
The electric field of an electromagnetic wave in free space is given by \[E = 10\sin \left( {{{10}^7}t + kx} \right)\hat jV/m\], where \[t\] and \[x\] are in seconds and metres respectively. It can be inferred that
(a) The wavelength \[\lambda \] is \[188.4m\]
(b) The wave number \[k\] is \[0.033rad/m\]
(c) The wave amplitude is 10 V/m
(d) The wave is propagating along +x direction
Which one of the following pairs of statements is correct?
(A) (a) and (b)
(B) (b) and (c)
(C) (a) and (c)
(D) (c) and (d)
Answer
165.3k+ views
Hint: The amplitude is the maximum value of the electric field, and sine of an angle is either equal to or greater than one. The speed of an electromagnetic wave in one form can be given as the angular frequency divided by the wave number.
Formula used: In this solution we will be using the following formulae;
\[c = \dfrac{\omega }{k}\] where \[c\] is the speed of light (also EM wave), \[\omega \] is the angular frequency, \[k\] is the wave number of the wave. \[k = \dfrac{{2\pi }}{\lambda }\] where \[\lambda \] is the wavelength of the wave.
\[y = A\sin \left( {kx + \omega t} \right)\] where \[y\] is the instantaneous displacement at a particular time \[t\] and position \[x\], \[A\] is the amplitude of the wave. The equation is an equation of progressive waves moving in the negative x direction.
Complete Step-by-Step Solution:
To identify which of the statements is true, we need to identify each of the quantities stated in the options. To do so, we shall recall the general equation of a progressive wave given by
\[y = A\sin \left( {kx + \omega t} \right)\] where \[y\] is the instantaneous displacement at a particular time \[t\] and position \[x\], \[A\] is the amplitude of the wave. This, though, is for a wave moving in the negative x direction.
By comparing with the given equation \[E = 10\sin \left( {{{10}^7}t + kx} \right)\hat jV/m\], we see that:
The wave must be travelling in the negative x direction, hence statement (d) is false.
The amplitude \[A = 10V/m\], hence statement (c) is correct
The angular frequency is \[{10^7}{s^{ - 1}}\]
To find the wavenumber and wavelength, we note that
\[c = \dfrac{\omega }{k}\] where \[c\] is the speed of light (also EM wave), \[\omega \] is the angular frequency, \[k\] is the wave number of the wave.
Hence,
\[3 \times {10^8} = \dfrac{{{{10}^7}}}{k}\]
\[ \Rightarrow k = \dfrac{{{{10}^7}}}{{3 \times {{10}^8}}} = 0.033{m^{ - 1}}\], hence, statement (b) is false.
The wavelength is
\[k = \dfrac{{2\pi }}{\lambda }\]
Hence,
\[0.033 = \dfrac{{2\pi }}{\lambda }\]
\[ \Rightarrow \lambda = \dfrac{{2\pi }}{{0.033}} = 188.49m\]
Hence, roughly, statement (a) is true.
Hence, the correct option is C
Note: For exam purposes, since other options but option (a) have been proven false, it is unnecessary to calculate the wavelength, because the options are given in pairs and the only possible answer would be statement (a) and (c), which is option C.
Formula used: In this solution we will be using the following formulae;
\[c = \dfrac{\omega }{k}\] where \[c\] is the speed of light (also EM wave), \[\omega \] is the angular frequency, \[k\] is the wave number of the wave. \[k = \dfrac{{2\pi }}{\lambda }\] where \[\lambda \] is the wavelength of the wave.
\[y = A\sin \left( {kx + \omega t} \right)\] where \[y\] is the instantaneous displacement at a particular time \[t\] and position \[x\], \[A\] is the amplitude of the wave. The equation is an equation of progressive waves moving in the negative x direction.
Complete Step-by-Step Solution:
To identify which of the statements is true, we need to identify each of the quantities stated in the options. To do so, we shall recall the general equation of a progressive wave given by
\[y = A\sin \left( {kx + \omega t} \right)\] where \[y\] is the instantaneous displacement at a particular time \[t\] and position \[x\], \[A\] is the amplitude of the wave. This, though, is for a wave moving in the negative x direction.
By comparing with the given equation \[E = 10\sin \left( {{{10}^7}t + kx} \right)\hat jV/m\], we see that:
The wave must be travelling in the negative x direction, hence statement (d) is false.
The amplitude \[A = 10V/m\], hence statement (c) is correct
The angular frequency is \[{10^7}{s^{ - 1}}\]
To find the wavenumber and wavelength, we note that
\[c = \dfrac{\omega }{k}\] where \[c\] is the speed of light (also EM wave), \[\omega \] is the angular frequency, \[k\] is the wave number of the wave.
Hence,
\[3 \times {10^8} = \dfrac{{{{10}^7}}}{k}\]
\[ \Rightarrow k = \dfrac{{{{10}^7}}}{{3 \times {{10}^8}}} = 0.033{m^{ - 1}}\], hence, statement (b) is false.
The wavelength is
\[k = \dfrac{{2\pi }}{\lambda }\]
Hence,
\[0.033 = \dfrac{{2\pi }}{\lambda }\]
\[ \Rightarrow \lambda = \dfrac{{2\pi }}{{0.033}} = 188.49m\]
Hence, roughly, statement (a) is true.
Hence, the correct option is C
Note: For exam purposes, since other options but option (a) have been proven false, it is unnecessary to calculate the wavelength, because the options are given in pairs and the only possible answer would be statement (a) and (c), which is option C.
Recently Updated Pages
Transistor as Amplifier: Working, Diagram, Uses & Questions

Moving Charges and Magnetism: Laws, Formulas & Applications

Environmental Chemistry Chapter for JEE Main Chemistry

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Get P Block Elements for JEE Main 2025 with clear Explanations

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Uniform Acceleration

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Wheatstone Bridge for JEE Main Physics 2025
