Chapter # 40

Ex1.

Electromangetic Waves

SOLVED EXAMPLES

A parallel-plate capacitor is being charged. Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires.

s i

Sol.

The electric field between the plates is

Q E= A 0 where Q is the charge accumulated at the positive plate. The flux of this field through the given area is

Q Q E = A × A = . 0 0 The displacement current is id = 0

Q d E dQ = 0 = . dt dt 0

dQ is the rate at which the charge is carried to the positive plate through the connecting wire. dt Thus , id = ie .

But

Ex2.

Sol.

The maximum electric field in a plane electromagnetic wave is 600 N/C . The wave is going in the xdirection and the elelctric field is in the y-direction. Find the maximum magnetic field in the wave and its direction. We have B0 =

600 N / C E0 = = 2 × 10–6 T. 3 10 8 m / s c

As E , B and the direction of propagation are mutually perpendicular,, B should be along the z-direction. Ex3.

Sol.

The electric field in an electromagnetic wave is given by E = (50 N/C) sin (t – x/c). Find the energy contained in a cylinder of cross-section 10 cm 2 and length 50 cm along the X-axis. The energy density is

1 1 0E02 = × (8.85 × 10–12 C2 /N-m 2) × (50 N/C)2 2 2 = 1.1 × 10–8 J/m 3 . The volume of the cylinder is V = 10 cm 2 × 50 cm = 5 × 10–4 m 3 . The energy contained in this volume is U = (1.1 × 10–8 J/m 3) × (5 × 10–4 m 3) = 5.5 × 10–12 J. uav =

Ex4. Sol.

Find the intensity of the wave discussed in example (40.3) ? The intensity is

1 E 2 c = (1.1 × 10–8 J/m 3) × (3 × 108 m/s) 2 0 0 = 3.3 W/m 2 .

I=

manishkumarphysics.in

Page # 1

Chapter # 40

QUESTIONS

Electromangetic Waves

FOR

SHORT

ANSWER

1.

In a microwave oven , the food is kept in a plastic container and the microwave is directed towards the food. The food is cooked without melting or igniting the plastic container. Explain.

2.

A metal rod is placed along the axis of a solenoid carrying a high-frequency alternating current. It is found that the rod gets heated. Explain why the rod gets heated.

3.

Can an electromagnetic wave be deflected by an electric field ? By a magnetic field ?

4.

A wire carries an alternating current i = i0 sint . Is there an electric field in the vicinity of the wire ?

5.

A capacitor is connected to an alternating-current source. Is there a magnetic field between the plates?

6.

Can an electromagnetic wave be polarized ?

7.

A plane electromagnetic waves is passing through a region. Consider the quantities (a) electric field , (b) magnetic field , (c) electrical energy in a small volume and (d) magnetic energy in a small volume . Construct pairs of the quantities that oscillate with equal frequencies.

Objective - I 1.

A magnetic field can be produced by (A) a moving charge (C) none of them

pqEcdh; {ks=k fuEu ds dkj.k mRiUu fd;k tk (A) xfr'khy vkos'k (C) buesa ls dksbZ ugha 2.

(B) a changing electric field (D*) both of them ldrk gS (B) ifjofrZr fo|qr {ks=k (D*) ;g nksuksa gh

A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle (A) does not delfect (B) deflects for a very short time and then comes back to the original position (C) deflects and remains deflecred as long as the battery is connected (D*) deflects and gradually comes to the original position in a time which is large compared to the time constant.

,d dqrcq uqek dh lqb]Z lekukarj iV~V la/kkfj=k dh IysVksa ds e/; fjDr LFkku esa j[kh x;h gSA la/kkfj=k dks ,d izfrjks/k ds lkFk cSVjh ls tksM+k tkrk gSA dqrcq uqek dh lqbZ (A) fo{ksfir ugha gksxhA (B) vYi le; ds fy;s fo{ksfir gksxh rFkk iqu% viuh izkjfEHkd fLFkfr esa vk tk;sxhA (C) fo{ksfir gksxh rFkk tc rd cSVjh la;ksftr jgsxh] fo{ksfir gh jgsxhA (D*) fo{ksfir gksxh rFkk /khjs&/khjs le;&fLFkjkad dh rqyuk esa ,d yEcs le; i'pkr~ ewy fLFkfr esa vk tk;sxhA 3.

Dimensions of 1/(00) is 1/(00) dh foek,a gS (A) L/T (B) T/L

(C*) L2/T2

(D) T2/L2

4.

Electromagnetic waves are produced by (A) a static charge (B) a moving charge (C*) an accleration charge (D) chargeless particles fo|qr&pqEcdh; rjaxsa fuEu mRiUu dh tkrh gS (A) ,d fLFkj vkos'k }kjk (B) ,d xfr'khy vkos'k }kjk (C*) ,d Rofjr vkos'k }kjk (D) vkos'k jfgr d.kksa }kjk

5.

