The magnetic field in a plane electromagnetic wave is $$\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}$$. The corresponding electric field will be :

<p>For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($E$) and the magnetic field ($B$) can be described using the equation:</p> <p>$E = cB$</p> <p>where</p> <ul> <li>$c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \text{m/s}$.</li> <li>$B$ is the magnitude of the magnetic field.</li> </ul> <p>Given the magnetic field $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}$, we can calculate the corresponding electric field magnitude using the formula above:</p> <p>$E = (3.0 \times 10^8) \times (3.5 \times 10^{-7})$</p> <p>$= 105 \, \text{Vm}^{-1}$</p> <p>Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, \text{Vm}^{-1}$. The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation. Given $B_y$, this means $E$ will have components in the $x-z$ plane. Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$-direction, the corresponding electric field component must lie in a plane perpendicular to the $y$-axis, which could be either the $x$ or the $z$ direction.</p> <p>However, knowing electromagnetic wave properties, if the wave is propagating along the $x$-axis and the magnetic field ($B_y$) is along the $y$-axis, then by right-hand rule, the electric field ($E$) must be along the $z$-axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.</p> <p>Therefore, the correct option is:</p> <p>$\text{Option A: } E_z = 105 \sin \left(1.5 \times 10^3 x + 0.5 \times 10^{11} t\right) \, \text{Vm}^{-1}$</p>