Electromagnetic Waves — Page 7 | JEE Physics

Q61

A plane electromagnetic wave of frequency

20 ~\mathrm{MHz}

propagates in free space along

\mathrm{x}

-direction. At a particular space and time,

\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}

. What is

\overrightarrow{\mathrm{B}}

at this point?

A

-2.2 \times 10^{-8} \hat{i} T

B

2.2 \times 10^{-8} \hat{i} T

C

2.2 \times 10^{-8} \hat{k} T

D

-2.2 \times 10^{-8} \hat{k} T

Correct Answer

Option C

Solution

In free space, the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:

B = \frac{E}{c}

where c is the speed of light in a vacuum, approximately equal to

3 \times 10^8 \mathrm{~m} / \mathrm{s}

. We are given that the electric field is

\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}

. To find the magnetic field, we can first calculate the magnitude of B:

B = \frac{6.6 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} = 2.2 \times 10^{-8} \mathrm{~T}

Now, we need to find the direction of the magnetic field.

Since the electromagnetic wave propagates in the x-direction, and the electric field is in the y-direction (j), the magnetic field should be in the z-direction (k) to satisfy the right-hand rule for electromagnetic waves.

The direction of the magnetic field will be positive k (counter-clockwise rotation from x to y).

So,

\overrightarrow{\mathrm{B}} = 2.2 \times 10^{-8} \hat{k} \mathrm{T}

.

Q62

The correct match between the entries in column I and column II are : .tg .tg I II Radiation Wavelength

List - I		List - II
(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)

Correct Answer

Option B

Solution

Decreasing order of wavelength, Radio wave > Microwave > Infrared > ROYGBIV(Visiable Resion) > Ultraviolet > X rays > $\gamma$ rays $\therefore$ Correct order is (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

Q63

The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is

6.0 \times 10^{-7} \mathrm{~T}

. The maximum value of electric field in the electromagnetic wave is

A

6.0 \times 10^{-7} ~\mathrm{Vm}^{-1}

B

5 \times 10^{14} ~\mathrm{Vm}^{-1}

C

180 ~\mathrm{Vm}^{-1}

D

2 \times 10^{15} ~\mathrm{Vm}^{-1}

Correct Answer

Option C

Solution

In an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation $E = cB$ where c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$ .

Given that the amplitude of the magnetic field B is $6.0 × 10^{-7} T$ , we can substitute these values into the equation to find the maximum value of the electric field: $E = (3 × 10^8 m/s) \times (6.0 × 10^{-7} T) = 180 V/m$ Therefore, 180 V/m is the correct answer.

Q64

For the plane electromagnetic wave given by

E=E_{0} \sin (\omega t-k x)

and

B=B_{0} \sin (\omega t-k x)

, the ratio of average electric energy density to average magnetic energy density is

A 1

B 4

C 2

D 1/2

Correct Answer

Option A

Solution

The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field.

This means that the average electric energy density is equal to the average magnetic energy density.

The average electric energy density ( $u_E$ ) is given by: $u_E = \dfrac{1}{2} \varepsilon_0 E^2$ and the average magnetic energy density ( $u_B$ ) is given by: $u_B = \dfrac{1}{2\mu_0} B^2$ where $\varepsilon_0$ is the permittivity of free space, $\mu_0$ is the permeability of free space, $E$ is the electric field strength, and $B$ is the magnetic field strength.

In an electromagnetic wave, $E$ and $B$ are related by the equation: $E = cB$ where $c$ is the speed of light in vacuum.

Substituting this into the energy density equations gives: $u_E = \dfrac{1}{2} \varepsilon_0 (cB)^2$ and $u_B = \dfrac{1}{2\mu_0} B^2$ Since $c = \dfrac{1}{\sqrt{\varepsilon_0 \mu_0}}$ , we can rewrite $u_E$ as: $u_E = \dfrac{1}{2} \dfrac{1}{\mu_0} B^2 = u_B$ Therefore, the ratio of the average electric energy density to the average magnetic energy density is 1.

Q65

The energy density associated with electric field

\vec{E}

and magnetic field

\vec{B}

of an electromagnetic wave in free space is given by

\left(\epsilon_{0}-\right.

permittivity of free space,

\mu_{0}-

permeability of free space)

A

U_{E}=\dfrac{\epsilon_{0} E^{2}}{2}, U_{B}=\dfrac{B^{2}}{2 \mu_{0}}

B

U_{E}=\dfrac{E^{2}}{2 \epsilon_{0}}, U_{B}=\dfrac{\mu_{0} B^{2}}{2}

C

U_{E}=\dfrac{\epsilon_{0} E^{2}}{2}, U_{B}=\dfrac{\mu_{0} B^{2}}{2}

D

U_{E}=\dfrac{E^{2}}{2 \epsilon_{0}}, U_{B}=\dfrac{B^{2}}{2 \mu_{0}}

Correct Answer

Option A

Solution

The energy density associated with the electric field

\vec{E}

and magnetic field

\vec{B}

of an electromagnetic wave in free space is given by: For the electric field:

U_{E} = \frac{1}{2} \epsilon_{0} E^2

For the magnetic field:

U_{B} = \frac{1}{2} \frac{B^2}{\mu_{0}}

These expressions match Option A:

U_{E} = \frac{\epsilon{0} E^{2}}{2}, U_{B} = \frac{B^{2}}{2 \mu{0}}

Q66

Given below are two statements: Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields. Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface. In the light of the above statements, choose the most appropriate answer from the options given below:

A Statement I is incorrect but Statement II is correct.

