Electromagnetic Waves
168 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
Match List - I with List - II.
Choose the correct answer from the options given below :
| List I | List II | ||
|---|---|---|---|
| (a) | Source of microwave frequency | (i) | Radioactive decay of nucleus |
| (b) | Source of infrared frequency | (ii) | Magnetron |
| (c) | Source of Gamma Rays | (iii) | Inner shell electrons |
| (d) | Source of X-rays | (iv) | Vibration of atoms and molecules |
| (v) | LASER | ||
| (vi) | RC circuit |
Choose the correct answer from the options given below :
A.
(a)-(vi), (b)-(v), (c)-(i), (d)-(iv)
B.
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
C.
(a)-(ii), (b)-(iv), (c)-(vi), (d)-(iii)
D.
(a)-(vi), (b)-(iv), (c)-(i), (d)-(v)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
The electric field in an electromagnetic wave is given by E = (50 NC$-$1) sin$\omega$ (t $-$ x/c)
The energy contained in a cylinder of volume V is 5.5 $\times$ 10$-$12 J. The value of V is _____________ cm3. (given $\in$0 = 8.8 $\times$ 10$-$12C2N$-$1m$-$2)
The energy contained in a cylinder of volume V is 5.5 $\times$ 10$-$12 J. The value of V is _____________ cm3. (given $\in$0 = 8.8 $\times$ 10$-$12C2N$-$1m$-$2)
Correct Answer: 500
Explanation:
$E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)$
Energy density = ${1 \over 2}{ \in _0}E_0^2$
Energy of volume $V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$
${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$
$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$
= .0005 $\times$ 106 (c.m)3
= 500 (c.m)3
Energy density = ${1 \over 2}{ \in _0}E_0^2$
Energy of volume $V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$
${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$
$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$
= .0005 $\times$ 106 (c.m)3
= 500 (c.m)3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be x $\times$ 10$-$8 T. The value of x is ___________.
Correct Answer: 2
Explanation:
$|B|\, = {{|E|} \over C} = {6 \over {3 \times {{10}^8}}}$
= 2 $\times$ 10$-$8 T
$\therefore$ x = 2
= 2 $\times$ 10$-$8 T
$\therefore$ x = 2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
The electric field in a plane electromagnetic wave is given by
$\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat j$. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is ${x \over {{{10}^9}}}{N \over {{m^2}}}$. Find the value of x .
$\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat j$. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is ${x \over {{{10}^9}}}{N \over {{m^2}}}$. Find the value of x .
Correct Answer: 354
Explanation:
E0 = 200
$I = {1 \over 2}{\varepsilon _0}E_0^2.C$
Radiation pressure
$P = {{2I} \over C}$
$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$
$ = {\varepsilon _0}E_0^2$
$ = 8.85 \times {10^{ - 12}} \times {200^2}$
$ = 8.85 \times {10^{ - 8}} \times 4$
$ = {{354} \over {{{10}^9}}}$
$I = {1 \over 2}{\varepsilon _0}E_0^2.C$
Radiation pressure
$P = {{2I} \over C}$
$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$
$ = {\varepsilon _0}E_0^2$
$ = 8.85 \times {10^{ - 12}} \times {200^2}$
$ = 8.85 \times {10^{ - 8}} \times 4$
$ = {{354} \over {{{10}^9}}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $\sqrt {{x \over 5}} $E. Where the value of x = ____________.
