674
14 Electromagnetism II
14.56 By prob. (14.55), ν = ν − ν = −2ν
v
c
ν = −
2 × 800 × 10 6
3 × 10 8
×
5
18
× 90
= −133 Hz
14.57 The relation between the current i and the magnetic field B expressed as
B · dl = μ 0 i is known as Ampere’s law. There are equal and opposite currents iin the conductors.
(a) r < a. The net current passing through the conductor bounded by the
closed path corresponds to that flowing through the inner conductor,
Fig. 14.7. Hence by Ampere’s theorem
Fig. 14.7 Magnetic field due
to current carrying coaxial
cylinder
B · dl = (B)(2πr ) = μ 0 i
(πr 2 )
πa 2
or B =
μ 0 ir
2πa 2
(b) a < r < b. Here the current through the outer conductor does not contribute to B.
B · dl = (B)(2πr ) = μ 0 i
or B =
μ 0 i
2πr
(c) b < r < c. Here currents through both the conductors contribute to B.
B · dl = (B)(2πr ) = μ 0 i − μ 0 i
π(r 2 − b 2 )
π(c 2 − b 2 )
or B =
μ 0 i
2πr
(c 2 − r 2 )
(c 2 − b 2 )
(d) r > c. As the net current flowing through the closed path is zero, B = 0.
14 Electromagnetism II
14.56 By prob. (14.55), ν = ν − ν = −2ν
v
c
ν = −
2 × 800 × 10 6
3 × 10 8
×
5
18
× 90
= −133 Hz
14.57 The relation between the current i and the magnetic field B expressed as
B · dl = μ 0 i is known as Ampere’s law. There are equal and opposite currents iin the conductors.
(a) r < a. The net current passing through the conductor bounded by the
closed path corresponds to that flowing through the inner conductor,
Fig. 14.7. Hence by Ampere’s theorem
Fig. 14.7 Magnetic field due
to current carrying coaxial
cylinder
B · dl = (B)(2πr ) = μ 0 i
(πr 2 )
πa 2
or B =
μ 0 ir
2πa 2
(b) a < r < b. Here the current through the outer conductor does not contribute to B.
B · dl = (B)(2πr ) = μ 0 i
or B =
μ 0 i
2πr
(c) b < r < c. Here currents through both the conductors contribute to B.
B · dl = (B)(2πr ) = μ 0 i − μ 0 i
π(r 2 − b 2 )
π(c 2 − b 2 )
or B =
μ 0 i
2πr
(c 2 − r 2 )
(c 2 − b 2 )
(d) r > c. As the net current flowing through the closed path is zero, B = 0.