Given the electric field of the electromagnetic wave:
\[ \vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right) \, \text{NC}^{-1} \]
In an electromagnetic wave, the magnetic field \(\vec{B}\) is perpendicular to both the electric field \(\vec{E}\) and the direction of propagation.
Since \(\vec{E}\) is along the \(\hat{i}\)-direction and the wave propagates along the \(\hat{k}\)-direction, the magnetic field \(\vec{B}\) must be along the \(\hat{j}\)-direction.
The relationship between the magnitudes of the electric and magnetic fields in an electromagnetic wave is given by:
\[ B = \frac{E}{c} \]
Substituting the given electric field magnitude:
\[ B = \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \]
Thus, the magnetic field is:
\[ \vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \]
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
The unit of $ \sqrt{\frac{2I}{\epsilon_0 c}} $ is: (Where $ I $ is the intensity of an electromagnetic wave, and $ c $ is the speed of light)
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Let $L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $\left| 5\alpha - 11\beta - 8\gamma \right|$ equals: