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Unformatted text preview: hat By = 0 at x = 10 cm implies the currents are in opposite directions. Thus, 1134 CHAPTER 29 By = μ0i1 μ0i2 μ0i2 ⎛ 4
2π ( L + x) 2π x 2π ⎝ L + x x ⎠
− using Eq. 29-4 and the fact that i1 = 4i2 . To get the maximum, we take the derivative with
2 respect to x and set equal to zero. This leads to 3x2 – 2Lx – L = 0, which factors and
becomes (3x + L)(x − L) = 0, which has the physically acceptable solution: x = L . This
produces the maximum By: μoi2/2πL. To proceed further, we must determine L.
Examination of the datum at x = 10 cm in Fig. 29-49(b) leads (using our expression
above for By and setting that to zero) to L = 30 cm.
(a) The maximum value of By occurs at x = L = 30 cm.
(b) With i2 = 0.003 A we find μo i2 /2πL = 2.0 nT.
(c) and (d) Figure 29-49(b) shows that as we get very close to wire 2 (where its field
strongly dominates over that of the more distant wire 1) By points along the –y direction.
The right-hand rule leads us to conclude that wire 2’s current is consequently is into the
page. We previously observed that the currents were in opposite directions, so wire 1’s
current is out of the page.
23. We assume the current flows in the +x direction and the particle is at some distance d
in the +y direction (away from the wire). Then, the magnetic field at the location of a
proton with charge q is B = ( μ0i / 2π d ) k. Thus,
F = qv × B = μ 0iq
2 pd ev × kj. ej In this situation, v = v − j (where v is the speed and is a positive value), and q > 0. Thus, μ 0iqv (( ) ) -7
ˆ × k = − μ 0iqv ˆ = − (4p ×10 T ⋅ m A)(0.350A)(1.60 ×10 C)(200m/s) ˆ
2π (0.0289 m)
= (−7.75 ×10 -23 N)i. 24. Initially, we have Bnet,y = 0 and Bnet,x = B2 + B4 = 2(μo i /2πd) using Eq. 29-4, where
d = 0.15 m . To obtain the 30º condition described in the problem, we must have
Bnet, y = Bnet, x tan(30°) ⎛ μi ⎞
⇒ B1′ − B3 = 2 ⎜ 0 ⎟ tan(30°)
⎝ 2π d ⎠ where B3 = μo...
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