QBManager

Electromagnetic Wave

155603 A radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(c =$ velocity of light)

1

\frac{E}{c}

2

\frac{2 E}{c}

3

\frac{2 E}{c^{2}}

4

\frac{E}{c^{2}}

Explanation:

B We know that energy of radiation,
$E = h v$
$E = \frac{hc}{λ}$
$(∵ v = c / λ)$
Momentum of incident radiation
$p = \frac{h}{λ} = \frac{E}{c} = p_{i}$
Momentum of reflected radiation
$p_{r} = - p_{i} = - E / c$
So,
The net momentum transformed to the surface
$Δ p = p_{i} - p_{r}$
$= E / c - (- E / c)$
$Δ p = \frac{2 E}{c}$

AIPMT- 2015

Electromagnetic Wave

155606 A plane electromagnetic wave of frequency 25 $MHz$ travels in free space along the $x$ -direction. If $\vec{E}$ at a particular point in space and time is $6.3 \hat{j} {Vm}^{- 1}, \vec{B}$ at that point is
(Given $C = {(ε_{0} μ_{0})}^{- 1 / 2}$

1

2.1 \times 10^{- 8} T

2

3.5 \times 10^{- 5} T

3

2.6 \times 10^{- 6} T

4

3.1 \times 10^{- 6} T

Explanation:

A Given that,
$\vec{E} = 6.3 j {Vm}^{- 1}$
In electromagnetic wave, the relation between electric and magnetic field is,
$B = \frac{E}{c} = \frac{6.3}{3 \times 10^{8}}$
$B = 2.1 \times 10^{- 8} T$
For direction of $\vec{B}$ using vector algebra, $E \times B$ should be $x$ - direction.
Since $(+ \hat{j}) \times (+ \hat{k}) = \hat{i}$
So, $B$ is along $+ \hat{k}$
Hence, magnetic field (B) is in z-direction.
$\vec{B} = 2.1 \times 10^{- 8} \hat{k T}$

Assam CEE-2021

Electromagnetic Wave

155607 For an EM Wave, the electric and magnetic fields are $300 V / m$ and $7.9 A / m$ respectively. The maximum rate of energy flow is

1

2730 \frac{watt}{m^{2}}

2

2790 \frac{watt}{m^{2}}

3

2370 \frac{watt}{m^{2}}

4

2390 \frac{watt}{m^{2}}

Explanation:

C Given that,
$E_{0} = 300 V / m, B_{0} = 7.9 A / m$
Maximum rate of energy flow,
$S = E_{0} \times B_{0}$
$S = 300 \times 7.9$
$S = 2370 watt / m^{2}$

TS EAMCET 04.08.2021

Electromagnetic Wave

155608 The magnetic field in a plane electromagnetic wave is given by $B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x +$ $3 π \times 10^{11} t) T$
Calculate the wavelength.

1

π \times 10^{3} m

2

2 \times 10^{- 3} m

3

2 \times 10^{3} m

4

π \times 10^{- 3} m

Explanation:

B Electromagnetic wave is given by,
$B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x + 3 π \times 10^{11} t)$
The general equation of magnetic .....(i) electromagnetic wave,
$B = B_{0} \sin (kx \pm ω t)$
On comparing equation (i) with equation (ii)
$k = π \times 10^{3}$
$k = \frac{2 π}{λ}$
$π \times 10^{3} = \frac{2 π}{λ}$
$λ = 2 \times 10^{- 3} m$

NEET (Oct.)- 2020

Electromagnetic Wave

155603 A radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(c =$ velocity of light)

1

\frac{E}{c}

2

\frac{2 E}{c}

3

\frac{2 E}{c^{2}}

4

\frac{E}{c^{2}}

Explanation:

B We know that energy of radiation,
$E = h v$
$E = \frac{hc}{λ}$
$(∵ v = c / λ)$
Momentum of incident radiation
$p = \frac{h}{λ} = \frac{E}{c} = p_{i}$
Momentum of reflected radiation
$p_{r} = - p_{i} = - E / c$
So,
The net momentum transformed to the surface
$Δ p = p_{i} - p_{r}$
$= E / c - (- E / c)$
$Δ p = \frac{2 E}{c}$

AIPMT- 2015

Electromagnetic Wave

155606 A plane electromagnetic wave of frequency 25 $MHz$ travels in free space along the $x$ -direction. If $\vec{E}$ at a particular point in space and time is $6.3 \hat{j} {Vm}^{- 1}, \vec{B}$ at that point is
(Given $C = {(ε_{0} μ_{0})}^{- 1 / 2}$

