Submitted by-Charan Singh(13105076)Q1. (B) Q2. (C) Q3. (C) Q4.
time.Q5. the time rate of change of magnetic flux linkage by the
circuit.Q6. (A) Q7. (B) Q8. (B) Q9. (A) Q10. (C) Q11. time rate
Q12. time and space.Q13. (A) True Q14.. Faraday discovered that the
induced emf Vemf (in volts), in any closed circuit is equal to the
time rate of change of the magnetic flux change by the circuit.
This is called Faradays Law and can be expressed as : Vemf = -d/dt
= -Nd/dt Where = N is the flux linkage, N is the number of turns in
the circuit, is the flux through each turn.The negative sign shows
that the induced voltage acts in such a way as to oppose the flux
producing it. This is known as Lenzs Law. And it emphasises that
the direction of current flow in the circuit is such that the
induced magnetic field produced by the induced current will oppose
the change in the original magnetic field.In Maxwells Form, this
law can be written as: Vemf = L E.dl =( -d/dt)S B.dS
Q15. The various Maxwells equations are:a) Gausss Law:
OR
b) Gausss Law for magnetism: OR
c) Maxwell-Faraday equation:(Faraday law of induction) OR
d) Amperes Circuital law:
OR
Q16. Vemf = (u x B).dl , dl = da , u = (d/dt) = a u x B = a x
Boaz = a Bo Vemf = 1=0 Boa . da = (Bo2)/2|l0 = (1/2)[Bol2 0] Vemf =
(1/2)Bol2
Q17. Vemf = -d/dt = -Nd/dt = B.dS Vemf = NBS(-d/dt) = d/dt =
130rad/s Vemf = -(130)(50)(0.06)(0.3x0.4) = - 46.8 VQ18.Vemf = ( u
x B ).l = uBlcos = (120 x 1000)/(3600 mls)(4.3x10-5 )(1.6)cos65o =
2.293cos65o Vemf = 0.97 mV
In free space, the following are the parameters: = 0, = o , = o
In general cases , wave equations or Helmholtz equations are:2Es
2Es =0 ........................................... (1)2Hs 2Hs =0
............................................ (2)Where 2=
j(+j)Putting the above values:2= jo(jo) = j22oo= j(oo) = j/c
Without loss of generality, if we assume that the wave
propagates along +az direction and that Es has only one
x-component, thenEs= Exs(z)axSubstituting into equation (1):(2-
2)Exs(z)ax =0This leads to: [d2/dz2 - 2]Exs(z) = 0Solution of above
equation is : Exs(z) = Eoe-z + EoezAs wave is propagating along +z
direction , Eo =0Inserting the time factor ejt into above equation
:E(z,t) = Re[Exs(z) ejtax ] = Re(Eoe(wt-(w/c)z)ax) E(z,t)=
Eocos(wt-(w/c)z)axSimilarly: H(z,t)= Re(Hoe(wt-(w/c)z)ay)
Q20. In conductors, the following are the parameters: , = o , =
or In general cases , wave equations or Helmholtz equations are:2Es
2Es =0 ........................................... (1)2Hs 2Hs =0
............................................ (2)Where 2=
j(+j)Putting the above values:2= jor( +jo) = +j= = (w/2) = (f)
Without loss of generality, if we assume that the wave propagates
along +az direction and that Es has only one x-component, thenEs=
Exs(z)axSubstituting into equation (1):(2- 2)Exs(z)ax =0This leads
to: [d2/dz2 - 2]Exs(z) = 0Solution of above equation is : Exs(z) =
Eoe-z + EoezAs wave is propagating along +z direction , Eo
=0Inserting the time factor ejt into above equation :E(z,t) =
Re[Exs(z) ejtax ] = Re(Eoe-ze(wt-()z)ax) E(z,t)= Eo
e-zcos(wt-()z)axSimilarly: H(z,t)= Re(Ho e-ze(wt-()z)ay)
Q21.Let us take the general form of a uniform plane wave: E(r,t)
= Eocos(k.r-wt) = Re[Eoej(k.r-wt)] .............(1)Where r = xax
+yay +zaz is the radius or position vector and k= kxax + kyay+ kzaz
is the wave number vector or the propagation vector; k is always in
the direction of wave propagation. The magnitude of k is related to
w according to the dispersion relation: k2 = k2x + k2y +k2z =
w2Thus, for lossless media, k is essentially the same as in the
preceding sections.With the general form of E as in eq 1, Maxwells
Equations for a source-free region reduce to k X E = wH k X H = -wE
k .H = 0 k .E = 0showing that E,H and k are mutually orthogonal ,
and E and H lie on the same plane. k.r = kxx + kyy + kzz =
constantThe H field corresponding to E field is: H = (1/w )k X E =
(ak X E)/Now, we consider the oblique incidence of a uniform plane
wave at a plane boundary as illustrated in the figure.The plane
defined by the propagation vector k and a unit vector normal to the
aN to the boundary is called the plane of incidence.The angle i
between k and aN is the angle of incidence.
Again, both the incident and reflected waves are in medium 1
while the transmitted wave is in medium 2.Let Ei = Eio cos(kixx +
kiyy + kizz wit)Er = Ero cos(krxx + kryy + krzz wrt)Et = Eto
cos(ktxx + ktyy + ktzz wtt)Where ki , kr and kt with their normal
and tangential components are shown in the figure.Since the
tangential component of E must be continuous across the boundary
z=0, Ei(z=0) + Er(z=0)+ Et(z=0)This boundary condition can be
satisfied by the waves in equations above for all x and y only if
1. wi = wr = wt =w2. kix = krx = ktx =kx 3. kiy = kry = kty =ky
Condition 1 implies that the frequency is unchanged. Conditions
2 and 3 require that the tangential components of the propagation
vectors be continuous(called the phase-matching conditions).This
means that the propagation vectors ki,kt,kr must all lie in the
plane of incidence. Thus, by condition 2 and 3, ki sini = kr sinr
And, ki sini = kt sint where r is the angle of reflection and t is
the angle of transmission.But for lossless media, ki=kr=1=w(11)
kt=2=w(22)
From the above equations, it is clear that : r= iSo that the
angle of reflection r equals the angle of incidence i , as in
optics. Also,Sin t/sin i = ki / kt = u2/u1 = [(11) /(22)] Where u=
w/k is the phase velocity. Above equation is well known Snells law
, which can be written as: n1 sin1 = n2 sin2where n1 = c(11) = c/u1
and n2 = c(22) = c/u2 are the refractive indices of the media.Based
on the general preliminaries on oblique incidence, i will show you
2 illustrations based on 2 special cases: one with the field E
perpendicular to the plane of incidence , the other with E parallel
to it.
Q22.(a) Let u= /w = loss tangent = {(/2)[(1+u2 )0.5 +1]}0.5 10 =
(w/c){(5x2)/2[(1+u2)0.5 +1]}0.5 10 =
[2x5x106(5)0.5]/(3x108){[(1+u2)0.5 + 1]}0.5 Which leads to : u = /w
= 1823 (b) =wu = 2 x 5 x 106x 1823 x 10-9/36 = 1.013 S/m
