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33.18. Model: The radius of the earth is much larger than the size of the current loop.
Solve: (a) From Equation 33.9, the magnetic field strength at the surface of the earth at the earth’s north pole is
7 22 2 0 5 3 6 3
2 10 T m/A 8.0 10 A m 6.2 10 T (^2) 6.38 10 m
B
z
μ μ
π
− −
× ×
= = = ×
×
This value is close to the value of 5 × 10
− 5 T given in Table 33.1.
(b) The current required to produce a dipole moment like that on the earth is
2
μ= AI = π R earth I ⇒8.0 × 10
22 A m
2 = π (6.38 × 10
6 m)
2
I ⇒ I = 6.3 × 10
8 A
Assess: This is an extremely large current to run through a wire around the equator.
33.24. Model: Assume that the solenoid is an ideal solenoid.
Solve: We can use Equation 33.16 to find the current that will generate a 3.0 mT field inside the solenoid:
0 solenoid
NI
B
l
solenoid
0
B l
I
μ N
Using l = 0.15 m and N = 0.15 m 0.0010 m =150,
3
7
3.0 10 T 0.15 m
2.4 A
4 10 T m/A 150
I
−
−
×
Assess: This is a reasonable current to pass through a good conducting wire of diameter 1 mm.
33.48. Model: Assume that the wire is infinitely long.
Visualize: Please refer to Figure P33.48. The wire, looped as it is, consists of a circular part and a linear part.
Solve: Using Equation 33.7 and Example 33.3, the magnetic field at P is
( ) (^ )
( ) (^ )
0 0 P loop center wire
7 7 4
4 10 T m/A 5.0 A 4 10 T m/A 5.0 A 4.1 10 T 2 0.010 m 2 0.010 m
I I
B B B
R R
− − −
= + = ×
