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Electromagnetic Waves

JEE Physics · 127 questions · Page 8 of 13 · Click an option or "Show Solution" to reveal answer

Q71
A plane EM wave is propagating along xx direction. It has a wavelength of 4 mm4 \mathrm{~mm}. If electric field is in yy direction with the maximum magnitude of 60 Vm−160 \mathrm{~Vm}^{-1}, the equation for magnetic field is :
A Bz=2×10−7sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\dfrac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}
B Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^T\mathrm{B}_z=2 \times 10^{-7} \sin \left[\dfrac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}
C Bz=60sin⁡[π2(x−3×108t)]k^T\mathrm{B}_z=60 \sin \left[\dfrac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T}
D Bx=60sin⁡[π2(x−3×108t)]i^T\mathrm{B}_x=60 \sin \left[\dfrac{\pi}{2}\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{i}} \mathrm{T}
Correct Answer
Option B
Solution

To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.

Firstly, the wavelength (λ \lambda ) of the wave is given as 4 mm=4×10−3 m4 \mathrm{~mm} = 4 \times 10^{-3} \mathrm{~m}.

The speed of light (and all electromagnetic waves in vacuum) is c=3×108 m/s c = 3 \times 10^8 \mathrm{~m/s} .

Using these values, we can find the frequency (f f ) of the wave using the relationship: c=λf c = \lambda f f=cλ=3×1084×10−3=75×109 Hz f = \dfrac{c}{\lambda} = \dfrac{3 \times 10^8}{4 \times 10^{-3}} = 75 \times 10^9 \mathrm{~Hz} The angular frequency (ω \omega ) which appears in wave equations is related to the frequency by ω=2πf \omega = 2\pi f .

However, in this context, what we need is the wave vector (k k ), which defines how the phase of the wave changes with space.

The wave vector k=2πλ k = \dfrac{2\pi}{\lambda} .

So, for this wave, k=2π4×10−3=π2×103 m−1 k = \dfrac{2\pi}{4 \times 10^{-3}} = \dfrac{\pi}{2} \times 10^3 \, \mathrm{m}^{-1} .

Knowing that the electric field (E \mathbf{E} ) and magnetic field (B \mathbf{B} ) are related as E=cB E = cB in a vacuum, where E E and B B are the magnitudes of the electric and magnetic fields, respectively, we can calculate the magnitude of the magnetic field using the provided maximum electric field magnitude (E=60 Vm−1 E = 60 \, \mathrm{Vm}^{-1} ).

B=Ec=603×108=2×10−7 T B = \dfrac{E}{c} = \dfrac{60}{3 \times 10^8} = 2 \times 10^{-7} \mathrm{~T} Therefore, the wave equation for the magnetic field, considering it propagates in the x x -direction and oscillates in a direction perpendicular to both the x x -direction and the direction of the electric field (thus, in the z z -direction if E \mathbf{E} is in the y y -direction), is: B=Bsin⁡(kx−ωt)k^ \mathbf{B} = B \sin(kx - \omega t) \hat{\mathbf{k}} B=2×10−7sin⁡(π2×103(x−3×108t))k^T \mathbf{B} = 2 \times 10^{-7} \sin \left(\dfrac{\pi}{2} \times 10^3(x - 3 \times 10^8t)\right) \hat{\mathbf{k}} \mathrm{T} This is represented by Option B: Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^T \mathrm{B}_z=2 \times 10^{-7} \sin \left[\dfrac{\pi}{2} \times 10^3\left(x-3 \times 10^8 \mathrm{t}\right)\right] \hat{\mathrm{k}} \mathrm{T} Therefore, the correct answer is Option B.

Q72
Average force exerted on a non-reflecting surface at normal incidence is 2.4×10−4 N2.4 \times 10^{-4} \mathrm{~N}. If 360 W/cm2360 \mathrm{~W} / \mathrm{cm}^2 is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:
A 20 m220 \mathrm{~m}^2
B 0.2 m20.2 \mathrm{~m}^2
C 0.1 m20.1 \mathrm{~m}^2
D 0.02 m20.02 \mathrm{~m}^2
Correct Answer
Option D
Solution

To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface.