B Both Statement I and Statement II are correct.

C Statement I is correct but Statement II is incorrect.

D Both Statement I and Statement II are incorrect.

Correct Answer

Option B

Solution

\begin{aligned} & \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{B}^2}{2 \mu_0} \\ & \because \mathrm{E}=\mathrm{CB} \text{ and } \mathrm{C}=\frac{1}{\mu_0 \varepsilon_0} \end{aligned}

Q67

The electric field in an electromagnetic wave is given by

\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}

. The magnetic field induction of this wave is (in SI unit) :

A

\overrightarrow{\mathrm{B}}=\hat{j} \dfrac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})

B

\overrightarrow{\mathrm{B}}=\hat{i} \dfrac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})

C

\vec{B}=\hat{j} 40 \cos \omega(t-z / c)

D

\overrightarrow{\mathrm{B}}=\hat{k} \dfrac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})

Correct Answer

Option A

Solution

To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field

\overrightarrow{\mathrm{E}}

and the magnetic field

\overrightarrow{\mathrm{B}}

in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:

\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}

where:

\overrightarrow{\mathrm{E}}

is the electric field.

\hat{\mathrm{k}}

is the unit vector in the direction of propagation of the wave.

\mathrm{c}

is the speed of light in a vacuum. Given the electric field:

\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}

The wave is propagating in the

z

-direction, so

\hat{\mathrm{k}} = \hat{z}

. The unit vector

\hat{\mathrm{i}}

represents the

x

-direction. The magnetic field induction is given by:

\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}

The cross product

\hat{\mathrm{i}} \times \hat{\mathrm{z}}

yields

\hat{\mathrm{j}}

(the unit vector in the

y

-direction):

\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}

Therefore, the magnetic field induction is:

\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})

The correct answer is: Option A:

\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})

Q68

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10^{10} \mathrm{~Hz}

and an amplitude of

50 \mathrm{~Vm}^{-1}

. The total average energy density of the electromagnetic field of the wave is : [Use

\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

]

A

4.425 \times 10^{-8} \mathrm{Jm}^{-3}

B

2.212 \times 10^{-10} \mathrm{Jm}^{-3}

C

2.212 \times 10^{-8} \mathrm{Jm}^{-3}

D

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

Correct Answer

Option D

Solution

The average energy density of an electromagnetic wave is given by the formula:

U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}

For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density.

Therefore, we can focus on just one part to find the total energy density.

The formula for the energy density due to the electric field is:

U_E = \frac{1}{2} \varepsilon_0 E^2

Given that the electric field amplitude ( $E$ ) is $50 \, \mathrm{Vm}^{-1}$ and the permittivity of free space ( $\varepsilon_0$ ) is $8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$ , we can calculate the energy density due to the electric field as follows:

U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 = 1.10625 \times 10^{-8} \, \mathrm{J/m^3}

Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.

Q69

The electric field of an electromagnetic wave in free space is represented as

\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}

. The corresponding magnetic induction vector will be :

A

\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}+\mathrm{k} z) \hat{j}

B

\overrightarrow{\mathrm{B}}=\dfrac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{j}

C

\overrightarrow{\mathrm{B}}=\mathrm{E}_0 \mathrm{C} \cos (\omega \mathrm{t}-\mathrm{k} z) \hat{j}

D

\overrightarrow{\mathrm{B}}=\dfrac{\mathrm{E}_0}{\mathrm{C}} \cos (\omega \mathrm{t}+\mathrm{kz}) \hat{j}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\ & \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\ & \hat{C}=\hat{E} \times \hat{B} \end{aligned}

Q70

The magnetic field in a plane electromagnetic wave is

\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}

. The corresponding electric field will be :

A

E_z=105 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

B

E_y=10.5 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

C

E_y=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

D

E_z=1.17 \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{Vm}^{-1}

Correct Answer

Option A

Solution

For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field (

E

) and the magnetic field (

B

) can be described using the equation: $E = cB$ where $c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \text{m/s}$ . $B$ is the magnitude of the magnetic field.

Given the magnetic field $B_y = (3.5 \times 10^{-7}) \sin (1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}$ , we can calculate the corresponding electric field magnitude using the formula above: $E = (3.0 \times 10^8) \times (3.5 \times 10^{-7})$ $= 105 \, \text{Vm}^{-1}$ Thus, the magnitude of the electric field associated with the given magnetic field is $105 \, \text{Vm}^{-1}$ .

The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation.

Given $B_y$ , this means $E$ will have components in the $x-z$ plane.

Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$ -direction, the corresponding electric field component must lie in a plane perpendicular to the $y$ -axis, which could be either the $x$ or the $z$ direction.

However, knowing electromagnetic wave properties, if the wave is propagating along the $x$ -axis and the magnetic field ( $B_y$ ) is along the $y$ -axis, then by right-hand rule, the electric field ( $E$ ) must be along the $z$ -axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.

Therefore, the correct option is: $\text{Option A: } E_z = 105 \sin \left(1.5 \times 10^3 x + 0.5 \times 10^{11} t\right) \, \text{Vm}^{-1}$