$\text{The electric field in a plane electromagnetic wave is given by}$
$E_z=60\cos (5x+1.5\times {10}^{9}t)\text{V/m.}$
$\text{Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):}$

To find the corresponding magnetic field of the plane electromagnetic wave, we can analyze the given equation for the electric field:

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

From this equation, we can identify the following parameters of the wave:

1. The wave propagates along the negative x-direction, since the phase term is of the form $kx+\omega t$ where the propagation vector is $\overrightarrow{k}=-5\hat{i}$ rad/m.
2. The angular frequency is $\omega =1.5\times {10}^9$ rad/s.
3. The amplitude of the electric field is $E_0=60$ V/m.
4. The electric field oscillates along the z-axis (indicated by the subscript in $E_z$), so its direction is $\overrightarrow{E}=E_z\hat{k}$.

Now, let's determine the direction of the magnetic field $\overrightarrow{B}$. For an electromagnetic wave, the direction of propagation is given by the cross product of the electric field unit vector and the magnetic field unit vector:
$\hat{c}=\hat{E}\times \hat{B}$

Substituting the unit vectors we know:
$-\hat{i}=\hat{k}\times \hat{B}$

Using vector cross-product rules, we know that $\hat{k}\times \hat{j}=-\hat{i}$. Therefore, the magnetic field vector must point along the positive y-direction ($\hat{j}$). This means the magnetic field has only a y-component, $B_y$.

Next, we calculate the amplitude of the magnetic field ($B_0$). The relationship between the electric field amplitude and the magnetic field amplitude in a medium is:
$v=\frac{\omega }{k}=\frac{E_0}{B_0}$

Let's find the speed of the wave $v$:
$v=\frac{1.5\times {10}^9}{5}=3\times {10}^8\text{ m/s}$

Now, calculate $B_0$:
$B_0=\frac{E_0}{v}=\frac{60}{3\times {10}^8}=2\times {10}^{-7}\text{ T}$

Since the magnetic field is in phase with the electric field, it shares the same cosine argument. Writing the expression for $B_y$ gives:
$B_y=2\times {10}^{-7}\cos (5x+1.5\times {10}^{9}t)\text{ T}$