Page 51 - Fiber Optic Communications Fund

32 Fiber Optic Communications

1.2 The magnetic field intensity at a distance of 1 mm from a long conductor carrying d.c. is 239 A/m. The
2
cross-section of the conductor is 2 mm . Calculate (a) the current and (b) the current density.
5
2
(Ans: (a) 1.5 A; (b) 7.5 ×10 A/m .)
1.3 The electric field intensity in a conductor due to a time-varying magnetic field is
5
E = 6 cos (0.1y) cos (10 t)x V/m (1.194)
Calculate the magnetic flux density. Assume that the magnetic flux density is zero at t = 0.
6
(Ans: B =−0.6sin (0.1y) sin (10 t)z μT.)
1.4 The law of conservation of charges is given by

∇ ⋅ J + = 0.
t
Show that Ampere’s law given by Eq. (1.42) violates the law of conservation of charges and Maxwell’s
equation given by Eq. (1.43) is in agreement with the law of conservation of charges.
Hint: Take the divergence of Eq. (1.42) and use the vector identity

∇ ⋅ ∇× H = 0.
1.5 The x-component of the electric field intensity of a laser operating at 690 nm is

E (t, 0)= 3 rect (t∕T ) cos (2f t) V/m, (1.195)
x
0
0
where T = 5 ns. The laser and screen are located at z = 0 and z = 5 m, respectively. Sketch the field
0
intensities at the laser and the screen in the time and frequency domain.
1.6 Starting from Maxwell’s equations (Eqs. (1.48)–(1.51)), prove that the electric field intensity satisfies
the wave equation
2
1 E
2
∇ E − = 0.
c t 2
2
Hint: Take the curl on both sides of Eq. (1.50) and use the vector identity
2
∇×∇× E =∇(∇ ⋅ E)−∇ E.
1.7 Determine the direction of propagation of the following wave:
[ ( √ )]
3 x
E = E = cos t − z + .
x x0
2c 2c
1.8 Show that
[ ]
( ) 2
Ψ=Ψ exp − t − k x − k y − k z (1.196)
y
z
0
x
2
2
2
2
2
is a solution of the wave equation (1.125) if = (k + k + k ).
z
x
y
Hint: Substitute Eq. (1.196) into the wave equation (1.125).