An electromagnetic wave going through vacuum is described by E =E0 sin(kx - t); B = B0 sin(kx - t). Then (A*) E0 k = B0 (B) E0B0 = k (C) E0 = Bo k manishkumarphysics.in

(D) none ot these

Page # 2

Chapter # 40

Electromangetic Waves

fuokZr esa xeu dj jgh fo|qr&pqEcdh; rjax fuEu lehdj.kksa }kjk O;Dr dh tkrh gS : rks

E =E0 sin(kx - t); B = B0 sin(kx - t).

(A*) E0 k = B0 6.

7.

(B) E0B0 = k

(C) E0 = Bo k

(D) buesa

ls dksbZ ugha

An electric field E and a magnetic field B exist in a region. The fields are not perpandicular to each other.. (A) This is not possible (B) No electromagnetic wave may be passing through the region (C*) An electromagnetic wave may be passing through the region. (D) An electromagnetic wave is certainly passing through the region fdlh LFkku ij fo|qr {ks=k E rFkk pqEcdh; {ks=k B mifLFkr gSA {ks=k ijLij yEcor~ ugha gS (A) ;g laHko ugha gSA (B) bl LFkku ls dksbZ fo|qr&pqEcdh; rjax ugha xqtj jgh gSA (C*) gks ldrk gS fd bl LFkku ls fo|qr&pqEcdh; rjax xqtj jgh gksA (D) bl LFkku ls fo|qr&pqEcdh; rjax fuf'pr :i ls xqtj jgh gSA Consider the following two statement regarding a linearly polarized, plane electromagnetic wave (A) The electric field and the magnetic field have equal average values (B) The electric energy and the magnetic energy have equal average values (A*) Both A and B are true (B) A is false but B is true (C) B is false but A is true (D) Both A and B are false

jSf[kd /kzqfor] lery fo|qr&pqEcdh; rjax ds lEca/k esa fuEu nks dFkuksa ij fopkj dhft;s % (a) fo|qr {ks=k rFkk pqEcdh; rjax ds vkSlr eku cjkcj gksrs gSAa (b) fo|qr&ÅtkZ rFkk pqEcdh; ÅtkZ ds vkSlr eku cjkcj gksrs gSaA (A*) a o b nksuksa lR; gSA (B) a vlR; gS fdUrq b lR; gSA (C) b vlR; gS] fdUrq a lR; gSA (D) a o b nksuksa vlR; gSA 8.

A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving (A*) along the electric field (B) along the magnetic field (C) along the direction of propagation of the wave (D) in a plane containing the magnetic field and the direction of propagation

,d lery fo|qr&pqEcdh; rjax ds ekxZ esa ,d eqDr bysDVªkWu j[k fn;k tkrk gSA bysDVªkWu xfr izkjEHk djsxk (A*) fo|qr&{ks=k dh fn'kk esAa (B) pqEcdh; {ks=k dh fn'kk esAa (C) rjax&xeu dh fn'kk esaA (D) ml ry esa ftlesa pqEcdh; {ks=k rFkk xeu dh fn'kk gksxAsa 9.

A plane ekectromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E. fdlh inkFkZ dh lrg ij ,d lery fo|qr pqEcdh; rjax vkifrr gSA rjax laosx p rFkk ÅtkZ E iznku djrh gS (A) p = 0, E 0 (B) p 0, E = 0 (C*) p 0, E 0 (D) p = 0, E = 0

Objective - II 1.

An electromagnetic wave going through vacuum is described by E = E0 sin(kx - t). Which of the following is/ are independent of the wavelength ? fuokZr esa xeu dj gh ,d fo|qr&pqEcdh; rjax fuEukuqlkj O;Dr dh tkrh gS : E = E0 sin(kx - t). fuEu esa ls [email protected] rjaxnS/;Z ij fuHkZj ugha djsxsa (A) k (B) (C*) k/ (D) k

2.

Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor (A*) increases (B*) decreases (C) does not change (D) is zero la/kkfj=k dh IysVksa ds e/; varjky esa foLFkkiu /kkjk xqtjrh gS] tc la/kkfj=k dk vkos'k (A*) c<+rk gS (B*) de gksrk gS (C) ifjofrZr ugha gksrk gSA (D) 'kwU; gksrk gSA

manishkumarphysics.in

Page # 3

Chapter # 40 Electromangetic Waves 3. Speed of electromagnetic waves is the same (A) for all wavelength (B) in all media (C*) for all intensities (D) for all frequencies fo|qr&pqEcdh; rjaxksa dh pky ,d leku jgrh gS (A) leLr rjaxnS/;ks± ds fy;s (B) leLr ek/;eksa esa (C*) leLr rhozrkvksa ds fy;s (D) leLr vko`fÙk;ksa

ds fy;s

4.

