13.3 Solutions
607
∴
B =
μ 0 i
4πr
(cos θ 1 + cos θ 2 )
(2)
(b) For infinite wire θ 1 → 0 and θ 2 → 0, in this limit (2) becomes
B =
μ 0 i
2πr
which is the expression for a long wire.
13.23 B =
μ 0 i
2 r
∴
i =
2Br
μ 0
=
2 × 10 −5 × 0.5
4 π × 10 −7 = 7.96 A
13.24 Magnetic field B 1 due to current i in one segment is (Fig. 13.23)
B 1 =
μ 0 i
4π R
(cos θ 1 + cos θ 2 )
Putting θ 1 = θ 2 = 45 ◦ and R = a/2
B 1 =
μ 0 i
√
2πa
Fields due to four sides are equal and additive. Therefore net field
B = 4B 1 =
2
√
2μ 0 i
π a
Fig. 13.23 Magnetic
induction at the centre of a
square conducting loop
13.25 The magnetic fields in the upper branch and lower branch act in the opposite
direction. The current in the upper branch is
4I
5
and in the lower branch is
I
5
. As the current is flowing through semicircles,
607
∴
B =
μ 0 i
4πr
(cos θ 1 + cos θ 2 )
(2)
(b) For infinite wire θ 1 → 0 and θ 2 → 0, in this limit (2) becomes
B =
μ 0 i
2πr
which is the expression for a long wire.
13.23 B =
μ 0 i
2 r
∴
i =
2Br
μ 0
=
2 × 10 −5 × 0.5
4 π × 10 −7 = 7.96 A
13.24 Magnetic field B 1 due to current i in one segment is (Fig. 13.23)
B 1 =
μ 0 i
4π R
(cos θ 1 + cos θ 2 )
Putting θ 1 = θ 2 = 45 ◦ and R = a/2
B 1 =
μ 0 i
√
2πa
Fields due to four sides are equal and additive. Therefore net field
B = 4B 1 =
2
√
2μ 0 i
π a
Fig. 13.23 Magnetic
induction at the centre of a
square conducting loop
13.25 The magnetic fields in the upper branch and lower branch act in the opposite
direction. The current in the upper branch is
4I
5
and in the lower branch is
I
5
. As the current is flowing through semicircles,