1

2.1 \times 10^{- 8} T

2

3.5 \times 10^{- 5} T

3

2.6 \times 10^{- 6} T

4

3.1 \times 10^{- 6} T

Explanation:

A Given that,
$\vec{E} = 6.3 j {Vm}^{- 1}$
In electromagnetic wave, the relation between electric and magnetic field is,
$B = \frac{E}{c} = \frac{6.3}{3 \times 10^{8}}$
$B = 2.1 \times 10^{- 8} T$
For direction of $\vec{B}$ using vector algebra, $E \times B$ should be $x$ - direction.
Since $(+ \hat{j}) \times (+ \hat{k}) = \hat{i}$
So, $B$ is along $+ \hat{k}$
Hence, magnetic field (B) is in z-direction.
$\vec{B} = 2.1 \times 10^{- 8} \hat{k T}$

Assam CEE-2021

Electromagnetic Wave

155607 For an EM Wave, the electric and magnetic fields are $300 V / m$ and $7.9 A / m$ respectively. The maximum rate of energy flow is

1

2730 \frac{watt}{m^{2}}

2

2790 \frac{watt}{m^{2}}

3

2370 \frac{watt}{m^{2}}

4

2390 \frac{watt}{m^{2}}

Explanation:

C Given that,
$E_{0} = 300 V / m, B_{0} = 7.9 A / m$
Maximum rate of energy flow,
$S = E_{0} \times B_{0}$
$S = 300 \times 7.9$
$S = 2370 watt / m^{2}$

TS EAMCET 04.08.2021

Electromagnetic Wave

155608 The magnetic field in a plane electromagnetic wave is given by $B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x +$ $3 π \times 10^{11} t) T$
Calculate the wavelength.

1

π \times 10^{3} m

2

2 \times 10^{- 3} m

3

2 \times 10^{3} m

4

π \times 10^{- 3} m

Explanation:

B Electromagnetic wave is given by,
$B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x + 3 π \times 10^{11} t)$
The general equation of magnetic .....(i) electromagnetic wave,
$B = B_{0} \sin (kx \pm ω t)$
On comparing equation (i) with equation (ii)
$k = π \times 10^{3}$
$k = \frac{2 π}{λ}$
$π \times 10^{3} = \frac{2 π}{λ}$
$λ = 2 \times 10^{- 3} m$

NEET (Oct.)- 2020

Electromagnetic Wave

155603 A radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(c =$ velocity of light)

1

\frac{E}{c}

2

\frac{2 E}{c}

3

\frac{2 E}{c^{2}}

4

\frac{E}{c^{2}}

Explanation:

B We know that energy of radiation,
$E = h v$
$E = \frac{hc}{λ}$
$(∵ v = c / λ)$
Momentum of incident radiation
$p = \frac{h}{λ} = \frac{E}{c} = p_{i}$
Momentum of reflected radiation
$p_{r} = - p_{i} = - E / c$
So,
The net momentum transformed to the surface
$Δ p = p_{i} - p_{r}$
$= E / c - (- E / c)$
$Δ p = \frac{2 E}{c}$

AIPMT- 2015

Electromagnetic Wave

155606 A plane electromagnetic wave of frequency 25 $MHz$ travels in free space along the $x$ -direction. If $\vec{E}$ at a particular point in space and time is $6.3 \hat{j} {Vm}^{- 1}, \vec{B}$ at that point is
(Given $C = {(ε_{0} μ_{0})}^{- 1 / 2}$

1

2.1 \times 10^{- 8} T

2

3.5 \times 10^{- 5} T

3

2.6 \times 10^{- 6} T

4

3.1 \times 10^{- 6} T

Explanation:

A Given that,
$\vec{E} = 6.3 j {Vm}^{- 1}$
In electromagnetic wave, the relation between electric and magnetic field is,
$B = \frac{E}{c} = \frac{6.3}{3 \times 10^{8}}$
$B = 2.1 \times 10^{- 8} T$
For direction of $\vec{B}$ using vector algebra, $E \times B$ should be $x$ - direction.
Since $(+ \hat{j}) \times (+ \hat{k}) = \hat{i}$
So, $B$ is along $+ \hat{k}$
Hence, magnetic field (B) is in z-direction.
$\vec{B} = 2.1 \times 10^{- 8} \hat{k T}$

Assam CEE-2021

Electromagnetic Wave

155607 For an EM Wave, the electric and magnetic fields are $300 V / m$ and $7.9 A / m$ respectively. The maximum rate of energy flow is

1

2730 \frac{watt}{m^{2}}

2

2790 \frac{watt}{m^{2}}

3

2370 \frac{watt}{m^{2}}

4

2390 \frac{watt}{m^{2}}

Explanation:

C Given that,
$E_{0} = 300 V / m, B_{0} = 7.9 A / m$
Maximum rate of energy flow,
$S = E_{0} \times B_{0}$
$S = 300 \times 7.9$
$S = 2370 watt / m^{2}$

TS EAMCET 04.08.2021

Electromagnetic Wave

155608 The magnetic field in a plane electromagnetic wave is given by $B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x +$ $3 π \times 10^{11} t) T$
Calculate the wavelength.