The pressure exerted by the light on a non-reflecting surface is given by the formula:

P=FAP = \frac{F}{A}

where

PP

is the pressure,

FF

is the force, and

AA

is the area. The pressure due to the light can also be related to the energy flux

II

by the relationship:

P=IcP = \frac{I}{c}

where

II

is the light energy flux and

cc

is the speed of light in a vacuum (

3×108 m/s3 \times 10^8 \mathrm{~m/s}

). Given that the average force

FF

is

2.4×10−4 N2.4 \times 10^{-4} \mathrm{~N}

and the light energy flux

II

is

360 W/cm2360 \mathrm{~W/cm}^2

, we first convert the flux to

W/m2\mathrm{W/m}^2

:

360 W/cm2=360×104 W/m2360 \mathrm{~W/cm}^2 = 360 \times 10^4 \mathrm{~W/m}^2

Now we can use the relation between pressure and energy flux:

P=Ic=360×1043×108=1.2 N/m2P = \frac{I}{c} = \frac{360 \times 10^4}{3 \times 10^8} = 1.2 \mathrm{~N/m}^2

Next, we use the pressure formula to find the area of the surface:

P=FA  ⟹  A=FPP = \frac{F}{A} \implies A = \frac{F}{P}

Substituting the values we have:

A=2.4×10−4 N1.2 N/m2=2×10−4 m2A = \frac{2.4 \times 10^{-4} \mathrm{~N}}{1.2 \mathrm{~N/m}^2} = 2 \times 10^{-4} \mathrm{~m}^2

The area of the surface is:

A=2×10−4 m2=0.02 m2A = 2 \times 10^{-4} \mathrm{~m}^2 = 0.02 \mathrm{~m}^2

Hence, the correct option is: Option D

0.02 m20.02 \mathrm{~m}^2
Q73
In the given electromagnetic wave Ey=600sin⁡(ωt−kx)Vm−1\mathrm{E}_{\mathrm{y}}=600 \sin (\omega t-\mathrm{kx}) \mathrm{Vm}^{-1}, intensity of the associated light beam is (in W/m2\mathrm{W} / \mathrm{m}^2 : (Given ϵ0=9×10−12C2 N−1 m−2\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} )
A 486
B 729
C 243
D 972
Correct Answer
Option A
Solution

To find the intensity of the given electromagnetic wave, we need to use the formula for the intensity of an electromagnetic wave:

I=12ϵ0cE02I = \frac{1}{2} \epsilon_0 c E_0^2

where:

II

is the intensity in

W/m2\mathrm{W} / \mathrm{m}^2
ϵ0\epsilon_0

is the permittivity of free space, given as

9×10−12C2 N−1 m−29 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}
cc

is the speed of light in vacuum, approximately

3×108 m/s3 \times 10^8 \mathrm{~m/s}
E0E_0

is the peak electric field, given as

600Vm−1600 \mathrm{Vm}^{-1}

Now, substitute these values into the formula:

I=12×9×10−12×3×108×(600)2I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times (600)^2

Simplify the expression step-by-step:

I=12×9×10−12×3×108×360000I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times 360000

First, calculate

9×3×3600009 \times 3 \times 360000

:

I=12×9.72×10−4×360000I = \frac{1}{2} \times 9.72 \times 10^{-4} \times 360000

Combine 9 and 3 into 27, giving you:

I=13.5×10−4×360000I = 13.5 \times 10^{-4} \times 360000

Then, calculate the multiplication:

I=13.5×36I = 13.5 \times 36

Finally, multiply the remaining values:

I=486×10−4I = 486 \times 10^{-4}

The final value is: The intensity,

II

, is 486

W/m2\mathrm{W} / \mathrm{m}^2

. Therefore, the correct option is: Option A: 486

Q74
A plane electromagnetic wave propagates along the + x direction in free space. The components of the electric field, E⃗\vec{E} and magnetic field, B⃗\vec{B} vectors associated with the wave in Cartesian frame are
A Ex,ByE_x, B_y
B Ey,BxE_y, B_x
C Ey,BzE_y, B_z
D Ez,ByE_z, B_y
Correct Answer
Option C
Solution

We know, for a plane electromagnetic wave,

E^×B^=C^\widehat E \times \widehat B = \widehat C

Given,

C^=i^\widehat C = \widehat i

(+ x direction) So,

E^×B^=i^\widehat E \times \widehat B = \widehat i

i.e.

j^×k^=i^\widehat j \times \widehat k = \widehat i

So,

E→=Ey\overrightarrow E = {E_y}
B→=B2\overrightarrow B = {B_2}

As,

i^, j^\widehat i,\,\widehat j

and

k^\widehat k

are unit vectors in the directions of x, y and z axes respectively. Hence, option 3 is correct.

Q75
The electric field of an electromagnetic wave in free space is E→=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)N/C\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C. The associated magnetic field in Tesla is
A B→=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\overrightarrow{\mathrm{B}}=\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})
B B→=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})
C B→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\overrightarrow{\mathrm{B}}=-\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})
D B→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=-\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})
Correct Answer
Option D
Solution

We start with the given electric field:

E⃗=57cos⁡[7.5×106 t−5×10−3(3x+4y)](4 i^−3 j^)(N/C)\vec{E} = 57 \cos\Bigl[7.5\times10^6\, t - 5\times10^{-3}(3x+4y)\Bigr](4\,\hat{i} - 3\,\hat{j}) \quad \text{(N/C)}

Since we are dealing with an electromagnetic wave in free space, the electric field, magnetic field, and the propagation direction are all mutually perpendicular.

In free space the magnetic field is related to the electric field by:

B⃗=1c n^×E⃗,\vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E},

where: •

c=3×108 m/sc = 3\times10^8 \, \text{m/s}

is the speed of light, and •

n^\hat{n}

is the unit vector in the direction of wave propagation.

Step 1.

Identify the propagation direction: The phase of the wave is written as

7.5×106 t−5×10−3(3x+4y).7.5\times10^6\,t - 5\times10^{-3}(3x+4y).

This tells us that the wave vector, up to a multiplicative constant, is

k⃗=5×10−3(3 i^+4 j^).\vec{k} = 5\times10^{-3}(3\,\hat{i} + 4\,\hat{j}).

The unit vector in the direction of propagation is then

n^=3i^+4j^32+42=3i^+4j^5.\hat{n} = \frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}+4\hat{j}}{5}.

Step 2. Compute the cross product: We have

E⃗∝(4 i^−3 j^),\vec{E} \propto (4\,\hat{i}-3\,\hat{j}),

so we need to calculate

n^×E⃗=15(3i^+4j^)×[57cos⁡(…)(4i^−3j^)].\hat{n} \times \vec{E} = \frac{1}{5}(3\hat{i}+4\hat{j}) \times \Bigl[57 \cos(\ldots)(4\hat{i}-3\hat{j})\Bigr].

Taking the constant factors out of the cross product gives

n^×E⃗=57cos⁡(…)5 [(3i^+4j^)×(4i^−3j^)].\hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,\Bigl[(3\hat{i}+4\hat{j}) \times (4\hat{i}-3\hat{j})\Bigr].

Recall that the cross product in the plane (where both vectors have zero

k^\hat{k}

components) will point in the

k^\hat{k}

(or

−k^-\hat{k}

) direction. To compute the cross product of vectors in the

xyxy

-plane, we use the formula

a⃗×b⃗=(axby−aybx) k^.\vec{a} \times \vec{b} = (a_xb_y - a_yb_x)\,\hat{k}.

