Electromagnetic Waves — Page 5 | NEET Physics

Q41

An electromagnetic wave is transporting energy in the negative

z

direction. At a certain point and certain time the direction of electric field of the wave is along positive

y

direction. What will be the direction of the magnetic field of the wave at that point and instant?

A Negative direction of

y

B Positive direction of

z

C Positive direction of

x

D Negative direction

x

Correct Answer

Option C

Solution

As, poynting vector

\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{H}}

Given energy transport $=$ negative $\mathrm{z}$ direction Electric field $=$ positive $\mathrm{y}$ direction $(-\hat{\mathrm{k}})=(+\hat{\mathrm{j}}) \times[\hat{\mathrm{i}}]$ Hence according to vector cross product magnetic field should be positive $\mathrm{x}$ direction.

Q42

Electromagnetic waves travel in a medium with speed of

1.5 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}

. The relative permeability of the medium is 2.0. The relative permittivity will be:

A 4

B 1

C 2

D 5

Correct Answer

Option C

Solution

\begin{aligned} & v=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \cdot \varepsilon_r}} \\ & \Rightarrow 1.5 \times 10^8=\frac{3 \times 10^8}{\sqrt{2 \cdot \varepsilon_r}} \Rightarrow \varepsilon_r=2 \end{aligned}

Q43

If microwaves, X rays, infracted, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum by M, X, I, G, U, R and V, the following is the arrangement in ascending order of wavelength :

A R, M, I, V, U, X and G

B M, R, V, X, U, G and I

C G, X, U, V, I, M and R

D I, M, R, U, V, X and G

Correct Answer

Option C

Solution

The arrangement in ascending order of wavelength will be

\lambda_{\mathrm{G}}<\lambda_{\mathrm{X}}<\lambda_{\mathrm{U}}<\lambda_{\mathrm{v}}<\lambda_{\mathrm{I}}<\lambda_{\mathrm{M}}<\lambda_{\mathrm{R}}

Q44

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 :

A 0.6 N

B 0.9 N

C 3 × 10–2 N

D 0.1 N

Correct Answer

Option A

Solution

Maximum electric field E = (B) (C)

\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)

\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)

Maximum force

{\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N

{F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N

(approx)

Q45

The electromagnetic waves travel in a medium at a speed of 2.0

\times

108 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :

A 2.25

B 4.25

C 6.25

D 8.25

Correct Answer

Option A

Solution

The speed of electromagnetic waves in a medium is given by the formula:

v = \frac{1}{\sqrt{\mu \varepsilon}}

where $\mu$ and $\varepsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.

Given that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$ , we can rewrite the above equation as :

v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}

which simplifies to :

v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_r \varepsilon_r}}

Substituting $c$ for $\dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$ (the speed of light in a vacuum), we get:

v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}

We can rearrange this equation to solve for $\varepsilon_r$ :

\varepsilon_r = \frac{c^2}{v^2 \mu_r} = \frac{(3 \times 10^8 m/s)^2}{(2 \times 10^8 m/s)^2 \times 1} = 2.25

So, the relative permittivity of the medium is 2.25.

Q46

The rms value of conduction current in a parallel plate capacitor is

6.9 \,\mu \mathrm{A}

. The capacity of this capacitor, if it is connected to

230 \mathrm{~V}

ac supply with an angular frequency of

600 \,\mathrm{rad} / \mathrm{s}

, will be :

A 5 pF

B 50 pF

C 100 pF

D 200 pF

Correct Answer

Option B

Solution

{Z_C} = {V \over I}

\Rightarrow {1 \over {\omega C}} = {{230} \over {6.9}}\,M\,\Omega

\Rightarrow C = {{6.9} \over {230\,\omega }}\,\mu F

= {{6.9} \over {230 \times 600}}\,\mu F

C = 50\,pF

Q47

The oscillating magnetic field in a plane electromagnetic wave is given by

B_{y}=5 \times 10^{-6} \sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T

. The amplitude of electric field will be :

A

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

B

5 \times 10^{-6} \,\mathrm{Vm}^{-1}

C

16 \times 10^{12} \,\mathrm{Vm}^{-1}

D

4 \times 10^{2} \,\mathrm{Vm}^{-1}

Correct Answer

Option D

Solution

In an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :

E = Bv

Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave.

In the given problem, the peak value of the magnetic field is given as :

B_0 = 5 \times 10^{-6} T

The wave speed (v) can be calculated from the given equation for the magnetic field, using the relationship between frequency (f), wave number (k), and speed :

v = \frac{\omega}{k} = \frac{1000 \pi \times 4 \times 10^8}{1000 \pi \times 5} = 8 \times 10^7 m/s

Substituting these values into the equation for the electric field gives the peak value of the electric field :

E_0 = B_0 v = 5 \times 10^{-6} T \times 8 \times 10^7 m/s = 4 \times 10^2 V/m

Q48

The energy of an electromagnetic wave contained in a small volume oscillates with

A double the frequency of the wave

B the frequency of the wave

C half the frequency of the wave

D zero frequency

Correct Answer

Option A

Solution

The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave.

Here's why: The electromagnetic wave consists of oscillating electric and magnetic fields.

The energy density of the wave is proportional to the square of the amplitude of these fields.

Since the square of a sinusoidal function oscillates with twice the frequency of the original function, the energy density of the electromagnetic wave also oscillates with twice the frequency of the wave.

Q49

A point source of

100 \mathrm{~W}

emits light with

5 \%

efficiency. At a distance of

5 \mathrm{~m}

from the source, the intensity produced by the electric field component is:

A

\dfrac{1}{40 \pi} \dfrac{W}{m^2}

B

\dfrac{1}{10 \pi} \dfrac{W}{m^2}

C

\dfrac{1}{20 \pi} \dfrac{W}{m^2}

D

\dfrac{1}{2 \pi} \dfrac{W}{m^2}

Correct Answer

Option A

Solution

A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency.

At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows: The intensity at a distance of $5$ meters can be calculated as: Intensity at 5 m

= \frac{5}{4\pi \times 5^2}\left(\frac{W}{m^2}\right)

= \frac{1}{20\pi}\left(\frac{W}{m^2}\right)

Since the intensity due to the electric field component is half of the total intensity: Intensity due to electric field

= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)

Thus, we arrive at our result:

= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)

Q50

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}