Electromagnetic Waves Questions — NEET Physics

Q1

The electric field in a plane electromagnetic wave is given by

E_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{V} / \mathrm{m}

Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

A

B z=60 \cos \left(5 x+1.5 \times 10^9 t\right) T

B

B_y=60 \sin \left(5 x+1.5 \times 10^9 t\right) T

C

B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T

D

B_x=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T

Correct Answer

Option C

Solution

In electromagnetic wave, $E$ and $B$ are in same phase and $B_0=\dfrac{E_0}{c}$ ; their planes are perpendicular to each other.

\begin{aligned} & \therefore B_y=\frac{60}{c} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & =\frac{60}{3 \times 10^8} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \\ & B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{T} \end{aligned}

Q2

The electromagnetic radiation which has the smallest wavelength are

A X-rays

B Gamma rays

C Ultraviolet rays

D Microwaves

Correct Answer

Option B

Solution

The electromagnetic radiation with the smallest wavelength among the given options is Gamma rays .

Let's go through each option to understand more clearly: Option A: X-rays X-rays have very short wavelengths, typically in the range of 0.01 to 10 nanometers.

They are used for medical imaging and other applications due to their ability to penetrate materials.

Option B: Gamma rays Gamma rays have the smallest wavelengths of all electromagnetic radiation, often less than 0.01 nanometers (or 0.1 angstroms).

They are highly energetic and are produced by nuclear reactions, radioactive decay, and other high-energy processes.

Option C: Ultraviolet rays Ultraviolet (UV) rays have wavelengths ranging from about 10 to 400 nanometers.

They are responsible for causing sunburns and are used in various scientific and industrial applications.

Option D: Microwaves Microwaves have much longer wavelengths, typically ranging from 1 millimeter to 1 meter.

They are commonly used in communication technologies and for heating food in microwave ovens.

Based on these explanations, the correct answer is: Option B: Gamma rays

Q3

If the ratio of relative permeability and relative permittivity of a uniform medium is

1: 4

. The ratio of the magnitudes of electric field intensity

(E)

to the magnetic field intensity

(H)

of an EM wave propagating in that medium is (Given that

\sqrt{\dfrac{\mu_0}{\varepsilon_0}}=120 \pi

):

A

30 \pi: 1

B

1: 120 \pi

C

60 \pi: 1

D

120 \pi: 1

Correct Answer

Option C

Solution

\frac{\mu_r}{E_r}=\frac{1}{4}

\begin{aligned} \frac{E}{H}=\frac{E \mu}{B} & =v \mu \\ =\frac{1}{\sqrt{\mu \varepsilon}} \mu & =\sqrt{\frac{\mu}{\varepsilon}}=\sqrt{\frac{\mu_0 \mu_r}{\varepsilon_0 \varepsilon_r}} \\ & =\sqrt{\frac{\mu_0}{\varepsilon_0}} \sqrt{\frac{\mu_r}{\varepsilon_r}} \\ & =120 \pi\left(\frac{1}{2}\right) \\ & =\frac{60 \pi}{1} \end{aligned}

Q4

The property which is not of an electromagnetic wave travelling in free space is that:

A They are transverse in nature

B The energy density in electric field is equal to energy density in magnetic field

C They travel with a speed equal to

\dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}

D They originate from charges moving with uniform speed

Correct Answer

Option D

Solution

Electromagnetic waves have several defining characteristics when they propagate through free space.

Let's evaluate each of the options provided: Option A: They are transverse in nature.

This is true .

Electromagnetic waves are transverse waves, meaning the directions of the electric field and magnetic field oscillations are perpendicular to the direction of wave propagation.

The electric field (E) and magnetic field (B) vectors are also perpendicular to each other and to the direction of propagation.

Option B: The energy density in the electric field is equal to the energy density in the magnetic field.

This is also true .

In electromagnetic waves, the energy density stored in the electric field is equal to the energy density stored in the magnetic field.

This is because the magnitudes of the electric and magnetic fields are related by

c = \frac{E}{B}

where c is the speed of light in vacuum. The energy density for each is given by

\frac{1}{2} \epsilon_0 E^2

for the electric field and

\frac{1}{2} \frac{B^2}{\mu_0}

for the magnetic field.

Given the relationship between E and B in a wave, these two energy densities are equal.

