Self Studies

Electromagnetic Waves Test - 16

Result Self Studies

Electromagnetic Waves Test - 16
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    Which wave characteristic describes the product of the frequency and the wavelength?
    Solution
    The relation between frequency $$f$$ and wavelength $$\lambda$$ is given by ,
                  $$v=f\lambda$$
    it is clear that velocity $$v$$ of wave describes the product of frequency and wavelength.
  • Question 2
    1 / -0
    Who first proposed the light as electromagnetic wave ?
    Solution
    In 1864, Maxwell predicted the existence of electromagnetic waves, the existence of which had not been confirmed before that time, and out of his prediction came the concept of light being a wave, or more specifically, a type of electromagnetic wave.

    It is a wide intuition that Albert Einstein proposed the dual nature theory but is not correct.
  • Question 3
    1 / -0
    In electromagnetic wave, according to Maxwell, changing electric field gives _______.
    Solution
    By integral form of modified Maxwell's equations, we have:
    $$\oint _{ \partial S } { B }\cdot d{ \ell  }=\mu _{ 0 }\int _{ S } ({ J }+\epsilon _{ 0 }\dfrac { \partial { E } }{ \partial t } )\cdot d{ S }\, $$

    Where the displacement current density is given by:
    $${ J }_{ D }={ \epsilon  }_{ 0 }\dfrac { \partial E }{ \partial t } $$

    The above integral states that, the line integral of magnetic field around any loop is equal to $${ \mu  }_{ 0 }$$ times the surface integral of the total current density (free current plus displacement current) over any open surface which has the loop as its boundary.
    Thus, it is clearly seen that a changing electric field gives rise to displacement current.
  • Question 4
    1 / -0
    In an electromagnetic wave
    Solution
    For electromagnetic waves $$\vec{E}$$ and $$\vec{B}$$ are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of $$\vec{E}\times \vec{B}$$.
    The direction of propagation of the wave is the direction of propagation of its energy and power.
    Hence, correct answer is option B.
  • Question 5
    1 / -0
    Solar radiation is
    Solution
    Solar radiation is radiant energy emitted by sun, particularly transverse electromagnetic energy wave. About half of the radiation is in the visible short-wave part of electromagnetic spectrum. The other half is mostly in the near-infrared part, with some in ultraviolet part of spectrum.
  • Question 6
    1 / -0
    In a parallel C-R network with a voltage source, the current flowing through resistor is _______ current and that flowing through the capacitor is ________ current.
    Solution
    In a parallel C-R circuit network with a voltage source, the current flowing through resistor is conduction current and that flowing through the capacitor is displacement current.
  • Question 7
    1 / -0
    The frequency of electromagnetic wave in free space is $$2$$ MHz. When it passes through a region of relative permittivity $$\varepsilon_r=4.0$$, then its wave length __________ & frequency ______________.
    Solution
    $$c=\dfrac{1}{\sqrt {\epsilon_o \mu_o}}$$
    In a dielectric
    $$v=\dfrac{1}{\sqrt {\epsilon_o \epsilon_r \mu_o \mu_r}}$$
    $$v=\dfrac{1}{\sqrt {4\epsilon_o \mu_o} }$$
    $$v=\dfrac{1}{2\sqrt {\epsilon_o \mu_o }}$$
    $$v=c/2$$
    Frequency does not change so wavelength should become halved to make velocity half fold.($$v=\nu \lambda$$)
    Hence the correct option is (D).
  • Question 8
    1 / -0
    $$X-$$ray falling on a material 
    Solution
    The emitted X-rays transfer energy to the material on which it is falling.
  • Question 9
    1 / -0
    An electromagnetic radiation has an energy of 13.2 keV. Then the radiation belongs to the region of
    Solution
    Given : E = 13.2 keV

    The wavelength of the radiation is given by;
    $$\lambda( in \overset{o}{A})= \dfrac{hc}{E(ineV)}$$

    $$= \dfrac{12400}{13.2 \times 10^3} = 0.939\overset{o}{A} \\\approx 1\overset{o}{A}$$

    X-rays cover wavelengths ranging from about $$10^{-8} m (10 nm )$$ to $$10^{-13} m (10^{-4}nm)$$.

    Electromagnetic radiation of energy 13.2 keV belongs to X-ray region of the electromagnetic spectrum.
  • Question 10
    1 / -0
    If the directions of electric and magnetic field vectors of a plane electromagnetic wave are along positive $$y$$-direction and positive $$z$$-direction respectively, then the direction of propagation of the wave is along:
    Solution
    $$e=E\times B$$, direction of propagation is always perpendicular to plane of $$E$$ and $$B$$. It will be positive in $$x$$-direction.
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now