Q 11 :

The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant k = 3 and permeability μ=2μ0, is

(μ0=permeability of vacuum)   [2026]

  • 6:1

     

  • 3 : 2

     

  • 36 : 1

     

  • 6 : 1

     

(1)

CV=μ=εrμr

CV=3×21=61



Q 12 :

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity,σ=10 mho/m. The ratio of maximum conduction current density to maximum displacement current density is ______.

[Take 14πε0=9×109 N m2/C2]           [2026]



(1800)

A

jc=σE

E⇒E0sin(ωt-kx)

jc=σE0sin(ωt-kx)

⇒(jc)max=σE0    ...(i)

Jd=idA=1A×ε0AdEdt

=ε0×E0ωcos(ωt-kx)

(jd)max=ε0E0ω    ...(ii)

(i)/(ii)

(jc)max(jd)max=σE0ε0ωE0⇒σε0ω

⇒10×4π×9×1092π×100×106

⇒1800



Q 13 :

The electric field of an electromagnetic wave travelling through a medium is given by E→(x,t)=25sin(2.0×1015t-107x)n^. Then the refractive index of the medium is _____ 

(All given measurement are in SI units.)         [2026]

  • 1.5

     

  • 1.7

     

  • 2

     

  • 1.2

     

(1)

ω=2×1015 rad/s

k=107 m-1

v=2πk·ω2π=ωk=2×1015107=2×108=c1.5

⇒μ=1.5



Q 14 :

Match List I With List II

  List I   List II
A) ∮B→·dl→=μ0ic+μ0ε0dϕEdt I. Gauss’ law for electricity
B) ∮E→·dl→=-dϕBdt II. Gauss’ law for magnetism
C) ∮E→·dA→=Qε0 III. Faraday law
D) ∮B→·dA→=0 IV. Ampere-Maxwell law

 

Choose the correct answer from the options given below:

  • A-IV, B-I, C-III, D-II

     

  • A-II, B-III, C-I, D-IV

     

  • A-IV, B-III, C-I, D-II

     

  • A-I, B-II, C-III, D-IV

     

(3)

Ampere-Maxwell law→∮B→·dl→=μ0ic+μ0ε0dϕEdt

Faraday law→∮E→·dl→=-dϕBdt

Gauss' law for electricity→∮E→·dA→=Qε0

Gauss' law for magnetism→∮B→·dA→=0



Q 15 :

A plane electromagnetic wave travelling in non-magnetic medium is given by

E=(9×108 NC-1)sin[(9×108 rad s-1)t-(6 m-1)x]

where x is in metre and t is in seconds. The dielectric constant of the medium is ________



(4)

V=ωk=9×1086=1.5×108 m/s,  

Refractive index μ=CV⇒K=3×1081.5×108

∴ K=4



Q 16 :

The electric field of an electromagnetic wave in free space is represented by E→=E0cos(ωt-kz)i^. The corresponding magnetic induction vector will be

  • B→=(E0C)cos(ωt-kz)j^

     

  • B→=(E0C)cos(ωt-kz)j^

     

  • B→=E0cos(ωt+kz)j^

     

  • B→=-(E0C)cos(ωt-kz)j^

     

(2)

E→×B→ gives direction of V→



Q 17 :

A plane electromagnetic wave of frequency 35 MHz travels in free space along positive X-direction. At a particular point (in space and time) E→=9.6 j^ V/m. The value of magnetic field at this point is:

  • 3.2×10-8 k^ T

     

  • 3.2×10-8 i^ T

     

  • 9.6 j^ T

     

  • 9.6×10-8 k^ T

     

(1)

EB=C

EB=3×108

B=E3×108=9.63×108

B=3.2×10-8T

B^=v^×E^

    =i^×j^=k^

so, B→=3.2×10-8k^ T