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13 Electromagnetism I
(b) In the limit n → ∞, tan
π
n
→
π
n
and (5) becomes
B =
μ 0 i
2a
an expression which is identical for B for a circular loop. This is reasonable since as n → ∞, polygon → circle.
13.29 For square l = 4a and for circle l = 2πr
∴
a
r
=
π
2
(1)
At the centre of the circle, B c =
μ 0 i
2r
.
At the centre of the square, B s =
2
√
2 μ 0 i
πa
via prob. (13.24).
∴
B c
B s
=
π
4
√
2
a
r
=
π 2
8
√
2
= 0.87
13.30 (a) The magnetic induction B at the centre of a circular wire is B =
μ 0 i
2r
.
Hence for the arc which subtends an angle θ at the centre
B =
μ 0 i
2r
θ
2π
=
μ 0 i θ
4π r
Induction at C due to the inner arc is
B 1 =
μ 0 iθ
4 π R 1
and due to the outer arc
B 2 = −
μ 0 iθ
4 π R 2
The negative sign arises due to the fact that the current has reversed. As
the radial part of the path points towards C, it does not contribute to B.
Therefore, the resultant induction is
B = B 1 + B 2 =
μ 0 iθ
4π
1
R 1
−
1
R 2
Note that for clockwise current we take B as positive and for counterclockwise we take B as negative.