Let •

a⃗=3i^+4j^,\vec{a} = 3\hat{i}+4\hat{j},

•

b⃗=4i^−3j^.\vec{b} = 4\hat{i}-3\hat{j}.

Then

a⃗×b⃗=(3⋅(−3)−4⋅4) k^=(−9−16) k^=−25 k^.\vec{a} \times \vec{b} = \Bigl(3\cdot (-3) - 4\cdot 4\Bigr)\,\hat{k} = (-9 - 16)\,\hat{k} = -25\,\hat{k}.

Thus,

n^×E⃗=57cos⁡(…)5 (−25 k^)=−57×255 cos⁡(…) k^.\hat{n} \times \vec{E} = \frac{57 \cos(\ldots)}{5}\,(-25\,\hat{k}) = -\frac{57 \times 25}{5}\, \cos(\ldots) \,\hat{k}.

Simplify the constant:

255=5,\frac{25}{5} = 5,

so we get

n^×E⃗=−57×51cos⁡(…) k^=−285cos⁡(…) k^.\hat{n} \times \vec{E} = -\frac{57 \times 5}{1} \cos(\ldots)\,\hat{k} = -285 \cos(\ldots) \,\hat{k}.

Step 3. Apply the relation for the magnetic field: Now, substituting back into the formula for

B⃗\vec{B}

,

B⃗=1c n^×E⃗=−2853×108cos⁡(…) k^.\vec{B} = \frac{1}{c}\,\hat{n} \times \vec{E} = -\frac{285}{3\times10^8} \cos(\ldots)\,\hat{k}.

Notice that

2853×108=57×53×108,\frac{285}{3\times10^8} = \frac{57 \times 5}{3\times10^8},

so we have

B⃗=−573×108cos⁡[7.5×106 t−5×10−3(3x+4y)](5 k^)(T).\vec{B} = -\frac{57}{3\times10^8} \cos\Bigl[7.5\times10^6\,t - 5\times10^{-3}(3x+4y)\Bigr](5\,\hat{k})\quad \text{(T)}.

This expression exactly matches Option D. Final Answer: Option D

Q76
A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x+x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3Vm−1\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}. Then, the magnetic field vector of the wave at that point is
A Bz=1.55×10−8 T\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}
B Bz=6.2×10−8 T\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}
C Bz=3.1×10−8 T\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}
D Bz=9.3×10−8 T\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}
Correct Answer
Option C
Solution

For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light,

c≈3×108 m/s,c \approx 3 \times 10^8 \, \mathrm{m/s},

through the equation

E=cB.E = cB.

Here's how to determine the magnetic field: We are given the electric field magnitude:

E=9.3 V/m.E=9.3 \, \mathrm{V/m}.

Use the relationship to find the magnetic field magnitude:

B=Ec=9.33×108 T.B = \frac{E}{c} = \frac{9.3}{3 \times 10^8} \, \mathrm{T}.

Calculating the value gives:

B≈3.1×10−8 T.B \approx 3.1 \times 10^{-8} \, \mathrm{T}.

Since the electric field is along the

yy

direction and the wave propagates along the

+x+x

direction, by the right-hand rule, the magnetic field is oriented along the

zz

direction. Thus, the magnetic field vector is

Bz=3.1×10−8 T,\mathrm{B}_z = 3.1 \times 10^{-8} \, \mathrm{T},

which corresponds to Option C.

