The electric field of an electromagnetic wave is given by $\vec{E}(x, t)=E_0 \hat{z} \cos (a x+b t)$...

The electric field of the electromagnetic wave is given by:$ \mathbf{E}(x, t) = E_0 \hat{z} \cos(ax + bt) $In this equation:$ a $ is the wave number, which relates to the wavelength $\lambda$ by the equation $ a = \frac{2\pi}{\lambda} $.$ b $ is the angular frequency, which is connected to the frequency $ f $ by the equation $ b = 2\pi f $.For an electromagnetic wave traveling through a vacuum, the speed of light $ c $ connects the wavelength $\lambda$ and the frequency $ f $ as follows:$ c = \lambda f $Expressing $ c $ in terms of $ a $ and $ b $, we have:$ c = \frac{2\pi}{a} \cdot \frac{b}{2\pi} $Solving for $ b $, we find:$ \begin{aligned} c &= \frac{b}{a} \\ b &= ac \end{aligned} $

Electromagnetic Waves - IAT-IISER Physics PYQs

The electric field of the electromagnetic wave is given by:

%5Cmathbf%7BE%7D(x%2C%20t)%20%3D%20E_0%20%5Chat%7Bz%7D%20%5Ccos(ax%20%2B%20bt)

In this equation:

a

is the wave number, which relates to the wavelength

%5Clambda

by the equation

a%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D

b

is the angular frequency, which is connected to the frequency

f

by the equation

b%20%3D%202%5Cpi%20f

.For an electromagnetic wave traveling through a vacuum, the speed of light

c

connects the wavelength

%5Clambda

and the frequency

f

as follows:

c%20%3D%20%5Clambda%20f

Expressing

c

in terms of

a

and

b

, we have:

c%20%3D%20%5Cfrac%7B2%5Cpi%7D%7Ba%7D%20%5Ccdot%20%5Cfrac%7Bb%7D%7B2%5Cpi%7D

Solving for

b

, we find:

%5Cbegin%7Baligned%7D%20c%20%26%3D%20%5Cfrac%7Bb%7D%7Ba%7D%20%5C%5C%20b%20%26%3D%20ac%20%5Cend%7Baligned%7D

Electromagnetic Waves

High-Yield Trend

Questions
1 MCQs

Official Solution

Electromagnetic Waves

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs