616
13 Electromagnetism I
Fig. 13.29 Magnetic field
due to a current-carrying
straight wire of finite length
B = 3B 1 =
27μ 0 I
2πl
∴
B (square)
B (triangle)
=
27/2
8
√
2
= 1.19
13.43 By prob. (13.36) B =
8N μ 0 I
5 3/2 R
=
(8)(100)(4π × 10 −7 )(2)
5 3/2 (0.2)
9 × 10 −4 T
13.44
B dl = μ 0 i
(Ampere’s law)
By prob. (13.43)
B =
μ 0 Ir
2π R 2 (r < R)
(1)
∴
For r =
R
2
, B =
μ 0 I
4π R
(2)
Further B =
μ 0 I
2πr
(r > R).
(3)
Equating (2) and (3) we find r = 2R. Thus at r = 2R, the magnetic field is
the same as at r = R/2.
13.45 (a) In the absence of magnetic material the number of flux linkages N φ B (N
being the number of turns) is proportional to the current
N φ B = Li
(1)
If n is the number of turns per unit length, A the cross-sectional area and
l the length of the solenoid and B the magnetic induction
N φ B = (nl)(B A)
(2)
By Ampere’s theorem
B = μ 0 ni
(3)
Combining (1), (2) and (3)
L = μ 0 n
2 Al
(4)
13 Electromagnetism I
Fig. 13.29 Magnetic field
due to a current-carrying
straight wire of finite length
B = 3B 1 =
27μ 0 I
2πl
∴
B (square)
B (triangle)
=
27/2
8
√
2
= 1.19
13.43 By prob. (13.36) B =
8N μ 0 I
5 3/2 R
=
(8)(100)(4π × 10 −7 )(2)
5 3/2 (0.2)
9 × 10 −4 T
13.44
B dl = μ 0 i
(Ampere’s law)
By prob. (13.43)
B =
μ 0 Ir
2π R 2 (r < R)
(1)
∴
For r =
R
2
, B =
μ 0 I
4π R
(2)
Further B =
μ 0 I
2πr
(r > R).
(3)
Equating (2) and (3) we find r = 2R. Thus at r = 2R, the magnetic field is
the same as at r = R/2.
13.45 (a) In the absence of magnetic material the number of flux linkages N φ B (N
being the number of turns) is proportional to the current
N φ B = Li
(1)
If n is the number of turns per unit length, A the cross-sectional area and
l the length of the solenoid and B the magnetic induction
N φ B = (nl)(B A)
(2)
By Ampere’s theorem
B = μ 0 ni
(3)
Combining (1), (2) and (3)
L = μ 0 n
2 Al
(4)