## A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives…

Q: A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity 5 × 10-16 W / m2. The maximum instantaneous potential difference across the two ends of the antenna is

(a) 1.23 μV

(b) 1.23 mV

(c) 1.23 V

(d) 12.3 μV

Ans: (a)

Sol: Energy Density $u = \frac{1}{2}\epsilon_0 E_0^2$

$\frac{I}{c} = \frac{1}{2}\epsilon_0 E_0^2$

Find E0 then , V0 = E0 d , where d= 2m

## The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density…

Q: The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is

(a) 6.37 × 10-9 J/m3

(b) 81.35 × 10-12 J/m3

(c) 3.3 × 10-3 J/m3

(d) 4.58 × 10-6 J/m3

Ans: (d)

Sol: Erms = 720 N/C

$\displaystyle u = \frac{1}{2} \epsilon_0 E_0^2$

## The electric field of a plane electromagnetic wave varies with time of amplitude 2 Vm-1 propagating along z-axis…

Q: The electric field of a plane electromagnetic wave varies with time of amplitude 2 Vm-1 propagating along z-axis. The average energy density of the magnetic field is (in Jm-3)

(a) 13.29 × 10-12

(b) 8.86 × 10-12

(c) 17.72 × 10-12

(d) 4.43 × 10-12

Ans: (b)

Sol: Avg mag. Energy density $\Large u_B = \frac{B_0^2}{4 \mu_0}$

## Light with an energy flux of 25 × 104 Wm-2 falls on a perfectly reflecting surface at normal incidence…

Q: Light with an energy flux of 25 × 104 Wm-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is

(a) 1.2 × 10-6 N

(b) 3.0 × 10-6 N

(c) 1.25 × 10-6 N

(d) 2.50 × 10-6 N

Ans: (d)

Sol: $\Large F = \frac{2 I A}{c}$

## A beam of plane polarized light falls normally on a polarizer of cross sectional area…

Q: A beam of plane polarized light falls normally on a polarizer of cross sectional area 3 × 10-4 m2. Flux of energy of incident ray in 10-3 W. The polarizer rotates with an angular frequency of 31.4 rad/s. The energy of light passing through the polarizer per revolution will be

(a) 10-4 Joule

(b) 10-3 Joule

(c) 10-2 Joule

(d) 10-1 Joule

Ans: (a)

Sol: $T = \frac{2 \pi}{\omega}$

$T = \frac{2 \times 3.14}{31.4} = 0.2 sec$

Hence, Energy = 0.5 × power × time

Energy = 0.5 × 10-3 × 0.2

= 10-4 Joule