Correct Answer: 3
Explanation:
$I = {1 \over 2}C{ \in _0}{E^2}$
${E^2} \propto I$
$I = {{Power} \over {Area}}$
${E^2} \propto {P \over A}$
$E \propto \sqrt P $
${{E'} \over E} = \sqrt {{{60} \over {100}}} $
$E' = \sqrt {{3 \over 5}} E$
So the value of x = 3
${E^2} \propto I$
$I = {{Power} \over {Area}}$
${E^2} \propto {P \over A}$
$E \propto \sqrt P $
${{E'} \over E} = \sqrt {{{60} \over {100}}} $
$E' = \sqrt {{3 \over 5}} E$
So the value of x = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Seawater at a frequency f = 9 $\times$ 102 Hz, has permittivity $\varepsilon $ = 80$\varepsilon $0 and resistivity $\rho$ = 0.25 $\Omega$m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0 sin(2$\pi$ft). Then the conduction current density becomes 10x times the displacement current density after time t = ${1 \over {800}}$s. The value of x is _____________. (Given : ${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$ Nm2C$-$2)
Correct Answer: 6
Explanation:
Given f = 9 $\times$ 102 Hz
$ \in $ = $ \in $0$ \in $r
$ \in $ = 80 $ \in $0
So $ \in $r = 80
$\rho $ = 0.25 $\Omega$m
V(t) = V0 sin (2$\pi$ft)
${I_d} = {{dq} \over {dt}} = {{cdv} \over {dt}}$
${I_d} = {{{ \in _0}{ \in _r}A} \over d}{d \over {dt}}({v_0}\sin (2\pi ft))$
${I_d} = {{{ \in _0}{ \in _r}A} \over d}{V_0}(2\pi f)\cos (2\pi ft)$ .......... (1)
& ${I_c} = {V \over R}$
${I_c} = {{{V_0}\sin (2\pi ft)} \over {\rho {d \over A}}} = {{A{v_0}\sin (2\pi ft)} \over {\rho d}}$ ....... (2)
divide equation (1) and (2)
${{{I_d}} \over {{I_c}}} = { \in _0}{ \in _r}2\pi f(\rho )\cot (2\pi ft)$
${{{I_d}} \over {{I_c}}} = {1 \over {4\pi \times 9 \times {{10}^9}}} \times 80 \times 2\pi \times 9 \times {10^2} \times (0.25) \times \cot (2\pi \times 9 \times {10^2} \times {1 \over {800}})$
$ = {{{{10}^3}} \over {{{10}^9}}}\left( {\cot \left( {{{9\pi } \over 4}} \right)} \right)$
$ = {{{{10}^3}} \over {{{10}^9}}}$
${{{I_d}} \over {{I_c}}} = {1 \over {{{10}^6}}}$
${I_c} = {10^6}{I_d}$
So x = 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If 2.5 $\times$ 10$-$6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm2. (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)
(Assume complete absorption and normal incidence conditions are there)
Correct Answer: 25
Explanation:
Pressure = ${{Intensity} \over C}$ (for absorbing surface)
I = P $\times$ C
I = ${{2.5 \times {{10}^{ - 6}}} \over {30c{m^2}}}$ N $\times$ 3 $\times$ 108 m/s
I = 25 W/cm2
I = P $\times$ C
I = ${{2.5 \times {{10}^{ - 6}}} \over {30c{m^2}}}$ N $\times$ 3 $\times$ 108 m/s
I = 25 W/cm2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x $\times$ 10$-$1 V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take ${\varepsilon _0} = 8.85 \times {10^{ - 12}}$ C2N$-$1 m$-$2, c = $3 \times {10^8}$ ms$-$1]
Correct Answer: 137
Explanation:
Intensity of electro magnetic wave is,
$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$
${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$
${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$
$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$
$E_0^2 = {{375} \over 2} = 187.5$
${E_0} = 13.69$
${E_0} \approx 137 \times {10^{ - 1}}$ v/m
$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$
${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$
${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$
$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$
$E_0^2 = {{375} \over 2} = 187.5$
${E_0} = 13.69$
${E_0} \approx 137 \times {10^{ - 1}}$ v/m
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is ${x \over {10}}\sqrt {{{{\mu _0}c} \over \pi }} {V \over m}$. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.
Correct Answer: 2
Explanation:
Firstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\pi r^2$ where r is the distance from the source. So,
$I = \frac{P}{4\pi r^2} \tag{1}$ ........(1)
This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :
$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$ .........(2)
where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space.
From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :
$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$
Since $ \varepsilon_0=\frac{1}{\mu_0 c^2} $
Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.
So, substituting $P = 0.8\, \text{W}$, $r = 10\, \text{m}$, $c = 3 \times 10^8\, \text{m/s}$, and $\mu_0 = 4\pi \times 10^{-7}\, \text{T m/A}$, we have :
$E_{\text{peak}} = \sqrt{\frac{2 \times 0.8 \times 4\pi \times 10^{-7} \times 3 \times 10^8}{4\pi \times 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$
Comparing with the expression in the question, we find that $x = 2$.
