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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:
1. $6\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>$ along z-direction
2. $6\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>$ along x-direction
3. $2\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>$ along z-direction
4. $2\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>$ along y-direction

The magnetic field of an electromagnetic wave is given by :-
$\overrightarrow{\mathrm{B}}=1.6\times {10}^{-6}<mspace\; width="1em"></mspace>\cos \left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(2\hat{i}+\hat{j}\right)\frac{Wb}{m^2}$
The associated electric field will be :-
1. $\overrightarrow{E}=4.8\times {10}^2<mspace\; width="1em"></mspace>\cos \left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(\hat{i}-2\hat{j}\right)\frac{V}{m}$
2. $\overrightarrow{E}=4.8\times {10}^2<mspace\; width="1em"></mspace>\cos \left(2\times {10}^{7}z\mathit{-}6\times {10}^{15}t\right)\left(2\hat{i}+\hat{j}\right)\frac{V}{m}$
3. $\overrightarrow{E}=4.8\times {10}^2<mspace\; width="1em"></mspace>\cos \left(2\times {10}^{7}z\mathit{-}6\times {10}^{15}t\right)\left(-2\hat{j}+\hat{i}\right)\frac{V}{m}$
4. $\overrightarrow{E}=4.8\times {10}^2<mspace\; width="1em"></mspace>\cos \left(2\times {10}^{7}z+6\times {10}^{15}t\right)\left(-\hat{i}+2\hat{j}\right)\frac{V}{m}$

50 W/m² intensity 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 1 m² surface area will be close to 
$\left(\mathrm{c}=3\times {10}^8<mspace\; width="1em"></mspace>\mathrm{m}/\mathrm{s}\right)$ :-
1. $15\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{N}$
2. $35\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{N}$
3. $10\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{N}$
4. $20\times {10}^{-8}<mspace\; width="1em"></mspace>\mathrm{N}$

The magnetic field of a plane electromagnetic wave is given by:
$\overrightarrow{\mathrm{B}}=B_0\hat{i}\left[\cos \left(kz-\omega t\right)\right]+B_1\hat{j}\left[\cos \left(kz\mathit{+}\omega t\right)\right]$ where ${\mathrm{B}}_0=3\times {10}^{-5}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{B}}_1=2\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{T}$. The rms value of the force experienced by a stationary charge Q = 10^-4 C at z  =0 closest to:
1. 0.9 N
2. 0.1 N
3. $3\times {10}^{-2}<mspace\; width="1em"></mspace>\mathrm{N}$
4. 0.6 N

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 $\overrightarrow{\mathrm{E}}=6.3<mspace\; width="1em"></mspace>\hat{j}<mspace\; width="1em"></mspace>V/m$. The corresponding magnetic field $\overrightarrow{\mathrm{B}}$, at that point will be 
1. $18.9\times {10}^{-8}<mspace\; width="1em"></mspace>\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\mathrm{T}$
2. $6.3\times {10}^{-8}<mspace\; width="1em"></mspace>\hat{\mathrm{k}}\mathrm{T}$
3. $2.1\times {10}^{-8}<mspace\; width="1em"></mspace>\hat{\mathrm{k}}\mathrm{T}$
4. $18.9\times {10}^{-8}<mspace\; width="1em"></mspace>\hat{\mathrm{k}}\mathrm{T}$

The energy associated with electric field is (U_E)
and with magnetic field is (U_B) for an
electromagnetic wave in free space. Then
1. ${\mathrm{U}}_{\mathrm{E}}=\frac{{\mathrm{U}}_{\mathrm{B}}}{2}$
2. ${\mathrm{U}}_{\mathrm{E}}<{\mathrm{U}}_{\mathrm{B}}$
3. ${\mathrm{U}}_{\mathrm{E}}={\mathrm{U}}_{\mathrm{B}}$
4. ${\mathrm{U}}_{\mathrm{E}}>{\mathrm{U}}_{\mathrm{B}}$
 

The electric field of a plane electromagnetic wave is given by
$\overrightarrow{\mathrm{E}}=E_0\hat{i}<mspace\; width="1em"></mspace>\cos \left(kz\right)\cos \left(\omega t\right)$
The corresponding magnetic field $\overrightarrow{\mathrm{B}}$ is then given by :
1. $\overrightarrow{\mathrm{B}}=\frac{E_0}{C}\hat{j}<mspace\; width="1em"></mspace>\sin \left(kz\right)\cos \left(\omega t\right)$
2. $\overrightarrow{\mathrm{B}}=\frac{E_0}{C}\hat{j}<mspace\; width="1em"></mspace>\cos \left(kz\right)\cos \left(\omega t\right)$
3. $\overrightarrow{\mathrm{B}}=\frac{E_0}{C}\hat{k}<mspace\; width="1em"></mspace>\sin \left(kz\right)\cos \left(\omega t\right)$
4. $\overrightarrow{\mathrm{B}}=\frac{E_0}{C}\hat{j}<mspace\; width="1em"></mspace>\cos \left(kz\right)\sin \left(\omega t\right)$

