Question

The electric field associated with an e. $m$. wave in vacuum is given by $\vec{ E }=\hat{ i } 40 \cos \left( kz -6 \times 10^{8} t \right)$, where $E , z$ and $t$ are in volt $/ m$, meter and seconds respectively. The value of wave vector $k$ is :

• A. $2\, m^{-1}$
• B. $0.5\, m^{-1}$
• C. $6\, m^{-1}$
• D. $3\, m^{-1}$

Answer 1

The Correct Option is A

To solve the problem, we need to determine the value of the wave vector \( k \) from the given electric field equation associated with an electromagnetic (e.m.) wave in vacuum.

The electric field is given by the equation:

\(\vec{E} = \hat{i} 40 \cos \left( kz - 6 \times 10^8 t \right) \)

In the standard form of a traveling wave, the wave's spatial and temporal components are represented as:

\(\cos(kz - \omega t) \)

where:

• \( k \) is the wave vector,
• \( \omega \) is the angular frequency.

By comparing the given equation:

\(\cos \left( kz - 6 \times 10^8 t \right) \)

to the standard form:

\(\cos(kz - \omega t) \)

We can identify that:

\(\omega = 6 \times 10^8 \) rad/s

For electromagnetic waves in vacuum, the relation between angular frequency \(\omega\), wave vector \(k\), and speed of light \(c\) is given by:

\(\omega = k \cdot c \)

Assuming the wave is traveling in a vacuum, the speed of light \(c\) is approximately \(3 \times 10^8 \, \text{m/s}\). Substituting into the equation we have:

\(\omega = k \cdot 3 \times 10^8 \)

Plug in the value of \(\omega\):

\(6 \times 10^8 = k \cdot 3 \times 10^8 \)

Solve for \(k\):

\(k = \frac{6 \times 10^8}{3 \times 10^8} = 2 \, m^{-1} \)

Therefore, the value of the wave vector \(k\) is \(2 \, m^{-1}\), which matches the correct option.

Option	Value	Correct
\(2 \, m^{-1}\)	Correct \(k\) for the equation	✔
\(0.5 \, m^{-1}\)	Incorrect based on calculation	✘
\(6 \, m^{-1}\)	Incorrect based on calculation	✘
\(3 \, m^{-1}\)	Incorrect based on calculation	✘