Electromagnetic waves with frequencies ranging from $2\,\text{MHz}$ to $30\,\text{MHz}$ are propagated by:

For a given electromagnetic wave,the magnitude of the electric field is $6.6 \,V \,m^{-1}$ at a point in space. The magnitude of the magnetic field at this point is . . . . . . $T$.

$A$ plane electromagnetic wave travels in free space along the $z$-axis. At a particular point in space, the electric field along the $x$-axis is $8.7 \ V \ m^{-1}$. The magnetic field along the $y$-axis is:

The electric field in an electromagnetic wave is given by $\overrightarrow{E} = \hat{i} 40 \cos \omega(t - \frac{z}{c}) \text{ N/C}$. The magnetic field induction of this wave is (in $SI$ units):

$A$ plane $EM$ wave travelling in vacuum along $z$-direction is given by $\vec E = E_0 \sin(kz - \omega t) \hat i$ and $\vec B = B_0 \sin(kz - \omega t) \hat j$.
$(i)$ Evaluate $\int \vec E \cdot d\vec l$ over the rectangular loop $1234$ shown in the figure.
$(ii)$ Evaluate $\int \vec B \cdot d\vec s$ over the surface bounded by loop $1234$.
$(iii)$ Use $\int \vec E \cdot d\vec l = -\frac{d\phi_E}{dt}$ to prove $\frac{E_0}{B_0} = c$.
$(iv)$ By using a similar process and the equation $\int \vec B \cdot d\vec l = \mu_0 I + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$,prove that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

$A$ radiation is emitted by a $1000 \, W$ bulb and it generates an electric field and magnetic field at point $P$, placed at a distance of $2 \, m$. The efficiency of the bulb is $1.25 \%$. The value of the peak electric field at $P$ is $x \times 10^{-1} \, V/m$. The value of $x$ is (rounded off to the nearest integer).
[Take $\varepsilon_{0} = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}, c = 3 \times 10^8 \, m/s$]