Which of the following have zero average value in a plane electromagnetic wave ? (A*) electric field (B*) magnetic field (C) electric energy (D) magnetic energy fdlh lery fo|qr&pqEcdh; rjax esa fuEu esa ls fdldk vkSlr eku 'kwU; gksrk gS (A*) fo|qr&{ks=k (B*) pqEcdh;&{ks=k (C) fo|qr&ÅtkZ (D) magnetic energy

5.

The energy contained in a small volume through which an electromagnetic wave is passing oscillates with (A) zero frequency (B) the frequency of the wave (C) half the frequency of the wave (D*) double the frequency of the wave ,d vR;Yi vk;ru esa fufgr ÅtkZ ftlls dksbZ fo|qr&pqEcdh; rjax xqtj jgh gks] nksyu djrh gS (A) 'kwU; vko`fÙk ls (B) rjax dh vko`fÙk ls (C) rjax dh vk/kh vko`fÙk ls (D*) rjax dh nqxuh vko`fÙk ls

WORKED OUT EXAMPLES 1.

Sol.

A parallel-plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area. Suppose the charge on the capacitor at time t is Q. The electric field between the plates of the capacitor

Q is E = A . The flux through the area considered is 0 Q A Q E = A . = 2 . 2 0 0 1 id = 0 d E = 0 2 0 dt

The displacemet current is

2.

dQ i dt = . 2

A plane electromagnetic wave propagating in the x-direction has a wavelength of 5.0 mm. The electric field is in the y-direction and its maximum magnitude is 30 V/m. Write suitable equations for the electric fields in the wave may be written as x E = E0 sin t c x B =m B0sin t c

We have , w = 2v =

Thus ,

2 c.

2 E = E0 sin (ct x )

2 = (30 V/m) sin 5.0 mm (ct x ) .

The maximum magnetic field is B0 =

So ,

3.

30 V / m E0 = = 10–7 T. 3 10 8 m / s c

2 2 B = B0 sin (ct x ) = (10–7 T) sin 5.0 mm(ct x) The magnetic field is along the Z-axis.

A light beam travelling in the x-direction is described by the eletric field Ey = (300 V/m) sin(t – x/c). An electron is constrained to move along the y-direction with a speed of 2.0 × 107 m/s. Find the maximum electric force and the maximum magnetic force on the electron. manishkumarphysics.in

Page # 4

Chapter # 40 Electromangetic Waves Sol. The maximum electric field is E0 = 300 V/m. The maximum magnetic field is B0 =

300 V / m E0 = = 10–6 T 3 10 8 m / s c

along the z-direction. The maximum electric force on the electron is Fe = qE0 = (1.6 × 10–19 C) × (300 V/m) = 4.8 × 10–17 N. The maximum magnetic force on the electron is F b = qv B = qvB0 max

= (1.6 × 10–19 C) × (2.0 × 107 m/s) × (10–6 T) 4. Sol.

= 3.2 × 10–18 N. Find the energy stored in a 60 cm length of a laser beam operating at 4 mW.

c 60 cm 60 cm = 2 × 10– c 9 s.The energy contained in the 60 cm length passes through a cross-section of the beam in 2×10 –9 s. But the energy passing through any cross-section in 2 ×10–9 s is

The time taken by the electaromagnetic wave to move through a distnce of 60 cm is t =

U = (4 mW) × (2 × 10–9 s) = (4 × 10–3 J/S) × (2 × 10–9 S) = 8 × 10–12 J. This is the energy contained in 60 cm length.

5.

Find the amplitude of the electric field in a parallel beam of light of intensity 2.0 W/m 2.

Sol.

The intensity of a plane electromagnetic wave is I = uavc = 0E02 C

or ,

E0 =

2I 0c

=

2 ( 2. 0 W / m 2 ) 2 8.85 10 12 C (3 10 8 m / s) 2 N m

= 38.8 N/C.

EXERCISE Q.1

Show that the dimensions of the displacement current o

O;Dr dhft;s fd foLFkkiu /kkjk o

dE dt

dE are that of an electric current. dt

dh foek,¡] fo|qr /kkjk ds leku gh gksrh gSA

Ans : Q.2

A point charge is moving along a straight line with a constant velocity v consider a small area A perpendicular to the direction of motion of the charge (figure 40= E1) Calculate the displacement current through the area when its distance from the charge is x .The value of x is not large so that the electronic given by coulomb’s field at any instant is essentially given by Coulomb’s law. manishkumarphysics.in

