13.3 Solutions
609
(b) The straight portions of the wire do not contribute to the field at O as the
current is directed towards C and makes an angle θ = 0 ◦ , for which the
Biot–Savart formula gives B = 0. Thus the entire induction comes from
the semicircular portion of the wire for which B =
μ 0 i
4R
.
13.28 (a) Let the angle θ be subtended at the centre by one side AC of a regular
n-sided polygon, Fig. 13.25. Then
Fig. 13.25 Magnetic
induction at the centre of a
current-carrying regular
n-sided polygon
θ =
2π
n
or
θ
2
=
π
n
(1)
The magnetic induction due to one side AC at the centre O is
B 1 =
μ 0 i
4πr
(cos α + cos α) =
μ 0 i cos α
2π r
(2)
where r is the distance of O from AC.
The field B due to n sides will be additive and is given by
B = n B 1 =
μ 0 ni cos α
2π r
(3)
Now r = a sin α, so that in (3)
cos α
r
=
cos α
a sin α
=
1
a
cot α =
1
a
tan
θ
2
=
1
a
tan
π
n
(4)
where we have used (1). Using (4) in (3)
B =
μ 0 ni
2 π a
tan
π
n
(5)
609
(b) The straight portions of the wire do not contribute to the field at O as the
current is directed towards C and makes an angle θ = 0 ◦ , for which the
Biot–Savart formula gives B = 0. Thus the entire induction comes from
the semicircular portion of the wire for which B =
μ 0 i
4R
.
13.28 (a) Let the angle θ be subtended at the centre by one side AC of a regular
n-sided polygon, Fig. 13.25. Then
Fig. 13.25 Magnetic
induction at the centre of a
current-carrying regular
n-sided polygon
θ =
2π
n
or
θ
2
=
π
n
(1)
The magnetic induction due to one side AC at the centre O is
B 1 =
μ 0 i
4πr
(cos α + cos α) =
μ 0 i cos α
2π r
(2)
where r is the distance of O from AC.
The field B due to n sides will be additive and is given by
B = n B 1 =
μ 0 ni cos α
2π r
(3)
Now r = a sin α, so that in (3)
cos α
r
=
cos α
a sin α
=
1
a
cot α =
1
a
tan
θ
2
=
1
a
tan
π
n
(4)
where we have used (1). Using (4) in (3)
B =
μ 0 ni
2 π a
tan
π
n
(5)