Therefore, the answer is $x = 2$.
$I = \frac{P}{4\pi r^2} \tag{1}$ ........(1)
This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :
$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$ .........(2)
where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space.
From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :
$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$
Since $ \varepsilon_0=\frac{1}{\mu_0 c^2} $
Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.
So, substituting $P = 0.8\, \text{W}$, $r = 10\, \text{m}$, $c = 3 \times 10^8\, \text{m/s}$, and $\mu_0 = 4\pi \times 10^{-7}\, \text{T m/A}$, we have :
$E_{\text{peak}} = \sqrt{\frac{2 \times 0.8 \times 4\pi \times 10^{-7} \times 3 \times 10^8}{4\pi \times 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$
Comparing with the expression in the question, we find that $x = 2$.
Therefore, the answer is $x = 2$.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
The wavelength of an X-ray beam is 10$\mathop A\limits^o $. The mass of a fictitious particle having the same energy as that of the X-ray photons is ${x \over 3}h$ kg. The value of x is __________. (h = Planck's constant)
Correct Answer: 10
Explanation:
Given wavelength of an x-ray beam = 10$\mathop A\limits^o $
$ \because $ E = ${{hc} \over \lambda } = m{c^2}$
$ \Rightarrow $ m = ${h \over {c\lambda }}$
The mass of a fictitious particle having the same energy as that of the x-ray photons = ${x \over 3}$h kg
$ \therefore $ ${x \over 3}h = {h \over {c\lambda }}$
$x = {3 \over {c\lambda }}$
$ = {3 \over {3 \times {{10}^8} \times 10 \times {{10}^{ - 10}}}}$
x = 10
$ \because $ E = ${{hc} \over \lambda } = m{c^2}$
$ \Rightarrow $ m = ${h \over {c\lambda }}$
The mass of a fictitious particle having the same energy as that of the x-ray photons = ${x \over 3}$h kg
$ \therefore $ ${x \over 3}h = {h \over {c\lambda }}$
$x = {3 \over {c\lambda }}$
$ = {3 \over {3 \times {{10}^8} \times 10 \times {{10}^{ - 10}}}}$
x = 10
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _________ $\times$ 10$-$2 cm.
Correct Answer: 667
Explanation:
Given, frequency of wave, f = 3 GHz = 3 $\times$ 109 Hz
Relative permittivity, $\varepsilon $r = 2.25
Since, f = C/$\lambda$
$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$ m
$\because$ $\lambda$m (wavelength of wave in a medium) = $\lambda$/$\mu$ and we know that, $\mu = \sqrt {{\mu _r}{\varepsilon _r}} $
As, dielectric is non-magnetic, $\mu$r = 1
$ \Rightarrow \mu = \sqrt {2.25} = 1.5$
$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$ m
= 6.67 cm = 667 $\times$ 10$-$2 cm
Relative permittivity, $\varepsilon $r = 2.25
Since, f = C/$\lambda$
$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$ m
$\because$ $\lambda$m (wavelength of wave in a medium) = $\lambda$/$\mu$ and we know that, $\mu = \sqrt {{\mu _r}{\varepsilon _r}} $
As, dielectric is non-magnetic, $\mu$r = 1
$ \Rightarrow \mu = \sqrt {2.25} = 1.5$
$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$ m
= 6.67 cm = 667 $\times$ 10$-$2 cm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is ____________ $\times$ 107 m/s.
Correct Answer: 15
Explanation:
Given, $\mu$r = $\varepsilon $r = 2
where, $\mu$r is relative permeability, $\varepsilon $r is relative permittivity.
Speed of electromagnetic wave v is given by
$v = {c \over n}$
where, n = refractive index = $\sqrt {{\mu _r}{\varepsilon _r}} = \sqrt 4 = 2$
$ \Rightarrow v = {{3 \times {{10}^8}} \over 2}$ = 15 $\times$ 107 m/s
$\because$ x $\times$ 107 = 15 $\times$ 107
$\Rightarrow$ x = 15
where, $\mu$r is relative permeability, $\varepsilon $r is relative permittivity.
