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Electromagnetic Waves Test - 25

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Electromagnetic Waves Test - 25
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  • Question 1
    1 / -0
    A long straight cable of lenght $$l$$ is placed symmetrically along z - axis has radius $$a(< < l)$$. The cable consists of a thin wire and a coaxial conducting tube. An alternating current $$I(t) = I_0 sin (2\pi\upsilon t) $$flows down the central thin wire and return along the coaxial conducting tube. The induced electric field at a distance s from the wire inside the cable is $$E(s, t) = \mu_oI_o\upsilon cos(2\pi \upsilon t)\ ln\left (\dfrac{s}{a} \right )\hat{k}$$. The displacement current density inside the cable is 
    Solution
    Given: The induced electric field at a distance from wire is $$E(s,t)=\mu_0I_0\upsilon\ cos(2\pi\upsilon t)ln\left(\dfrac{s}{a}\right)\hat k$$

    Displacement current density is given by:
    $$\vec{J}_d=\epsilon_0\dfrac {d\bar{E}}{dt}$$

    $$\Rightarrow\epsilon_0\mu_0I_0 v\dfrac{\partial}{\partial(t)}(cos2\pi v t)(\dfrac{s}{a})\hat{k}$$

    Substitute $$c$$ for $$\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$$.

    $$\Rightarrow\dfrac{1}{c^2}I_0 2 \pi v^2(-sin(2\pi v t)\ ln(\dfrac{s}{a})\hat{k}$$

    $$\Rightarrow(\dfrac{v}{c})^2 \, 2\pi I_0 sin (2\pi \,v \,t)\ ln(\dfrac{a}{s})\hat{k}$$

    $$\implies\dfrac{2\pi}{\lambda^2}I_0\ ln(\dfrac{a}{s})sin(2\pi v t) \hat {k}$$

    Option $$(A)$$ is correct.
  • Question 2
    1 / -0
    Choose the correct answer from the alternatives given.
    Which among the following does not represent Maxwell's equation?
    Solution
    Maxwell's equation are as follows 
    (i) $$\phi \overline{E}.\overline{dA} = \dfrac{q}{\varepsilon _0}$$ (Gauss's law of electricity)
    (ii) $$\phi \overline{B}.\overline{dA} = 0$$ (Gauss's law of magnetism)
    (iii) $$\phi \overline{E}.\overline{dl} = -\dfrac{d\phi E}{dt}$$ (Saraday's law)
    (iv) $$\phi \overline{B}.\overline{dl} =\mu_0I_C + \mu_0 \varepsilon_0 \dfrac{d\phi E}{dt}$$ (Ampere - Maxwell law)
  • Question 3
    1 / -0
    The ultra-high frequency band of radio waves in electromagnetic wave is used as in:
    Solution
    The ultra high-frequency band of radio waves in FM waves is used in cellular phone communication. It is of the order of $$10^9Hz$$
  • Question 4
    1 / -0
    Choose the correct answer from the alternatives given.
    Which of the following is not true for electromagnetic waves?
    Solution
    They travel at different speed in air depending on their frequency. At constant as the speed will be same irrespective of frequency. Also frequency is source dependent and doesn't controls speed.
  • Question 5
    1 / -0
    An electromagnetic wave radiates outwards from a dipole antenna, with $$E_0$$ as the amplitude of its electric field vector. The electric field $$B_0$$ which transports significant energy from the source falls off as
    Solution
    An antenna that produces the Electromagnetic wave are radiated outwards. The amplitude of electric field vector $$(E_0)$$. This electric field vector transports the energy from the source through the medium.

    The electric field intensity of the wave from the source at a distance is inversely proportional to the distance between the source and the point.
    $$E_0=\dfrac 1r$$

    Option $$(C)$$ is correct.
  • Question 6
    1 / -0
    The part of the spectrum of the electromagnetic radiation used to cook food is then
    Solution
    Microwaves are used to cook food. Microwave oven is a domestic application of these waves.
  • Question 7
    1 / -0
    The electric field of an electromagnetic wave traveling through the vacuum is given by the equation $$E=E_0\ sin (Kx-\omega t).$$ The quantity that is independent of wavelength is:
    Solution
    To find: The quantity that is independent of the wavelength.

    The angular frequency $$\omega$$ is given by:
    $$\omega \, = \, 2\pi \nu$$
    The frequency of a wave varies with the wavelength. So, angular frequency is dependent on wavelength.

    The quantity $$k$$ is defined as the wavenumber and it is given by:
    $$k = \dfrac{2\pi}{\lambda}$$
    It shows that $$k$$ is dependent on wavelength.

    The value of $$\dfrac{k}{\omega}$$ can be obtained as:
    $$\dfrac {k}{\omega} \, = \, \dfrac{2\pi / \lambda}{2\pi \nu}\\\implies \, \dfrac{1}{\nu \lambda} \, = \, \dfrac{1}{c}\,\,\ \ \ \ \ \ \ \ \ \ \ \ \  (\because \, c \, = \, \nu \lambda)$$
    where c is the speed of electromagnetic wave in vacuum. It is a constant whose value is $$3 \, \times \, 10^8 \, ms^{-1}$$.

    So, option $$(B)$$ is correct.
  • Question 8
    1 / -0
    The matter-wave picture of electromagnetic wave/radiation elegantly incorporated the:
    Solution
    The matter-wave picture of electromagnetic wave/radiation elegantly incorporated  the Heisenberg uncertainty principle.
    The correct option is A.
  • Question 9
    1 / -0
    Which of the following electromagnetic waves is used in medicine to destroy cancer cells?
    Solution
    Gamma rays has property to kill cancer cell because the energy released by gamma ray is perfect to kill and leave out the healthy ones.
  • Question 10
    1 / -0
    The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is:
    Solution
    The intensity of an electromagnetic wave is given by:
    $$I=U_{av}c$$

    here, $$U_{av}$$ is the average energy and $$c$$ is the speed of light.

    The average energy in terms of the electric field is written as:
    $$U_{av}=\dfrac 12\varepsilon_oE_o^2$$

    Average energy in terms of magnetic field:
    $$U_{av}=\dfrac 12\dfrac{B_o^2}{\mu_o}$$

    We know that, $$E_o=cB_o$$ and $$c=\dfrac{1}{\sqrt{\mu_o\varepsilon_o}}$$

    $$\therefore U_{av\ (electric\ field)}=\dfrac 12\varepsilon_o\times\dfrac{1}{\mu_o\varepsilon_o}B_o^2$$

    $$\Rightarrow\dfrac 12\dfrac{B_o^2}{\mu_o}$$

    Since the energy due to electric and magnetic field is same. so, both will have the same contribution to the intensity.
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