A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is E_y = 9.3 Vm^–1. Then, the magnetic field vector of the wave at that point is-

To determine the magnetic field vector of the electromagnetic wave, we use the relation between the electric field ($E$) and the magnetic field ($B$) in a plane wave traveling in free space.

Step 1: Use the Relation Between $E$ and $B$

For an electromagnetic wave traveling in free space, the magnitudes of the electric and magnetic fields are related by:

$$B = \frac{E}{c}$$

where:

$c = 3.0 \times 10^8$ m/s (speed of light in free space),

$E_y = 9.3$ V/m (given electric field magnitude),

$B$ is the magnitude of the magnetic field.

Step 2: Calculate the Magnetic Field Magnitude

$$B = \frac{9.3}{3.0 \times 10^8}$$

 $$B = 3.1 \times 10^{-8} \ \text{T}$$

Step 3: Determine the Direction of the Magnetic Field

The wave propagates along the $+x$-direction.

The given electric field ($E_y$) is along the $+y$-axis.

Using the right-hand rule for electromagnetic waves: $$\vec{E} \times \vec{B} \text{ gives the direction of wave propagation}.$$

Since $\vec{E}$ is along $+y$ and the wave propagates along $+x$, $\vec{B}$ must be along the $+z$-direction.

Final Answer:

$$B_z = 3.1 \times 10^{-8} \text{ T}$$

So, the magnetic field vector at that point is:

$$\mathbf{B} = (3.1 \times 10^{-8}) \hat{z} \ \text{T}$$