Speed of electromagnetic wave v is given by
$v = {c \over n}$
where, n = refractive index = $\sqrt {{\mu _r}{\varepsilon _r}} = \sqrt 4 = 2$
$ \Rightarrow v = {{3 \times {{10}^8}} \over 2}$ = 15 $\times$ 107 m/s
$\because$ x $\times$ 107 = 15 $\times$ 107
$\Rightarrow$ x = 15
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
For a plane electromagnetic wave, the magnetic field at a point x and time t is
$\overrightarrow B \left( {x,t} \right)$ = $\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$ T
The instantaneous electric field $\overrightarrow E $ corresponding to $\overrightarrow B $ is :
(speed of light c = 3 × 108 ms–1)
$\overrightarrow B \left( {x,t} \right)$ = $\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$ T
The instantaneous electric field $\overrightarrow E $ corresponding to $\overrightarrow B $ is :
(speed of light c = 3 × 108 ms–1)
A.
$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right]$ ${V \over m}$
B.
$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}$
C.
$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$
D.
$\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The correct match between the entries in
column I and column II are :
| I | II |
|---|---|
| Radiation | Wavelength |
| (a) Microwave | (i) 100 m |
| (b) Gamma rays | (ii) 10–15 m |
| (c) A.M. radio waves | (iii) 10–10 m |
| (d) X-rays | (iv) 10–3 m |
A.
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
B.
(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
C.
(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
D.
(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
An electron is constrained to move along
the y-axis with a speed of 0.1 c (c is the
speed of light) in the presence of
electromagnetic wave, whose electric
field is
$\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right)$ V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 $ \times $ 108 ms–1 and electron charge = 1.6 $ \times $ 10–19 C)
$\overrightarrow E = 30\widehat j\sin \left( {1.5 \times {{10}^7}t - 5 \times {{10}^{ - 2}}x} \right)$ V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 $ \times $ 108 ms–1 and electron charge = 1.6 $ \times $ 10–19 C)
A.
4.8 $ \times $ 10–19 N
B.
2.4 $ \times $ 10–18 N
C.
3.2 $ \times $ 10–18 N
D.
1.6 $ \times $ 10–18 N
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The electric field of a plane electromagnetic wave is given by
$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$
Its magnetic field will be given by :
$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$
Its magnetic field will be given by :
A.
${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$
B.
${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$
C.
${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$
D.
${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Choose the correct option relating wave lengths of different parts of electromagnetic wave spectrum:
A.
$\lambda $radio waves > $\lambda $micro waves > $\lambda $visible > $\lambda $x-rays
B.
$\lambda $visible > $\lambda $x-rays > $\lambda $radio waves > $\lambda $micro waves
C.
$\lambda $visible < $\lambda $micro waves < $\lambda $radio waves < $\lambda $x-rays
D.
$\lambda $x-rays
< $\lambda $micro waves < $\lambda $radio waves < $\lambda $visible
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$.
The magnetic field $\overrightarrow B $ , at the moment t = 0 is :
$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$.
The magnetic field $\overrightarrow B $ , at the moment t = 0 is :
A.
$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j$
B.
$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k$
C.
$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k$
D.
$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The magnetic field of a plane electromagnetic wave is
$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$ T
where c = 3 $ \times $ 108 ms–1 is the speed of light. The corresponding electric field is :
$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$ T
where c = 3 $ \times $ 108 ms–1 is the speed of light. The corresponding electric field is :
A.
$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$ V/m
B.
$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$ V/m
C.
$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$ V/m
D.
$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
In a plane electromagnetic wave, the directions
of electric field and magnetic field are
represented by $\widehat k$ and $2\widehat i - 2\widehat j$, respectively.
What is the unit vector along direction of
propagation of the wave?
A.
${1 \over {\sqrt 5 }}\left( {\widehat i + 2\widehat j} \right)$
B.
${1 \over {\sqrt 5 }}\left( {2\widehat i + \widehat j} \right)$
C.
${1 \over {\sqrt 2 }}\left( {\widehat i + \widehat j} \right)$
D.
