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

◦

.