane electromagnetic wave, the direction of wave propagation is given by the cross product of the electric field and the magnetic field:
$\hat{n}=\hat{e}\times \hat{b}$
where $\hat{e}$ is the unit vector of the electric field, and $\hat{b}$ is the unit vector of the magnetic field.
Substituting the known unit vectors:
$-\hat{i}=\hat{k}\times \hat{b}$
Using the vector cross-product relations ($\hat{k}\times \hat{j}=-\hat{i}$), we find that:
$\hat{b}=\hat{j}$
This means the magnetic field is directed along the positive y-axis ($B_y$).

3. Determining the Amplitude of the Magnetic Field:
The relation between the amplitudes of the electric field ($E_0$) and the magnetic field ($B_0$) in free space is given by:
$B_0=\frac{E_0}{c}$
where:
• $E_0=60\text{V/m}$ is the amplitude of the electric field.
• $c$ is the speed of light in vacuum. We can calculate $c$ from the wave parameter ratio $c=\frac{\omega }{k}$.
From the given expression, the angular frequency is $\omega =1.5\times {10}^9{\text{rad s}}^{-1}$ and the wave number is $k=5{\text{m}}^{-1}$:
$c=\frac{1.5\times {10}^9}{5}=3\times {10}^8\text{m/s}$
Now, calculate the magnetic field amplitude $B_0$:
$B_0=\frac{60}{3\times {10}^8}=2\times {10}^{-7}\text{T}$

4. Writing the Final Expression:
The magnetic field must oscillate in phase with the electric field. Therefore, it will have the same cosine phase factor:
$B_y=2\times {10}^{-7}\cos (5x+1.5\times {10}^{9}t)\text{T}$