Energy Density & Intensity

A point light source emits E.M. waves in free space. A detector, placed at a distance of $L\text{ m}$, measures the intensity as $I_o$. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as $45^\circ$. The measured intensity at new location will be ______.

A laser beam has intensity of $4.0 \times 10^{14}\text{ W/m}^2$. The amplitude of magnetic field associated with beam is ______ T.

(Take $\epsilon_o = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2$ and $c = 3 \times 10^{8}\text{ m/s}$)

The equation of the electric field of an electromagnetic wave propagating through free space is given by: $E = \sqrt{377} \sin\left(6.27 \times 10^{3} t - 2.09 \times 10^{-5} x\right)\text{ N/C}$

The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right)\text{ W/m}^2$. The value of $\alpha$ is $\underline{\qquad}\,.\ \Big($ Take $\sqrt{\frac{\mu_0}{\varepsilon_o}} = 377$ in SI units $\Big)$

Due to presence of an em-wave whose electric component is given by $E = 100 \sin(\omega t - kx)\text{ N C}^{-1}$ a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

The unit of $\sqrt{\frac{2I}{\varepsilon_0 c}}$ is:

(I = intensity of an electromagnetic wave, c = speed of light)

In the given electromagnetic wave $E_y = 600 \sin(\omega t - kx)\text{ V m}^{-1}$, intensity of the associated light beam is (in $\text{W/m}^2$): (Given $\epsilon_0 = 9 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$)

A plane electromagnetic wave propagating in $x$-direction is described by

$E_y = \left(200\text{ V m}^{-1}\right) \sin\left[1.5 \times 10^{7} t - 0.05 x\right]$;

The intensity of the wave is:

(Use $\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$)

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10}\text{ Hz}$ and an amplitude of $50\text{ V m}^{-1}$. The total average energy density of the electromagnetic field of the wave is: [Use $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2$ ]

The energy density associated with electric field $\vec{E}$ and magnetic field $\vec{B}$ of an electromagnetic wave in free space is given by $(\epsilon_0 -$ permittivity of free space, $\mu_0 -$ permeability of free space)

For the plane electromagnetic wave given by $E = E_0 \sin(\omega t - kx)$ and $B = B_0 \sin(\omega t - kx)$, the ratio of average electric energy density to average magnetic energy density is

The ratio of average electric energy density and total average energy density of electromagnetic wave is:

The electric field in an electromagnetic wave is given as

$\vec{E} = 20 \sin \omega \left(t - \frac{x}{c}\right) \hat{j}\ \text{N C}^{-1}$

where $\omega$ and $c$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $5 \times 10^{-4}\text{ m}^{3}$ will be

(Given $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2$)

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

A point source of $100\text{ W}$ emits light with $5\%$ efficiency. At a distance of $5\text{ m}$ from the source, the intensity produced by the electric field component is:

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

(Given: $\varepsilon$0 = 8.85 $\times$ 10$-$12C2N$-1\,\text{m}-$2)

The intensity of the light from a bulb incident on a surface is 0.22 W/m2. The amplitude of the magnetic field in this light-wave is ______________ $\times$ 10$-$9 T.

(Given: Permittivity of vacuum $\varepsilon$0 = 8.85 $\times$ 10$-$12 C2 N${}^{-1}$-m${}^{-2}$, speed of light in vacuum c = 3 $\times$ 108 ms${}^{-1}$)

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

Nearly 10% of the power of a $110\text{ W}$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1\text{ m}$ from the bulb to a distance of $5\text{ m}$ is $a \times 10^{-2}\text{ W/m}^2$. The value of 'a' will be _________.

For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (Ue) and magnetic (Um) fields is:

The electric field in an electromagnetic wave is given by E = (50 NC${}^{-1}$) sin$\omega$ (t $-$ x/c) The energy contained in a cylinder of volume V is 5.5 $\times$ 10$-$12 J. The value of V is _____________ cm3. (given $\varepsilon$0 = 8.8 $\times$ 10$-$12C2N$-1\,\text{m}-$2)

Intensity of sunlight is observed as 0.092 Wm${}^{-2}$ at a point in free space. What will be the peak value of magnetic field at the point?

($\varepsilon_0 = 8.85 \times 10^{-12} C^2 N^{-1} m^{-2}$)

The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $\sqrt{\frac{x}{5}}$E. Where the value of x = ____________.

A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x $\times$ 10$-$1 V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take $\varepsilon_0 = 8.85 \times 10^{-12}$ C2N${}^{-1}$ m${}^{-2}$, c = $3 \times 10^{8}$ ms${}^{-1}$]

The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is $\frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}} \frac{V}{m}$. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.

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

A plane electromagnetic wave, has

frequency of 2.0 $\times$ 1010 Hz and its energy density is 1.02 $\times$ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to

( $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9} \frac{Nm^2}{C^2}$ and speed of light

= 3 $\times$ 108 ms–1)

Suppose that intensity of a laser is $\frac{315}{\pi}$ W/m2.

The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.

$\varepsilon$0 = 8.86 × 10–12 C2 Nm–2; c = 3 × 108 ms–1)

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then:

The mean intensity of radiation on the surface of the Sun is about 108 W/m2. The rms value of the corresponding magnetic field is closet to:

A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by:

[Given permittivity of space $\varepsilon$0 = 9 $\times$ 10–12 SI units, Speed of light c = 3 $\times$ 108 m/s]

A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be:

A red $LED$ emits light at $0.1$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $1\ m$ from the diode is:

During the propagation of electromagnetic waves in a medium:

An electromagnetic wave of frequency 1 $\times$ 1014 hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If $\varepsilon$0 = 8.8 $\times$ 10$-$12 C2/N-m2, then average energy density of electric field will be:

A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly:

The $rms$ value of the electric field of the light coming from the Sun is $720\text{ N/C}.$ The average total energy density of the electromagnetic wave is