608
13 Electromagnetism I
B 1 =
μ 0
4a
4I
5
=
μ 0 I
5a
B 2 =
μ 0
4a
I
5
=
μ 0 I
20 a
Net field B = B 1 − B 2 =
3μ 0 I
20 a
13.26 Field B at the centre is due to three-fourths of the circle (B 1 ) added to that
due to the straight segment (B 2 )
B 1 =
3
4
×
μ 0 i
2R
B 2 =
μ 0 i
4π d
(cos θ 1 + cos θ 2 )
From the geometry of Fig. 13.24, θ 1 = θ 2 = 45 ◦ and d =
R
√
2
.
Then
B = B 1 + B 2 =
μ 0 i
2 π R
1 +
3π
4
Fig. 13.24 Magnetic
induction at the centre of a
current-carrying wire made of
three-fourths of a circle and a
chord
13.27 (a) The magnetic induction due to straight wires is
B 1 =
μ 0 i
4π R
+
μ 0 i
4π R
=
μ 0 i
2π R
(1)
because straight wires are of infinite length only on left side.
Induction at 0 due to semicircular portion is
B 2 =
μ 0 i
4R
(2)
Total magnetic induction
B = B 1 + B 2 =
μ 0 i
2π R
+
μ 0 i
4R
=
μ 0 i
4π R
(2 + π)