Option C: They travel with a speed equal to

\frac{1}{\sqrt{\mu_0 \varepsilon_0}}

. This statement is true . The speed of electromagnetic waves in vacuum is given by

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}

, where

\mu_0

is the magnetic permeability of free space and

\varepsilon_0

is the electric permittivity of free space.

This relationship derives from Maxwell’s equations in a vacuum.

Option D: They originate from charges moving with uniform speed.

This statement is false .

Electromagnetic waves are not generally produced by charges moving with a uniform speed; rather, they are produced by charges that are accelerating.

Uniform motion (where velocity is constant and acceleration is zero) does not result in radiation of electromagnetic waves.

If a charge is accelerating – changing either the speed or direction of its motion – it emits electromagnetic radiation.

Therefore, the correct answer is Option D , as it is the property that is not true for electromagnetic waves traveling in free space.

Q5

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of

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

and amplitude

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

. Then the amplitude of oscillating magnetic field is : (Speed of light in free space

=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}

)

A

1.6 \times 10^{-8} \mathrm{~T}

B

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

C

1.6 \times 10^{-6} \mathrm{~T}

D

1.6 \times 10^{-9} \mathrm{~T}

Correct Answer

Option B

Solution

C=\frac{E_{0}}{B_{0}}

\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}

=\frac{48}{3 \times 10^{8}}

=1.6 \times 10^{-7} \mathrm{~T}

Q6

The magnetic field of a plane electromagnetic wave is given by

\overrightarrow B = 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat j

, then the associated electric field will be :

A

9\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat k

V/m

B

3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i

V/m

C

3 \times {10^{ - 8}}\sin (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i

V/m

D

9\sin (1.6 \times {10^3}x - 48 \times {10^{10}}t)\widehat k

V/m

Correct Answer

Option A

Solution

For electromagnetic wave,

|\overrightarrow B | = {{|\overrightarrow E |} \over c}

Here

\overrightarrow B

is magnetic field associated with EM wave

\overrightarrow E

is electric field associated with EM wave c is the speed of EM wave

\Rightarrow |\overrightarrow E | = c|\overrightarrow B |

= 3 \times {10^8} \times 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)

V/m Direction can be determined from Poynting vector

= {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}

\overrightarrow E = 9\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat k

V/m

Q7

The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability

{\mu _0}

and permittivity

{\varepsilon _0}

is (Given that c - velocity) of light in free space

A

{{\sqrt {{\mu _0}{\varepsilon _0}} } \over c}

B c

C

{1 \over c}

D

{c \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

Correct Answer

Option C

Solution

We know,

{{{E_0}} \over {{B_0}}}

(in vacuum) = c $\therefore$

{{{B_0}} \over {{E_0}}} = {1 \over c}

or,

{{{B_0}} \over {{E_0}}} = \sqrt {{\mu _0}{\varepsilon _0}}

Q8

When light propagates through a material medium of relative permittivity

\varepsilon

r and relative permeability

\mu

r , the velocity of light, v is given by (c-velocity of light in vacuum)

A v = c

B

v = \sqrt {{{{\mu _r}} \over {{\varepsilon _r}}}}

C

v = \sqrt {{{{\varepsilon _r}} \over {{\mu _r}}}}

D

v = {c \over {\sqrt {{\varepsilon _r}{\mu _r}} }}

Correct Answer

Option D

Solution

v = {1 \over {\sqrt {{\varepsilon _m}{\mu _m}} }}

v = {1 \over {\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}

Since

c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }}

\Rightarrow v = {c \over {\sqrt {{\varepsilon _r}{\mu _r}} }}

Q9

For a plane electromagnetic wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?

A

- \widehat j + \widehat k, - \widehat j + \widehat k

B

\widehat j + \widehat k, \widehat j + \widehat k

C

- \widehat j + \widehat k, - \widehat j - \widehat k

D

\widehat j + \widehat k, - \widehat j - \widehat k

Correct Answer

Option C

Solution

Wave in x direction C =

{\overrightarrow E }

$\times$

{\overrightarrow B }

\left( { - \widehat j + \widehat k} \right) \times \left( { - \widehat j - \widehat k} \right)

= \widehat i + \widehat i

= 2\widehat i

Q10

The ratio of contributions made by the electric field and magnetic field components to the intensity of and electromagnetic wave is : (c = speed of electromagnetic waves)

A 1 : 1

B 1 : c

C 1 : c 2

D c : 1

Correct Answer

Option A

Solution

The total energy of an electromagnetic wave is equally shared by E and B. So the ratio is 1 : 1.