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Unformatted text preview: ng Eq. 29-9 (with φ = π) and the right-hand rule,
we find that the current in the semicircular arc H J contributes μ 0i 4 R1 (into the page) to
the field at C. Also, arc D A contributes μ 0i 4 R2 (out of the page) to the field there. Thus,
the net field at C is
B= μ 0i ⎛ 1 ⎜
4 ⎝ R1 − 1 ⎞ (4p ×10 -7 T ⋅ m A)(0.281A) ⎛
⎟ = 1.67 ×10 T.
⎝ 0.0315m 0.0780m ⎠ (b) The direction of the field is into the page.
9. (a) The currents must be opposite or antiparallel, so that the resulting fields are in the
same direction in the region between the wires. If the currents are parallel, then the two
fields are in opposite directions in the region between the wires. Since the currents are the
same, the total field is zero along the line that runs halfway between the wires.
(b) At a point halfway between they have the same magnitude, μ0i/2πr. Thus the total
field at the midpoint has magnitude B = μ0i/πr and
i= prB μ0 = p ( 0.040 m ) ( 300 ×10−6 T ) 4p × 10 -7 T ⋅ m A = 30 A. 10. (a) Recalling the straight sections discussion in Sample Problem — “Magnetic field
at the center of a circular arc of current,” we see that the current in the straight segments
collinear with C do not contribute to the field at that point.
Equation 29-9 (with φ = π) indicates that the current in the semicircular arc contributes
μ 0i 4 R to the field at C. Thus, the magnitude of the magnetic field is
B= μ 0i
4R = (4p ×10 -7 T ⋅ m A)(0.0348A)
= 1.18 ×10 -7 T.
4(0.0926m) (b) The right-hand rule shows that this field is into the page.
11. (a) BP1 = μ0i1 / 2π r1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm,
and BP2 = μ0i2 / 2π r2 where r2 = d2 = 1.5 cm. From BP1 = BP2 we get
⎛ 1.5 cm ⎞
i2 = i1 ⎜ 2 ⎟ = ( 6.5 A ) ⎜
⎟ = 4.3A.
⎝ 2.25 cm ⎠
⎝ r1 ⎠ 1130 CHAPTER 29 (b) Using the right-hand rule, we see that the current i2 carried by wire 2 must be out of
12. (a) Since they carry current in the same direction, then (by the righ...
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