58.
(a) The contribution to B
C
from the (infinite) straight segment of the wire is
B
C1
=
µ
0
i
2πR
.
The contribution from the circular loop is
B
C2
=
µ
0
i
2R
.
Thus,
B
C
= B
C1
+ B
C2
=
µ
0
i
2R
1 +
1
π
.
B
C
points out of the page.
(b) Now
B
C1
⊥
B
C2
so
B
C
=
B
2
C1
+ B
2
C2
=
µ
0
i
2R
1 +
1
π
2
,
and
B
C
points at an angle (relative to the plane of the paper) equal to
tan
−1
B
C1
B
C2
= tan
−1
1
π
=
18
◦
.