## A plane EM wave travelling along z– direction is described by E = E0 sin(kz – ωt)j . Show that

Q: A plane EM wave travelling along z– direction is described by $\displaystyle E = E0 sin(kz – ωt)\hat{j}$ Show that

(i) The average energy density of the wave is given by $u_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}$

(ii) The time averaged intensity of the wave is given by $I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$

Solution : We know that energy density associated with E.F.

$\displaystyle u_E = \frac{1}{2} \epsilon_0 E^2$

We know that energy density associated with M.F

$\displaystyle u_B = \frac{B^2}{2 \mu_0}$

Thus total energy of EM wave is given by

$\displaystyle u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2$

$\displaystyle u = \frac{1}{2} \epsilon_0 E^2 + \frac{B^2}{2 \mu_0}$

putting the values of E.F. and M.F.

$\displaystyle u = \frac{1}{2} \epsilon_0 E_0^2 sin^2 (kz -\omega t) + \frac{B_0^2}{2 \mu_0} sin^2 (kz -\omega t)$

If we take the average value over a long period of time $\displaystyle sin^2 (kz -\omega t) = \frac{1}{2}$

The above equation can also be written as

$\displaystyle u_{avg} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{B_0^2}{4 \mu_0}$

But uE = uB

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

or , $\displaystyle u_{avg} = \frac{B_0^2}{2 \mu_0}$

### Intensity :

Energy crossing per unit area per unit time perpendicular to the direction of propogation is called intensity of wave.
Energy contained in the volume :
$\displaystyle U = u_{avg} \times Volume$

$\displaystyle U = \frac{1}{2} \epsilon_0 E_0^2 \times (A \times c \Delta t)$

$\displaystyle Intensity = \frac{U}{A \Delta t} = \frac{1}{2} \epsilon_0 E_0^2 c$

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

## Sea water at frequency v = 4 × 108 Hz has permittivity ε ≈ 80 ε0 , permeability μ ≈ μ0 and resistivity ρ = 0.25 m . imagine a parallel plate capacitor …

Q: Sea water at frequency v = 4 × 108 Hz has permittivity ε ≈ 80 ε0 , permeability μ ≈ μ0 and resistivity ρ = 0.25 m . imagine a parallel plate capacitor immersed in sea water and driven by an alternation voltage source V(t) = V0 sin (2πvt). What fraction of the conduction current density is the displacement current density ?

Sol: Suppose d =separation between plates of capacitor immersed in sea water & applied voltage across the plates  is V(t) = V0 sin (2πvt)

Electric field between the plates , $E = \frac{V(t)}{d}$

$E = \frac{V_0 }{d}sin2\pi \nu t$

Conduction current density , $\large J^c = \frac{E}{\rho} = \frac{V_0 }{\rho d}sin2\pi \nu t$

$\large J^c = J_0^c sin2\pi \nu t$ ; Where $\large J_0^c = \frac{V_0 }{\rho d}$

Displacement current density , $\large J^d = \epsilon \frac{dE}{dt}$

$\large J^d = \epsilon \frac{d}{dt}(\frac{V_0 }{d}sin2\pi \nu t)$

$\large J^d = \frac{\epsilon 2\pi \nu V_0}{d} cos (2\pi \nu t )$

$\large J^d = J_0^d cos (2\pi \nu t )$ ; Where $\large J_0^d = \frac{\epsilon 2\pi \nu V_0}{d}$

$\large \frac{J_0^d }{J_0^c} = \frac{\frac{\epsilon 2\pi \nu V_0}{d}}{\frac{V_0 }{\rho d}}$

$\large \frac{J_0^d }{J_0^c} = 2 \pi \nu \epsilon \rho$

$\large \frac{J_0^d }{J_0^c} = 2 \pi \nu (80\epsilon_0 ) \rho$

$= 2 \pi \nu (80\epsilon_0 ) \times 0.25 = 4 \pi \epsilon_0 \nu \times 10$

$= \frac{4 \times 10^8 \times 10}{9 \times 10^9} = \frac{4}{9}$

## Show that the magnetic field B at a point in between the plates of a parallel plate capacitor during charging is …

Q: Show that the magnetic field B at a point in between the plates of a parallel plate capacitor during charging is $\frac{\mu_0 \epsilon_0 r}{2} \frac{dE}{dt}$ (symbols having usual meaning). Sol: Magnetic field at a didtance r from the axis of plate

$\displaystyle B = \frac{\mu_0 I}{2 \pi r}= \frac{\mu_0}{2 \pi r}I_d$

$\displaystyle B = \frac{\mu_0}{2 \pi r} (\epsilon_0 \frac{d\phi_E}{dt})$

$\displaystyle B = \frac{\mu_0}{2 \pi r} (\epsilon_0 \frac{d}{dt}(E \pi r^2))$

$\displaystyle B = \frac{\mu_0 \epsilon_0 r}{2} \frac{dE}{dt}$

## The average Energy flux of sunlight is 1.0 kWm^-2 . This energy of radiation is falling normally on the metal plate surface …

Q: The average Energy flux of sunlight is 1.0 kW/mm2 . This energy of radiation is falling normally on the metal plate surface of area 10 cm^2 which completely absorbs the energy . How much force is exerted  on the plate if it is exposed to sunlight for 10 minutes .