Q77
Due to presence of an em-wave whose electric component is given by E=100sin⁡(ωt−kx)NC−1E=100 \sin (\omega t-k x) \mathrm{NC}^{-1} a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as
A 50sin⁡(ωt−kx)NC−150 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}
B 400sin⁡(ωt−kx)NC−1400 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}
C 200sin⁡(ωt−kx)NC−1200 \sin (\omega t-k x) \mathrm{NC}^{-1}
D 25sin⁡(ωt−kx)NC−125 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}
Correct Answer
Option C
Solution

We know, Energy density

=12ε0E2C= {1 \over 2}{\varepsilon _0}{E^2}C
⇒EnergyVolume=12ε0E2C\Rightarrow {\text{Energy} \over \text{Volume}} = {1 \over 2}{\varepsilon _0}{E^2}C
⇒Energy=12ε0Ec2x vol\Rightarrow \text{Energy}= {1 \over 2}{\varepsilon _0}E_c^2x\,vol

Given that,

(Energy)1=(Energy)2{\left( {\text{Energy}} \right)_1} = {\left( {\text{Energy}} \right)_2}
⇒12ε0E12cxπR12L1=12ε0E22cxπR22L2\Rightarrow {1 \over 2}{\varepsilon _0}E_1^2cx\pi R_1^2{L_1} = {1 \over 2}{\varepsilon _0}E_2^2cx\pi R_2^2{L_2}
⇒E12R12L1=E22R22L1 (as L1=L2)\Rightarrow E_1^2R_1^2{L_1} = E_2^2R_2^2{L_1}\,(as\,{L_1} = {L_2})
⇒E12R12=E22R22\Rightarrow E_1^2R_1^2 = E_2^2R_2^2
⇒E1R1=E2R2\Rightarrow {E_1}{R_1} = {E_2}{R_2}
⇒E1R1=E2(R12) (As R2=R12)\Rightarrow {E_1}{R_1} = {E_2}\left( {{{{R_1}} \over 2}} \right)\,\left( {As\,{R_2} = {{{R_1}} \over 2}} \right)
⇒E2=2E1=2×100\Rightarrow {E_2} = 2{E_1} = 2 \times 100
⇒E2=200 N/C\Rightarrow {E_2} = 200\,N/C

Hence, option (3) is correct.

Q78
The unit of 2Iε0c\sqrt{\dfrac{2I}{\varepsilon_0 c}} is :(I = intensity of an electromagnetic wave, c = speed of light)
A Vm
B NC-1
C NC
D Nm
Correct Answer
Option B
Solution

The unit of the expression 2Iε0c\sqrt{\dfrac{2I}{\varepsilon_0 c}} can be determined by exploring the relationship between intensity (II) and the electric field (E0E_0) in an electromagnetic wave.

The intensity of an electromagnetic wave is given by: I=12ε0E02×c I = \dfrac{1}{2} \varepsilon_0 E_0^2 \times c From this equation, we can solve for the electric field E0E_0: E0=2Iε0c E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} The symbol E0E_0 represents the electric field, which is measured in units of Newtons per Coulomb (N/C).

Therefore, the unit of the expression 2Iε0c\sqrt{\dfrac{2I}{\varepsilon_0 c}} is also N/CN/C.

Thus, the expression 2Iε0c\sqrt{\dfrac{2I}{\varepsilon_0 c}} has units of N/CN/C.

Q79
Match List I with List II .tg .tg LIST I LIST II A. Microwaves I. Physiotherapy B. UV rays II. Treatment of cancer C. Infra-red light III. Lasik eye surgery D. X-ray IV. Aircraft navigation Choose the correct answer from the options given below:
A A - IV, B - III, C - I, D - II
B A - II, B - IV, C - III, D - I
C A - III, B - II, C - I, D - IV
D A - IV, B - I, C - II, D - III
Correct Answer
Option A
Solution

A. Microwave → IV B. UV rays → III C. Infra-red → I D. X-ray → II

Q80

Match with : .tg .tg

List - IList - II
(a) UV rays (i) Diagnostic tool in medicine
(b) X-rays (ii) Water purification
(c) Microwave (iii) Communication, Radar
(d) Infrared wave (iv) Improving visibility in foggy days
A (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
B (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
D (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Correct Answer
Option B
Solution

UV - Water purification X-rays - Diagnostic tool in medicine Microwave - Communication, Radar Infrared wave - Improving visibility in foggy days.

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