${1 \over {\sqrt 2 }}\left( {\widehat j + \widehat k} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
A plane electromagnetic wave, has
frequency of 2.0 $ \times $ 1010 Hz and its energy density is 1.02 $ \times $ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( ${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$ and speed of light
= 3 $ \times $ 108 ms–1)
frequency of 2.0 $ \times $ 1010 Hz and its energy density is 1.02 $ \times $ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( ${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$ and speed of light
= 3 $ \times $ 108 ms–1)
A.
190 nT
B.
150 nT
C.
160 nT
D.
180 nT
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
A plane electromagnetic wave is propagating
along the direction
${{\widehat i + \widehat j} \over {\sqrt 2 }}$
, with its polarization
along the direction $\widehat k$ . The correct form of the
magnetic field of the wave would be (here B0
is an appropriate constant) :
A.
${B_0}{{\widehat i - \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$
B.
${B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$
C.
${B_0}{{\widehat j - \widehat i} \over {\sqrt 2 }}\cos \left( {\omega t + k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$
D.
${B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The electric fields of two plane electromagnetic
plane waves in vacuum are given by
$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$ and
$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$
At t = 0, a particle of charge q is at origin with
a velocity $\overrightarrow v = 0.8c\widehat j$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$ and
$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$
At t = 0, a particle of charge q is at origin with
a velocity $\overrightarrow v = 0.8c\widehat j$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
A.
${E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)$
B.
${E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)$
C.
${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$
D.
${E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
A plane electromagnetic wave of frequency
25 GHz is propagating in vacuum along the
z-direction. At a particular point in space and
time, the magnetic field is given by $\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT$. The corresponding electric field $\overrightarrow E $ is (speed of light c = 3 × 108 ms–1)
A.
15 $\widehat i$V / m
B.
-15 $\widehat i$V / m
C.
1.66 × 10–16 $\widehat i$V / m
D.
-1.66 × 10–16 $\widehat i$V / m
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The electric field of a plane electromagnetic wave is given by
$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$
At t = 0, a positively charged particle is at the point (x, y, z) = $\left( {0,0,{\pi \over k}} \right)$.
If its instantaneous velocity at (t = 0) is ${v_0}\widehat k$ , the force acting on it due to the wave is :
$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$
At t = 0, a positively charged particle is at the point (x, y, z) = $\left( {0,0,{\pi \over k}} \right)$.
If its instantaneous velocity at (t = 0) is ${v_0}\widehat k$ , the force acting on it due to the wave is :
A.
parallel to $\widehat k$
B.
parallel to ${{\widehat i + \widehat j} \over {\sqrt 2 }}$
C.
antiparallel to ${{\widehat i + \widehat j} \over {\sqrt 2 }}$
D.
zero
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
If the magnetic field in a plane electromagnetic wave is given by
$\overrightarrow B $ = 3 $ \times $ 10-8 sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat j$ T, then what will be expression for electric field ?
$\overrightarrow B $ = 3 $ \times $ 10-8 sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat j$ T, then what will be expression for electric field ?
A.
$\overrightarrow E $ = (9sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat k$ V/m)
B.
$\overrightarrow E $ = (60sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat k$ V/m)
C.
$\overrightarrow E $ = (3 $ \times $ 10-8 sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat i$ V/m)
D.
$\overrightarrow E $ = (3 $ \times $ 10-8 sin(1.6 $ \times $ 103x + 48 $ \times $ 1010t)$\widehat j$ V/m)
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
Suppose that intensity of a laser is ${{315} \over \pi }$ W/m2.
The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.
$ \in $0 = 8.86 × 10–12 C2 Nm–2; c = 3 × 108 ms–1)
The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.
