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The maxwell's equation:

$\oint \vec{B} . \vec{d l} = μ_{0} (i + ε_{0} . \frac{d ϕ_{E}}{d t})$ is a statement of :

Faraday's law of induction

Modified Ampere's law

Gauss's law of electricity

Gauss's law of magnetism

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Explanation

The given equation is a statement of modified ampere's law.

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The charge of a parallel plate capacitor is varying as $q = q_{0} \sin ω t$ . Then find the magnitude of displacement current through the capacitor. (Plate Area = A, separation of plates = d)

$q_{0} \cos (ω t)$

$q_{0} ω \sin ω t$

$q_{0} ω \cos ω t$

$\frac{q_{0} A ω}{d} \cos ω t$

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Explanation

$E = \frac{q}{A ε_{0}} φ = E A = \frac{q}{ε_{0}} C u r r e n t t h r o u g h c a p a c i t o r, i = ε_{0} \frac{d ϕ}{d t} = ε_{0} \frac{d}{d t} (\frac{q}{ε_{0}}) = \frac{d q}{d t} = i T h u s d i s p l a c e m e n t c u r r e n t = \frac{d q}{d t} = ω q_{0} \cos ω t$

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The rate of change of the voltage of a parallel plate capacitor if the instantaneous displacement current of 1 A is established between the two plates of a 1 $μ$ F parallel plate capacitor:

$10^{6} V / s$

10 V/s

$10^{8} V / s$

$10^{- 6} V / s$

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Explanation

$I_{D} = C \frac{d V}{d t} \frac{d V}{d t} = \frac{1}{1 \times 10^{- 6}} = 10^{6} V / s$

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The relation between electric field E and magnetic field induction B in an electromagnetic waves

$E = \sqrt{\frac{μ_{0}}{ε_{0}}} B$

E = cB

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

$E = \frac{B}{c^{2}}$

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Explanation

$E = cB$ is the correct relation.

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An electromagnetic wave is propagating along Y-axis. Then:

The oscillating electric field is along X-axis and the oscillating magnetic field is along Y-axis.

The oscillating electric field is along Z-axis and the oscillating magnetic field is along X-axis.

Both oscillating electric and magnetic fields are along Y-axis, but the phase difference between them is $90^{°}$

Both oscillating electric and magnetic fields are mutually perpendicular in arbitrary directions.

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Explanation

The oscillating electric field is along Z-axis and the oscillating magnetic field is along X-axis.

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In electromagnetic wave the phase difference between electric and magnetic field vectors $\vec{E} a n d \vec{B}$ is :

$π / 2$

$π$

$π / 4$

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Explanation

The electric and magnetic field components of a linearly polarized electromagnetic wave oscillate in such a way that they peak at the same time and they become zero at the same time but they point to different directions in space, separated by an angle of $90^{0} .$

Since there is no time difference between the peaks of the electric and magnetic oscillations the phase difference between the electric and magnetic field vectors of a linearly polarized electromagnetic wave is zero.

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An electromagnetic wave going through the vacuum is described by $E = E_{0} \sin (k x - ω t)$

Which is the following is/are independent of the wavelength?

$k / ω$

$k ω$

$ω$

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Explanation

$k = \frac{2 π}{λ} a n d ω = \frac{2 π c}{λ} H e n c e, \frac{k}{2 π} = \frac{ω}{2 π c} \Rightarrow \frac{k}{ω} i n d e p e n d e n t o f w a v e l e n g t h .$

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In a plane E.M. wave, the electric field oscillates sinusoidally at a frequency of 2.5 $\times 10^{10}$ Hz and amplitude 480 V/m. The amplitude of the oscillating magnetic field will be:

$1.52 \times 10^{- 8} W b / m^{2}$

$1.52 \times 10^{- 7} W b / m^{2}$

$1.6 \times 10^{- 6} W b / m^{2}$

$1.6 \times 10^{- 7} W b / m^{2}$

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Explanation

$(3) A m p l i t u d e o f M a g n e t i c f i e l d = \frac{E_{0}}{c} = \frac{480}{3 \times 10^{8}} = 1.6 \times 10^{- 6} W b / m^{2}$

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The energy density of the electromagnetic wave in vacuum is given by the relation

$\frac{1}{2} . \frac{E^{2}}{ε_{0}} + \frac{B^{2}}{2 μ_{0}}$

$\frac{1}{2} ε_{0} E^{2} + \frac{1}{2} μ_{0} B^{2}$

$\frac{E^{2} + B^{2}}{C}$

$\frac{1}{2} ε_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}$

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Explanation

$E l e c t r i c e n e r g y d e n s i t y = \frac{1}{2} ε_{0} E^{2} M a g n e t i c e n e r g y d e n s i t y = \frac{B^{2}}{2 μ_{0}} S o, t o t a l e n e r g y = \frac{1}{2} ε_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}$

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A lamp radiates power $P_{0}$ uniformly in all directions, the amplitude of electric field strength $E_{0}$ at a distance r from it is:

$E_{0} = \frac{P_{0}}{2 {πε}_{0} {cr}^{2}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{2 {πε}_{0} c r^{2}}\}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{4 {πε}_{0} {cr}^{2}}\}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{8 {πε}_{0} {cr}^{}}\}}$

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Explanation

$S_{a v} = \frac{P}{4 {πr}^{2}} = \frac{1}{2} ε_{0} {E_{0}}^{2} c E_{0} = \sqrt{\frac{P}{2 π ε_{0} c r^{2}}}$

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