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Electromagnetic Waves

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2 months ago

Electromagnetic Waves Class 12th

If E and H represents the intensity of electric field and magnetizing field respectively, then the unit of E/H will be:

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Vishal Baghel

Contributor-Level 10

$\frac{E}{H} = \frac{V o l t / m e t r e}{A m p e r e / m e t r e} = \frac{V o l t}{A m p e r e} = o h m$

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2 months ago

Electromagnetic Waves Class 12th

The electric field in a plane electromagnetic wave is given by

$\vec{E} = 200 c o s [(\frac{0.5 * 1 0^{3}}{m}) x (1.5 * 1 0^{11} \frac{r a d}{s} * t)] \frac{V}{m} \hat{j}$

If this wave falls normally on a perfectly reflecting surface having an area of 100 cm². If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $\frac{x}{1 0^{9}} \frac{N}{m^{2}} .$ Find the value of x

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Vishal Baghel

Contributor-Level 10

$P = \frac{2 l}{C} P \to$ radiation pressure

$= \frac{2}{C} (\frac{1}{2} ε_{0} E_{0}^{2} C)$

$\begin{array}{l} = ε_{0} E_{0}^{2} \\ = (8.85 * 1 0^{- 12}) * {(200)}^{2} \\ = 354 * 1 0^{- 9} N / m^{2} \end{array}$

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2 months ago

Electromagnetic Waves Class 12th

The electric field in an electromagnetic wave is given by E = 56.5 sin $ω (t - x / c) N C^{- 1} .$ Find the intensity of the wave if it is propagating along x-axis in the free space.

(Given $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ )

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Vishal Baghel

Contributor-Level 10

$E = 56.5 s i n (w (\frac{t - x}{c})) N / c$

$E_{0} = 56.5$

$ε_{0} = 8.85 * 1 0^{- 12}$

C = 3 * 108

l = $\frac{1}{2} ε_{0} E_{0}^{2} C$

$= \frac{1}{2} * 8.85 * 1 0^{- 12} * {(56.5)}^{2} * 3 * 1 0^{8}$

$= 42377.11 * 1 0^{- 4}$

$l = 4.24 w m^{- 2}$

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New answer posted

2 months ago

Electromagnetic Waves Class 12th

Match List – I with List – II :

List – I List – II

(A) Ultravoilet &nbs

...more

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Vishal Baghel

Contributor-Level 10

Ultraviolet rays – sterilizing surgical instrument

Microwaves – Radar system

Infrared waves – Greenhouse effect

X-rays – Study crystal structure

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2 months ago

Electromagnetic Waves Class 12th

A beam of light travelling along X-axis is described by the electric field E_y = 900sin (t – x/c). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3 * 10⁷ ms^-1 will be:

(Given speed of light = 3 * 10⁸ ms^-1)

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alok kumar singh

Contributor-Level 10

$E_{y} = 900 s i n ω (t - \frac{x}{c})$

$\frac{F_{E}}{F_{B}} = \frac{q E}{| q \bar{V} * \bar{B} |} = \frac{C}{V} = \frac{3 * 1 0^{8}}{3 * 1 0^{7}} = 10$

$\Rightarrow \frac{F_{E}}{F_{B}} = \frac{10}{1}$

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2 months ago

Electromagnetic Waves Class 12th

An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of ________ GHz.

(Given µ_r = 1 for dielectric medium)

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Payal Gupta

Contributor-Level 10

$v = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8}}{\sqrt[]{6.25 * 1}} = \frac{3 * 1 0^{8}}{2.5} = 1.25 * 1 0^{8} m / s e c$

$A s f λ = f (5 * 1 0^{- 3} * 4) = 1.25 * 1 0^{8} \Rightarrow f = 6.25 G H z$

So, $f \approx 6 .$

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2 months ago

Electromagnetic Waves Class 12th

An electric bulb is rated as 200W. What will be the peak magnetic field at 4m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

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Payal Gupta

Contributor-Level 10

I_R = I_B (due to magnetic field)

(due to Radiation)

$\frac{P}{A} * E f f i c i e n c y = \frac{B_{0}^{2}}{2 μ_{0}} c$

$\Rightarrow \frac{200}{4 π * 4^{2}} * \frac{3.5}{100} = \frac{B_{0}^{2}}{2 * 4 π * 1 0^{- 7}} * 3 * 1 0^{8}$

$\Rightarrow B_{0} = 1.71 * 1 0^{- 8} T$

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New answer posted

2 months ago

Electromagnetic Waves Class 12th

A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 * 10^-2 Am^-1 at a point, what will be the approximate magnitude of electric field intensity at that point ?

(Given : Permeability of free space µ₀ = 4p * 10^-⁷ NA^-², speed of light in vacuum c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

Velocity of wave in a medium = $\frac{1}{\sqrt[]{μ_{m} \in_{m}}}$ $\frac{}{_{}_{}}$

$\begin{array}{l} v = \frac{1}{\sqrt[]{μ_{0} μ_{r} \in_{r} \in_{0}}} \\ v = \frac{3 * 1 0^{8}}{\sqrt[]{1.61 * 6.44}} = 0.931 * 1 0^{8} m / s e c \end{array}$

$\begin{array}{l} B = μ_{m} H \\ = μ_{r} μ_{0} H \\ = 1.61 * 4 π * 1 0^{- 7} * 4.5 * 1 0^{- 2} \\ = 1.61 * 4 π * 4.5 * 1 0^{- 9} \end{array}$

$\frac{E}{B} = v$

E = vB = 0.931 * 10⁸ * 1.61 * 4p * 4.5 * 10^-⁹

= [0.931 * 1.61 * 4p * 4.5] * 10^-¹

= 8.476

» 8.476 v m^-¹

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New answer posted

2 months ago

Electromagnetic Waves Class 12th

The displacement current of 4.425 $μ A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs^-¹. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x * 10^-³. The value of x is,

(Permittivity of free space, $E_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$

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Payal Gupta

Contributor-Level 10

According to definition of displacement current, we can write

$I_{d} = ε_{0} \frac{d ?}{d t} = ε_{0} \frac{d (E S)}{d t} = ε_{0} \frac{d (\frac{V}{l} S)}{d t} = \frac{ε_{0} S}{l} (\frac{d V}{d t})$

$\Rightarrow l = \frac{8.85 * 1 0^{- 12} * 40 * 1 0^{- 4} * 1 0^{6}}{4.425 * 1 0^{- 6}} 8 * 1 0^{- 3} m$

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New answer posted

2 months ago

Electromagnetic Waves Class 12th

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

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alok kumar singh

Contributor-Level 10

Given, $λ = 8 m m, v = 8 m m, v = 3 * 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 * 1 0^{- 3}} = \frac{π}{4} * 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{j} v / m$

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

B0 = $\frac{60}{3 * 1 0^{8}} = \frac{20}{1 0^{8}} = 2 * 1 0^{- 7}$ $\vec{B} = 2 * 1 0^{- 7} [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{k} T$

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