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Unformatted text preview: noid (which does not make use of the coil diameter) is the preferred method:
B = μ 0in = μ 0i
where i = 0.30 A, FG N IJ
HK = 0.25 m, and N = 200. This yields B = 3.0 × 10−4 T . 52. We find N, the number of turns of the solenoid, from the magnetic field
B = μ 0in = μ oiN / : N = B / μ 0i. Thus, the total length of wire used in making the
solenoid is 1145 2 πrN = c hc hb g 2 π 2.60 × 10−2 m 23.0 × 10−3 T 130 m
.
2 πrB
= 108 m.
=
μ 0i
2 4 π × 10−7 T ⋅ m / A 18.0 A c hb g 53. The orbital radius for the electron is
r= mv
mv
=
eB eμ 0ni which we solve for i: ( 9.11×10−31 kg ) ( 0.0460 ) ( 3.00 ×108 m s )
mv
i=
=
eμ 0 nr (1.60 ×10−19 C ) ( 4π×10−7 T ⋅ m A ) (100 0.0100m ) ( 2.30 ×10−2 m )
= 0.272A.
54. As the problem states near the end, some idealizations are being made here to keep
the calculation straightforward (but are slightly unrealistic). For circular motion (with
speed, v⊥, which represents the magnitude of the component of the velocity perpendicular
to the magnetic field [the field is shown in Fig. 2919]), the period is (see Eq. 2817)
T = 2πr/v⊥ = 2πm/eB.
Now, the time to travel the length of the solenoid is t = L / v where v is the component
of the velocity in the direction of the field (along the coil axis) and is equal to v cos θ
where θ = 30º. Using Eq. 2923 (B = μ0in) with n = N/L, we find the number of
revolutions made is t /T = 1.6 × 106.
55. (a) We denote the B fields at point P on the axis due to the solenoid and the wire as
Bs and Bw , respectively. Since Bs is along the axis of the solenoid and Bw is
perpendicular to it, Bs ⊥ Bw . For the net field B to be at 45° with the axis we then must
have Bs = Bw. Thus,
μi
Bs = μ 0is n = Bw = 0 w ,
2 πd
which gives the separation d to point P on the axis:
d= iw
6.00 A
=
= 4.77 cm.
2π is n 2π ( 20.0 ×10−3 A ) (10 turns cm ) (b) The magnetic field strength is
B = 2 Bs = 2 ( 4π ×10−7 T ⋅ m A ) ( 20.0 ×103 A ) (10 turns 0.0100 m ) = 3.55 ×10−5 T. 1146...
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 Fall '08
 schuller
 Magnetism, Work

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