1

π \times 10^{3} m

2

2 \times 10^{- 3} m

3

2 \times 10^{3} m

4

π \times 10^{- 3} m

Explanation:

B Electromagnetic wave is given by,
$B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x + 3 π \times 10^{11} t)$
The general equation of magnetic .....(i) electromagnetic wave,
$B = B_{0} \sin (kx \pm ω t)$
On comparing equation (i) with equation (ii)
$k = π \times 10^{3}$
$k = \frac{2 π}{λ}$
$π \times 10^{3} = \frac{2 π}{λ}$
$λ = 2 \times 10^{- 3} m$

NEET (Oct.)- 2020

Electromagnetic Wave

155603 A radiation of energy ' $E$ ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(c =$ velocity of light)

1

\frac{E}{c}

2

\frac{2 E}{c}

3

\frac{2 E}{c^{2}}

4

\frac{E}{c^{2}}

Explanation:

B We know that energy of radiation,
$E = h v$
$E = \frac{hc}{λ}$
$(∵ v = c / λ)$
Momentum of incident radiation
$p = \frac{h}{λ} = \frac{E}{c} = p_{i}$
Momentum of reflected radiation
$p_{r} = - p_{i} = - E / c$
So,
The net momentum transformed to the surface
$Δ p = p_{i} - p_{r}$
$= E / c - (- E / c)$
$Δ p = \frac{2 E}{c}$

AIPMT- 2015

Electromagnetic Wave

155606 A plane electromagnetic wave of frequency 25 $MHz$ travels in free space along the $x$ -direction. If $\vec{E}$ at a particular point in space and time is $6.3 \hat{j} {Vm}^{- 1}, \vec{B}$ at that point is
(Given $C = {(ε_{0} μ_{0})}^{- 1 / 2}$

1

2.1 \times 10^{- 8} T

2

3.5 \times 10^{- 5} T

3

2.6 \times 10^{- 6} T

4

3.1 \times 10^{- 6} T

Explanation:

A Given that,
$\vec{E} = 6.3 j {Vm}^{- 1}$
In electromagnetic wave, the relation between electric and magnetic field is,
$B = \frac{E}{c} = \frac{6.3}{3 \times 10^{8}}$
$B = 2.1 \times 10^{- 8} T$
For direction of $\vec{B}$ using vector algebra, $E \times B$ should be $x$ - direction.
Since $(+ \hat{j}) \times (+ \hat{k}) = \hat{i}$
So, $B$ is along $+ \hat{k}$
Hence, magnetic field (B) is in z-direction.
$\vec{B} = 2.1 \times 10^{- 8} \hat{k T}$

Assam CEE-2021

Electromagnetic Wave

155607 For an EM Wave, the electric and magnetic fields are $300 V / m$ and $7.9 A / m$ respectively. The maximum rate of energy flow is

1

2730 \frac{watt}{m^{2}}

2

2790 \frac{watt}{m^{2}}

3

2370 \frac{watt}{m^{2}}

4

2390 \frac{watt}{m^{2}}

Explanation:

C Given that,
$E_{0} = 300 V / m, B_{0} = 7.9 A / m$
Maximum rate of energy flow,
$S = E_{0} \times B_{0}$
$S = 300 \times 7.9$
$S = 2370 watt / m^{2}$

TS EAMCET 04.08.2021

Electromagnetic Wave

155608 The magnetic field in a plane electromagnetic wave is given by $B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x +$ $3 π \times 10^{11} t) T$
Calculate the wavelength.

1

π \times 10^{3} m

2

2 \times 10^{- 3} m

3

2 \times 10^{3} m

4

π \times 10^{- 3} m

Explanation:

B Electromagnetic wave is given by,
$B_{y} = 2 \times 10^{- 7} \sin (π \times 10^{3} x + 3 π \times 10^{11} t)$
The general equation of magnetic .....(i) electromagnetic wave,
$B = B_{0} \sin (kx \pm ω t)$
On comparing equation (i) with equation (ii)
$k = π \times 10^{3}$
$k = \frac{2 π}{λ}$
$π \times 10^{3} = \frac{2 π}{λ}$
$λ = 2 \times 10^{- 3} m$

NEET (Oct.)- 2020

Question Bank Series

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