Page # 5

Chapter # 40

Electromangetic Waves

,d fcUnq vkos'k fu;r pky v ls ,d ljy js[kk ds vuqfn'k xfr'khy gSA vkos'k dh xfr dh fn'kk ds yEcor~ ,d vR;Yi {ks=kQy ij fopkj dhft;sA tc {ks=kQy vkos'k ls x nwjh ij gS rks blls xqtjus okyh foLFkkiu /kkjk dh x.kuk dhft;sA x dk eku bruk vf/kd ugha gS fd fdlh {k.k ij fo|qr {ks=k vko';d :i ls dwyke ds fu;e }kjk O;Dr fd;k tk;sA q Au

Ans:

2x 3

Q.3

A parallel-plate capacitor having plate area A and plate separation d is joined to a battery of emf e and internal resistance R at t = 0 consider a plane surface of area A/2 parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. ,d lekukarj iV~V la/kkfj=k dh IysV dk {ks=kQy A rFkk IysV&varjky d gS] bldks t = 0 ij fo-ok-cy E rFkk vkarfjd izfrjks/k R okyh cSVjh ls tksM+k x;k gSA lery lrg ftldk {ks=kQy A/2 gS] ij fopkj dhft;s tks IysVksa ds chp lefer

:i ls fLFkr gS rFkk IysVksa ds lekukarj gSA bl lrg ls xqtjus okyh foLFkkiu /kkjk le; ds Qyu :i esa Kkr dhft;sA Ans: Q.4

Q.5

td e 2R AR

Consider the situation of the previous problem Define displacement resistance R d = V/ id of the space between the plates where V is the potential difference between the plates and is the displacement currently show that Rd varies with time as Rd = R (e t/ – 1) fiNys iz'u esa of.kZr fLFkfr ij fopkj dhft;sA IysVksa ds chp ds LFkku ds fy;s foLFkkiu izfrjks/k Rd = V/ id ifjHkkf"kr dhft;s] tgk¡ V IysVksa ds chp foHkokarj rFkk Id foLFkkiu /kkjk gSA O;Dr dhft;s fd Rd le; ds lkFk fuEukuqlkj ifjofrZr gksrk gS : Rd = R (e t/ – 1) Using B = µ0 H find the ratio Eo/ Ho for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance This ratio is a universal constant called the impedance of free space. B = µ0 H dk mi;ksx djds] fuokZr esa xeu dj jgh lery fo|qr&pqEcdh; rjax ds fy;s vuqikr Eo / Ho Kkr dhft;sA

iznf'kZr dhft;s fd bldh foek,¡ fo|qr izfrjks/k ds leku gSA ;g vuqikr ,d lkoZf=kd fu;rkad gS rFkk eqDrkdk'k dh izfrck/kk dgykrk gSA Ans:

377

Q.6

The sunlight reaching the earth has maximum electric field of 810 V/m .What is the maximum magnetic field in this light ? i`Foh ij igqp a us okys lw;&Z izdk'k esa fo|qr {ks=k dk vf/kdre eku 810 [email protected] gksrk gSA bl izdk'k esa pqEcdh; {ks=k dk

vf/kdre eku fdruk gksrk gS\ Ans:

2.7 µ T

Q.7

The magnetic field in a plane electromagnetic wave is given by B =( 200 µ T ) sin [(4.0 × 10 15 s – 1) (t – x /c]. Find the maximum electric field and the average energy density corresponding to the electric field ,d lery fo|qr&pqEcdh; rjax esa pqEcdh; {ks=k fuEu lehdj.k }kjk O;Dr fd;k tkrk gS : B =( 200 µ T ) sin [(4.0 × 10 15 s – 1) (t – x /c]

vf/kdre fo|qr {ks=k ,oa fo|qr {ks=k ds laxr vkSlr ÅtkZ ?kuRo Kkr dhft;sA Ans:

6 × 10 4 N/C , 0.016 J/m 3

Q.8

A laser beam has intensity 2.5 × 20 14 W/m 2 Find the amplitudes of electric and magnetic fields in the beam. ,d ystj iqt a dh rhozrk 2.5 × 2014 okWV/eh2 gSA iqt a esa fo|qr rFkk pqEcdh; {ks=kksa ds vk;ke Kkr dhft;sA 8 4.3 × 10 N/C 1.44 T

Ans: Q.9

The intensity of the sunlight reaching the earth is 1380 W/m2 Assume this light to be a plane monochromatic wave Find the amplitudes of electric and magnetic fields in this wave. i`Foh ij igqp a us okys lkSj&izdk'k dh rhozrk 1380 okWV/eh2 gSA eku yhft;s fd ;g izdk'k ,d o.khZ lery rjax gSA bl

rjax esa fo|qr ,oa pqEcdh; {ks=kksa ds vk;ke Kkr dhft;sA Ans:

1.02 × 10 3 N/C , 3.40 × 10

–6

T

manishkumarphysics.in

Page # 6

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close