Chapter 15: Problem 2154
The frequency of light wave of wavelength \(5000 \mathrm{~A}\) is \(\mathrm{Hz}\) (A) \(6 \times 10^{14}\) (B) \(1.5 \times 10^{-2}\) (C) \(1.5\) (D) \(6 \times 10^{1}\)
Chapter 15: Problem 2154
The frequency of light wave of wavelength \(5000 \mathrm{~A}\) is \(\mathrm{Hz}\) (A) \(6 \times 10^{14}\) (B) \(1.5 \times 10^{-2}\) (C) \(1.5\) (D) \(6 \times 10^{1}\)
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Get started for freeEnergy density of an electromagnetic wave of intensity \(0.02 \mathrm{Wm}^{-2}\) is (A) \(6.67 \times 10^{-11} \mathrm{Jm}^{-3}\) (B) \(6 \times 10^{6} \mathrm{Jm}^{-3}\) (C) \(1.5 \times 10^{10} \mathrm{Jm}^{-3}\) (D) none of the above
Unit of \(\mu_{0} \mathrm{c}\) is same as that of (A) current (B) resistance (C) electric charge (D) velocity
The amplitude of the magnetic field part of an electromagnetic wave in vacuum is \(\mathrm{Bm}=510 \mathrm{nT}\). Then the amplitude of the electric part of the wave is (A) \(1.53 \times 10^{11} \mathrm{~V} / \mathrm{m}\) (B) \(1.53 \mathrm{~V} / \mathrm{m}\) (C) \(1.53 \times 10^{2} \mathrm{~V} / \mathrm{m}\) (D) \(1.53 \times 10^{8} \mathrm{~V} / \mathrm{m}\)
According to Maxwell, a changing electric field produces (A) emf (B) Electric current (C) magnetic field (D) radiation pressure
The potential difference between the plates of a parallel plate capacitor is charging at the rate of \(10^{6} \mathrm{Vs}^{-1}\). If the capatance is $2 \mu \mathrm{F}$. The displacement current in the dielectric of the capacitor will be (A) \(4 \mathrm{~A}\) (B) \(3 \mathrm{~A}\) (C) \(2 \mathrm{~A}\) (D) \(1 \mathrm{~A}\)
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