The electric field in a plane electromagnetic wave is given by

$E_z = 60 \cos (5x + 1.5 \times 10^9 t)V/m$.

Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):

Solution:

The given electric field of a plane electromagnetic wave is $E_z = 60 \cos (5x + 1.5 \times 10^9 t)V/m$.

From this equation, we can identify:

1. Amplitude of the electric field, $E_0 = 60 V/m$.

2. Wave number, $k = 5 rad/m$.

3. Angular frequency, $\omega = 1.5 \times 10^9 rad/s$.

For a plane electromagnetic wave, the electric field ($\vec{E}$), magnetic field ($\vec{B}$), and the direction of propagation ($\vec{v}$) are mutually perpendicular. The direction of propagation is given by $\vec{v} \propto \vec{E} \times \vec{B}$.

The wave is given by a function of $(kx + \omega t)$, which indicates that the wave is propagating in the negative x-direction. So, $\vec{v}$ is along $-\hat{i}$.

The electric field is along the z-direction, i.e., $\vec{E} = E_z \hat{k}$.

We need to find the direction of $\vec{B}$ such that $\hat{k} \times \vec{B}$ is along $-\hat{i}$.

Using the right-hand rule or vector cross product rules:

$\hat{k} \times \hat{j} = -\hat{i}$

Therefore, the magnetic field must be along the y-direction, i.e., $B_y$.

Now, let's find the amplitude of the magnetic field, $B_0$. The relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave is $E_0 = c B_0$, where $c$ is the speed of light in vacuum ($c = 3 \times 10^8 m/s$).

$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 20 \times 10^{-8} = 2 \times 10^{-7} T$.

Since the electric and magnetic fields in an electromagnetic wave are in phase, the functional form (cosine) and the argument $(5x + 1.5 \times 10^9 t)$ will be the same for the magnetic field.

Combining the amplitude, direction, and phase, the expression for the magnetic field is:

$B_y = 2 \times 10^{-7} \cos (5x + 1.5 \times 10^9 t)T$.

Comparing this with the given options, option (1) matches our derived expression.