## The pressure exerted by an electromagnetic wave of Intensity I on a non-reflecting surface is

Q: The pressure exerted by an electromagnetic wave of Intensity I(W/m2) on a non-reflecting surface is (c = velocity of light)

(a) I c

(b) I c2

(c) I/c

(d) I/c2

Sol: $\large Intensity = \frac{Energy}{Area \times time}$

$\large I = \frac{F \times s}{A \times t} = \frac{F \times c}{A}$

$\large \frac{I}{c} = \frac{F}{A}$

$\large \frac{I}{c} = P$ ( Pressure = Force/Area )

Ans: (c)

## The amplitude of the sinusoidally oscillating electric field of a plane wave is 60 V/m . Then the amplitude of the magnetic field is

Q: The amplitude of the sinusoidally oscillating electric field of a plane wave is 60 V/m . Then the amplitude of the magnetic field is

(a) 2 × 107 T

(b) 6 × 107 T

(c) 6 × 10-7 T

(d) 2 × 10-7 T

Sol: $\large c = \frac{E_0}{B_0}$

$\large 3 \times 10^8 = \frac{60}{B_0}$

B0 = 2 × 10-7 T

## The speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permittivity 4 is

Q: The speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permittivity 4 is

(a) 1 × 108 m/s

(b) 2.5 × 108 m/s

(c) 2 × 108 m/s

(d) 1.5 × 108 m/s

Sol: $\large c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

$\large v = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}}$

$\large v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$

$\large v = \frac{3 \times 10^8}{\sqrt{4 \times 2.25}}$

v = 1 × 108 m/s

## The r.m.s value of electric field of light coming from the sun is 720 N/C . The average total energy density of the electromagnetic wave is

Q: The r.m.s value of electric field of light coming from the sun is 720 N/C . The average total energy density of the electromagnetic wave is

(a) 3.3 × 10-3 Jm-3

(b) 4.58 × 10-6 Jm-3

(c) 6.37 × 10-9 Jm-3

(d) 81.35 × 10-12 Jm-3

Sol: Total average energy density of EM wave is

$\large u = u_E + u_B = 2 u_E$ (since , uE = uB)

$\large u = 2 (\frac{1}{2}\epsilon_0 E_{rms}^2) = \epsilon_0 E_{rms}^2$

= (8.85 × 10-12)× (720)2

= 4.58 × 10-6 Jm-3

## Sodium lamps are used in foggy conditions because

Q: Sodium lamps are used in foggy conditions because

(a) Yellow light is scattered less by the fog particles

(b) Yellow light is scattered more by fog particles

(c) yellow is unaffected during its passage through the fog

(d) wavelength of yellow light is the mean of the visible part of

Sol: Yellow light is less scattered by the fog particles

Ans: (a)

## The mean electric energy density between the plates of charged capacitor is

Q: The mean electric energy density between the plates of charged capacitor is (here , q = charge on capacitor , A = Area of capacitor plate )

(a) $\large \frac{q^2}{2 \epsilon_0 A^2}$

(b) $\large \frac{q}{2 \epsilon_0 A^2}$

(c) $\large \frac{q^2}{2 \epsilon_0 A}$

(d) None of above

Sol: $\large Energy \; density = \frac{Energy}{volume} = \frac{U}{V}$

$\large u = \frac{\frac{q^2}{2 C}}{A d} = \frac{q^2}{2 \frac{\epsilon_0 A}{d} A d}$

$\large = \frac{q^2}{2 \epsilon_0 A^2}$

Ans: (a)

## The sun radiates electromagnetic energy at the rate of 3.9 × 1026 W . Its radius is 6.96 × 108 m . The intensity of sun light (in W/m2) at the solar surface will be

Q: The sun radiates electromagnetic energy at the rate of 3.9 × 1026 W . Its radius is 6.96 × 108 m . The intensity of sun light (in W/m2) at the solar surface will be

(a) 5.6 × 106

(b) 5.6 × 107

(c) 4.2 × 106

(d) 4.2 × 107

Sol: $\large Intensity = \frac{P}{A}= \frac{3.9 \times 10^{26}}{4 \pi r^2}$

$\large = \frac{3.9 \times 10^{26}}{4 \times 3.14 \times (6.99 \times 10^8)^2}$

= 5.6 × 107 W/m2