13.3 Solutions
615
B t l = (B) (2π r ) = μ 0 i 0 = μ 0 i
r 2
R 2
∴
B =
μ 0 ir
2π R 2
(r < R)
13.41 Let the radius of the inner arc of the loop be r and that of the outer arc 2r .
Both the arcs are quarter of a circle. The straight portions do not contribute
to B as their directions pass through P. As the currents in the two arcs flow
in the opposite sense, B will be down due to current in the outer arc and up
due to the current in the inner wire:
∴
B net =
1
4
μ 0 I
2r
−
1
4
μ 0 I
2 × 2r
=
μ 0 I
16 r
down
13.42 By prob. (13.22) at P, Fig. 13.29
B =
μ 0 I
4π x
(cos θ 1 + cos θ 2 ),
(1)
put θ 1 = θ 2 then
cos θ 1 = cos θ 2 =
L/2
(L/2)
2
+ x 2
So that (1) becomes
B =
μ 0 I
4π x
L
(L/2)
2
+ x 2
(2)
(a) Let the square be of side a = l/4. The distance of the centre of square
from the side is a/2. Put x = a/2 and L = a = l/4 in (2) to find for one
side
B 1 =
2
√
2μ 0 I
πl
As there are four equal sides, B = 4B 1 =
8
√
2μ 0 I
πl
·
The B-field will be perpendicular to the plane containing the wire and
the field point.
(b) Let each side of the equilateral triangle be a = l/3. Distance of the
centre of the triangle from any side is x = a/2
√
3. Put L = a = l/3
and x = a/2
√
3 in (1) to find for one side B 1 =
9μ 0 I
2πl
. Hence for three
sides
615
B t l = (B) (2π r ) = μ 0 i 0 = μ 0 i
r 2
R 2
∴
B =
μ 0 ir
2π R 2
(r < R)
13.41 Let the radius of the inner arc of the loop be r and that of the outer arc 2r .
Both the arcs are quarter of a circle. The straight portions do not contribute
to B as their directions pass through P. As the currents in the two arcs flow
in the opposite sense, B will be down due to current in the outer arc and up
due to the current in the inner wire:
∴
B net =
1
4
μ 0 I
2r
−
1
4
μ 0 I
2 × 2r
=
μ 0 I
16 r
down
13.42 By prob. (13.22) at P, Fig. 13.29
B =
μ 0 I
4π x
(cos θ 1 + cos θ 2 ),
(1)
put θ 1 = θ 2 then
cos θ 1 = cos θ 2 =
L/2
(L/2)
2
+ x 2
So that (1) becomes
B =
μ 0 I
4π x
L
(L/2)
2
+ x 2
(2)
(a) Let the square be of side a = l/4. The distance of the centre of square
from the side is a/2. Put x = a/2 and L = a = l/4 in (2) to find for one
side
B 1 =
2
√
2μ 0 I
πl
As there are four equal sides, B = 4B 1 =
8
√
2μ 0 I
πl
·
The B-field will be perpendicular to the plane containing the wire and
the field point.
(b) Let each side of the equilateral triangle be a = l/3. Distance of the
centre of the triangle from any side is x = a/2
√
3. Put L = a = l/3
and x = a/2
√
3 in (1) to find for one side B 1 =
9μ 0 I
2πl
. Hence for three
sides