$ \in $0 = 8.86 × 10–12 C2 Nm–2; c = 3 × 108 ms–1)
Correct Answer: 194
Explanation:
I = ${1 \over 2}$$\varepsilon $0$E_0^2$c
$ \Rightarrow $ E0 = $\sqrt {{{2I} \over {{\varepsilon _0}c}}} $
$ \therefore $ Erms = ${{{E_0}} \over {\sqrt 2 }}$ = $\sqrt {{I \over {{\varepsilon _0}c}}} $
= $\sqrt {{{{{315} \over \pi }} \over {8.86 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} $
= 194
$ \Rightarrow $ E0 = $\sqrt {{{2I} \over {{\varepsilon _0}c}}} $
$ \therefore $ Erms = ${{{E_0}} \over {\sqrt 2 }}$ = $\sqrt {{I \over {{\varepsilon _0}c}}} $
= $\sqrt {{{{{315} \over \pi }} \over {8.86 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} $
= 194
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in
free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable
magnetic field component in the electromagnetic wave ?
A.
$\overrightarrow B $ = 2 × 10–7
sin(1.5 × 102
x + 0.5 × 1011t) $\widehat j$
B.
$\overrightarrow B $ = 60
sin(0.5 × 103x + 0.5 × 1011t) $\widehat k$
C.
$\overrightarrow B $ = 2 × 10–7
sin(0.5 × 103
z + 1.5 × 1011t) $\widehat i$
D.
$\overrightarrow B $ = 2 × 10–7
sin(0.5 × 103
z - 1.5 × 1011t) $\widehat i$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
An electromagnetic wave is represented by the electric field $\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]$
. Taking unit
vectors in x, y and z directions to be $\widehat i,\widehat j,\widehat k$
, the direction of propagation $\widehat s$, is :
A.
$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$
B.
$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$
C.
$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$
D.
$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface
has an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be :
A.
6.3 × 10–4 Ns
B.
5.0 × 10–3 Ns
C.
1.4 × 10–6 Ns
D.
3.5 × 10–6 Ns
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
The electric field of a plane electromagnetic
wave is given by
$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$
The corresponding magnetic field $\overrightarrow B $ is then given by
$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$
The corresponding magnetic field $\overrightarrow B $ is then given by
A.
$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)$
B.
$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)$
C.
$\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)$
D.
$\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
50 W/m2 energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1m2 surface area will be close to
(c = 3 × 108 m/s) :-
A.
20 × 10–8 N
B.
35 × 10–8 N
C.
10 × 10–8 N
D.
15 × 10–8 N
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The magnetic field of a plane electromagnetic
wave is given by :
$$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
$$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
A.
0.6 N
B.
0.9 N
C.
3 × 10–2 N
D.
0.1 N
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The magnetic field of an electromagnetic wave
is given by :-
$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$
The associated electric field will be :-
$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$
The associated electric field will be :-
A.
$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$
B.
$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$
C.
$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}$
D.
$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
A plane electromagnetic wave travels in free
space along the x-direction. The electric field
component of the wave at a particular point of
space and time is E = 6 V m–1 along y-direction.
Its corresponding magnetic field component,
B would be :
A.
2 × 10–8 T along y-direction
B.
6 × 10–8 T along z-direction
C.
2 × 10–8 T along z-direction
D.
6 × 10–8 T along x-direction
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The mean intensity of radiation on the surface of the Sun is about 108 W/m2 . The rms value of the corresponding magnetic field is closet to :
A.
102 T
B.
10$-$2 T
C.
10$-$4 T
D.
1 T
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
A.
6 V/m
B.
10 V/m
C.
30 V/m
D.
24 V/m
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :
[Given permittivity of space $ \in $0 = 9 $ \times $ 10–12 SI units, Speed of light c = 3 $ \times $ 108 m/s]
[Given permittivity of space $ \in $0 = 9 $ \times $ 10–12 SI units, Speed of light c = 3 $ \times $ 108 m/s]
A.
2 kV/m
B.
1 kV/m
C.
1.4 kV/m
D.
0.7 kV/m
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:
A.
$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$
B.
$\left( {\sqrt n ,\sqrt n } \right)$
C.
$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$
D.
$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$ The magnetic field $\overrightarrow B $(x,z, t) is given by $-$ (c is the velocity of light)
A.
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
B.
${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
C.
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$
D.
${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6
sin $\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$ then the maximum electric field associated with it is -
A.
4.5 $ \times $ 104 N/C
B.
4 $ \times $ 104 N/C
C.
6 $ \times $ 104 N/C
D.