Light is incident normally on a completely absorbing surface with an energy flux of 25 $\mathrm{W}<mspace\; width="1em"></mspace>{\mathrm{cm}}^{-2}$, if the surface has an area of 25 ${\mathrm{cm}}^{-2}$, the momentum transferred to the surface in 40 min time duration will be :
1. $5.0\times {10}^{-3}<mspace\; width="1em"></mspace>\mathrm{Ns}$
2. $3.5\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{Ns}$
3. $1.4\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{Ns}$
4. $6.3\times {10}^{-4}<mspace\; width="1em"></mspace>\mathrm{Ns}$

The electric field of a plane polarized electromagnetic wave in free space at time t= 0 is given by an expression
$\overrightarrow{\mathrm{E}}\left(x,y\right)=10\hat{j}\cos \left[\left(6x+8z\right)\right]$
The magnetic field $\overrightarrow{\mathrm{B}}\left(x,z,t\right)$ is given by : (c is the velocity of light)
1. $\frac{1}{\mathrm{c}}\left(6\hat{k}+8\hat{i}\right)\cos \left[\left(6x-8z+10ct\right)\right]$
2. $\frac{1}{\mathrm{c}}\left(6\hat{k}-8\hat{i}\right)\cos \left[\left(6x\mathit{+}8z\mathit{-}10ct\right)\right]$
3. $\frac{1}{\mathrm{c}}\left(6\hat{k}+8\hat{i}\right)\cos \left[\left(6x\mathit{+}8z\mathit{-}10ct\right)\right]$
4. $\frac{1}{\mathrm{c}}\left(6\hat{k}-8\hat{i}\right)\cos \left[\left(6x\mathit{+}8z+10ct\right)\right]$

If the magnetic field of a plane electromagnetic wave
is given by (The speed of light = 3 × 10⁸/m/s)
$\mathrm{B}=100\times {10}^{-6}<mspace\; width="1em"></mspace>\sin \left[2\mathrm{\pi }\times 2\times {10}^{15}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)\right]$ then the maximum electric field associated with it is :
1. 4 × 10⁴ N/C
2. 4.5 × 10⁴ N/C
3. 6 × 10⁴ N/C
4. 3 × 10⁴ N/C

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 × 10–12 SI units, Speed of light
c = 3 × 10⁸ m/s]:-

1. 1 kV/m
2. 2 kV/m
3. 1.4 kV/m
4. 0.7 kV/m 

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 electrric fields, and the ratio of the
magnitudes of magnetic fields of the wave
before and after entering into the medium are
respectively, given by :
1. $\left(\frac{1}{\sqrt{\mathrm{n}}},\frac{1}{\sqrt{\mathrm{n}}}\right)$
2. $\left(\sqrt{\mathrm{n}},\frac{1}{\sqrt{\mathrm{n}}}\right)$
3. $\left(\sqrt{\mathrm{n}},\sqrt{\mathrm{n}}\right)$
4. $\left(\frac{1}{\sqrt{\mathrm{n}}},\sqrt{\mathrm{n}}\right)$

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 magnetic wave ?
1. $\overrightarrow{\mathrm{B}}=2\times {10}^{-7}<mspace\; width="1em"></mspace>\sin \left(0.5\times {10}^{3}z+1.5\times {10}^{11}t\right)\hat{i}$
2. $\overrightarrow{\mathrm{B}}=2\times {10}^{-7}<mspace\; width="1em"></mspace>\sin \left(1.5\times {10}^2\mathrm{x}+0.5\times {10}^{11}t\right)\hat{j}$
3. $\overrightarrow{\mathrm{B}}=2\times {10}^{-7}<mspace\; width="1em"></mspace>\sin \left(0.5\times {10}^{3}z\mathit{-}1.5\times {10}^{11}t\right)\hat{i}$
4. $\overrightarrow{\mathrm{B}}=60<mspace\; width="1em"></mspace>\sin \left(0.5\times {10}^3\mathrm{x}+1.5\times {10}^{11}t\right)\hat{k}$

An electromagnetic wave is represented by the electric field $\overrightarrow{\mathrm{E}}=E_0\hat{n}\sin \left[\omega t+\left(6y-8z\right)\right]$. Taking unit vectors in x,y and z directions to be $\hat{\mathrm{i}},\hat{j},\hat{k}$, the direction of propagation $\hat{\mathrm{s}}$, is:
1. $\hat{\mathrm{s}}=\frac{4\hat{j}-3\hat{k}}{5}$
2. $\hat{\mathrm{s}}=\frac{3\hat{i}-4\hat{j}}{5}$
3. $\hat{\mathrm{s}}=\left(\frac{-3\hat{j}+4\hat{k}}{5}\right)$
4. $\hat{\mathrm{s}}=\frac{-4\hat{k}+3\hat{j}}{5}$

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 :
1. 10 V/m
2. 24 V/m
3. 30 V/m
4. 6 V/m