Sol: Energy flux = 1.0 kW/mm2 = 1.0 × 103 W/m2

A = 10 cm2 = 10 × 10-4 m2

t = 10 min = 10 × 60 sec = 600 sec

Total Energy falling on the plate ,

U = energy flux × Area × time

= 1.0 × 103 × 10 × 10-4 × 600

= 600 J

Momentum of the radiation $\displaystyle p = \frac{U}{c} = \frac{600}{3 \times 10^8}$

= 2 × 10-6 kg m/s

Force on the plate $\displaystyle F = \frac{\Delta p}{t} = \frac{2 \times 10^{-6}}{600}$

F = 3.3 × 10-9 N

## If a capacitor of 2 μF is charged  to 20 V and then suddenly short-circuited by a coil of negligible resistance …

Q: If a capacitor of 2 μF is charged  to 20 V and then suddenly short-circuited by a coil of negligible resistance and of inductance 8.0 μH . Calculate the maximum amplitude and the frequency of the resulting current oscillations .

Sol: C = 2 × 10-6 F ; V = 20 volt

L = 8 × 10-6 H

The equation for oscillatory discharge of capacitor through an inductance is given by

$\displaystyle q = q_0 sin\omega t$ ; Where $\omega = \frac{1}{\sqrt{L C}}$

The resulting oscillatory current will be

$\displaystyle I = \frac{dq}{dt} = q_0 \omega cos\omega t$

$\displaystyle I = I_0 cos\omega t$ ;Where q0 ω = I0 is amplitude of oscillatory current

Amplitude of oscillatory current

I0 = q0 ω $\displaystyle = C V \times \frac{1}{\sqrt{L C}}$

$\displaystyle I_0 = V \times \sqrt{\frac{C}{L}}$

$\displaystyle I_0 = 20 \times \sqrt{\frac{2 \times 10^{-6}}{8 \times 10^{-6}}}$

= 10 A

## A parallel plate capacitor of area 50 cm^2 and plate separation 3.0 mm is charged initially to 80 μ C ….

Q: A parallel plate capacitor of area 50 cm2 and plate separation 3.0 mm is charged initially to 80 μ C . Due to a radioactive source nearby the medium between the plates gets slightly conducting and the plates loses the charge initially at the rate of 1.5 × 10^-8 C/s .

(i)  what is the magnitude and direction of displacement current ?

(ii) What is the magnetic field between the plates ?

Solution : As conduction current within the plates is from positive plate to negative plate .

Displacement current ,

$\displaystyle I_d = \epsilon_0 \frac{d \phi_e}{dt}$

$\displaystyle I_d = \epsilon_0 \frac{d}{dt}(E A)$

$\displaystyle I_d = \epsilon_0 A \frac{d E}{dt}$

$\displaystyle I_d = \epsilon_0 A \frac{1}{\epsilon_0 A} \frac{d Q}{dt}$

$\displaystyle I_d = \frac{d Q}{dt} = 1.5 \times 10^{-8} A$

(ii) Since charge is decreasing with time so, $\frac{dQ}{dt}$ and hence $\frac{dE}{dt} < 0$ . It shows that the direction of Id is opposite to that of electric field . Thus Id has same magnitude (= 1.5 x 10-8 A) as the conduction current but opposite direction to conduction current .

The total current I’ = I + Id = 0

According to Ampere’s circuital law ,

$\displaystyle \oint \vec{B}.\vec{dl} = \mu_0 I’ = 0$

Hence magnetic field within the plates is zero at all points .

## An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis …

Q: An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (figure). the wire is set into motion along its length with uniform velocity $v = v \hat{k}$  . Calculate the pointing vector $\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$ Sol: Electric field due to infinitely long wire is $\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 a} \hat{j}$

Magnetic field due to long wire carrying current is $\vec{B} = \frac{\mu_0 I}{2 \pi a} \hat{i}$

Equivalent current flowing through the wire is , I = λ v

$\large \vec{B} = \frac{\mu_0 \lambda v}{2 \pi a} \hat{i}$

$\large \vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$

$\large \vec{S} = \frac{1}{\mu_0}(\frac{\lambda}{2 \pi \epsilon_0 a} \hat{j} \times \frac{\mu_0 \lambda v}{2 \pi a} \hat{i})$

$\large \vec{S} = – \frac{\lambda^2 v}{4 \pi^2 \epsilon_0 a^2 } \hat{k}$

## Light with an energy flux of 25 × 10^4 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: $\Large F = \frac{2 I A}{c}$

## An EM waves is propagating in a medium with a velocity v→ = v î . The instantaneous oscillating electric field …

Q: An EM waves is propagating in a medium with a velocity v = v î . The instantaneous oscillating electric field of this EM waves is along +Y axis . Then the direction of oscillating magnetic field of the EM waves will be along

(a) -Y axis

(b) -X direction

(c) +Z direction

(d) -Z direction

Ans: (c)

Sol: For EM waves

$\displaystyle \vec{E} \times \vec{B} || \vec{v}$

$\displaystyle E \hat{j} \times \vec{B} || v \hat{i}$

$\displaystyle \vec{B} = B \hat{k}$

## Dimension of 1/μoεo is

Q: Dimension of $\frac{1}{\mu_0 \epsilon_0}$ is

(a) L2/T2

(b) T2/L2

(c) T/L

(d) L/T

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

$\large \frac{1}{\mu_0 \epsilon_0} = c^2 = [L^2T^{-2}]$

Ans: (a)