3 $ \times $ 104 N/C
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
A.
${U_E} = {{{U_B}} \over 2}$
B.
${U_E} > {U_B}$
C.
${U_E} < {U_B}$
D.
${U_E} = {U_B}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $\overrightarrow E = 6.3\widehat j\,V/m.$ The corresponding magnetic field $\overrightarrow {B,} $ at that point will be :
A.
18.9 $ \times $ 10$-$8 $\widehat k$T
B.
2.1 $ \times $ 10$-$8 $\widehat k$T
C.
6.3 $ \times $ 10$-$8 $\widehat k$T
D.
18.9 $ \times $ 108 $\widehat k$T
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
A plane electromagnetic wave of wavelength $\lambda $ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by :
A.
B.
C.
D.
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
An EM wave from air enters a medium. The electric fields are
$\overrightarrow {{E_1}} $ = ${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$ in air and
$\overrightarrow {{E_2}} $ = ${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$ in medium,
where the wave number k and frequency $\nu $ refer to their values in air. The medium is non-magnetic. If ${\varepsilon _{{r_1}}}$ and ${\varepsilon _{{r_2}}}$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
$\overrightarrow {{E_1}} $ = ${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$ in air and
$\overrightarrow {{E_2}} $ = ${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$ in medium,
where the wave number k and frequency $\nu $ refer to their values in air. The medium is non-magnetic. If ${\varepsilon _{{r_1}}}$ and ${\varepsilon _{{r_2}}}$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
A.
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$
B.
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$
C.
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$
D.
${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :
A.
${{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$
B.
${{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$
C.
${{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$
D.
${1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}$
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
A monochromatic beam of light has a frequency $v = {3 \over {2\pi }} \times {10^{12}}Hz$ and is propagating along the direction ${{\widehat i + \widehat j} \over {\sqrt 2 }}.$
It is polarized along the $\widehat k$ direction. The acceptable form for the magnetic field is :
It is polarized along the $\widehat k$ direction. The acceptable form for the magnetic field is :
A.
B.
C.
D.
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
The electric field component of a monochromatic radiation is given by
$\overrightarrow E $ = 2 E0 $\widehat i$ cos kz cos $\omega $t
Its magnetic field $\overrightarrow B $ is then given by :
$\overrightarrow E $ = 2 E0 $\widehat i$ cos kz cos $\omega $t
Its magnetic field $\overrightarrow B $ is then given by :
A.
${{2{E_0}} \over c}$ $\widehat j$ sin kz cos $\omega $t
B.
$-$ ${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega $t
C.
${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega $t
D.
${{2{E_0}} \over c}$ $\widehat j$ cos kz cos $\omega $t
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 8th April Morning Slot
Magnetic field in a plane electromagnetic wave is given by
$\overrightarrow B $ = B0 sin (k x + $\omega $t) $\widehat j\,T$
Expression for corresponding electric field will be :
Where c is speed of light.
$\overrightarrow B $ = B0 sin (k x + $\omega $t) $\widehat j\,T$
Expression for corresponding electric field will be :
Where c is speed of light.
A.
$\overrightarrow E $ = B0 c sin (k x + $\omega $t) $\widehat k$ V/m
B.
$\overrightarrow E $ = ${{{B_0}} \over c}$ sin (k x + $\omega $t) $\widehat k$ V/m
C.
$\overrightarrow E $ = $-$ B0 c sin (kx +$\omega $t) $\widehat k$ V/m
D.
$\overrightarrow E $ = B0 c sin (kx $-$$\omega $t) $\widehat k$ V/m
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
Consider an electromagnetic wave propagating in vacuum. Choose the correct
statement :
A.
For an electromagnetic wave propagating in +x direction the electric field is $\vec E = {1 \over {\sqrt 2 }}{E_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y - \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)$
B.
For an electromagnetic wave propagating in +x direction the electric field is
$\vec E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$
C.
For an electromagnetic wave propagating in + y direction the electric field is
$\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat y$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$
and the magnetic field is $\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$
D.
For an electromagnetic wave propagating in + y direction the electric field is
$\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$
and the magnetic field is $\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat y$
and the magnetic field is $\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat y$