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Unformatted text preview: i /2πd and B1′ = μ0i / 2π d ′. Since tan(30º) = 1/ 3 , this leads to 1137
(a) We now make the assumption that wire #2 must be at −π/2 rad (−90º, the bottom of
the cylinder) since it would pose an obstacle for the motion of wire #1 (which is needed
to make these graphs) if it were anywhere in the top semicircle.
(b) Looking at the θ1 = 90º datum in Fig. 2957(b)), where there is a maximum in Bnet x
(equal to +6 μT), we are led to conclude that B1x = 6.0 μ T − 2.0 μ T = 4.0 μ T in that
situation. Using Eq. 294, we obtain
i1 = 2π RB1x μ0 = 2π (0.200 m)(4.0 ×10−6 T)
= 4.0 A .
4π ×10−7 T ⋅ m/A (c) The fact that Fig. 2957(b) increases as θ1 progresses from 0 to 90º implies that wire
1’s current is out of the page, and this is consistent with the cancellation of Bnet y at
θ1 = 90° , noted earlier (with regard to Fig. 2957(c)).
(d) Referring now to Fig. 2957(b) we note that there is no xcomponent of magnetic field
from wire 1 when θ1 = 0, so that plot tells us that B2x = +2.0 μT. Using Eq. 294, we find
the magnitudes of the current to be
i2 = 2π RB2 x μ0 2π (0.200 m)(2.0 × 10−6 T)
=
= 2.0 A .
4π × 10−7 T ⋅ m/A (e) We can conclude (by the righthand rule) that wire 2’s current is into the page.
31. (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 P do not contribute to the field at that point. We use the result of Problem
2921 to evaluate the contributions to the field at P, noting that the nearest wire segments
(each of length a) produce magnetism into the page at P and the further wire segments
(each of length 2a) produce magnetism pointing out of the page at P. Thus, we find (into
the page)
2 ( 4p × 10 7 T ⋅ m A ) (13 A )
⎛ 2 μ 0 i ⎞ ⎛ 2 μ 0i ⎞
2 μ 0i
BP = 2 ⎜
⎜ 8pa ⎟ − 2 ⎜ 8p ( 2a ) ⎟ = 8pa =
⎟⎜
⎟
8π ( 0.047 m )
⎝
⎠⎝
⎠
= 1.96 × 10 5 T...
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 Fall '08
 schuller
 Magnetism, Work

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