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826_PartUniversity Physics Solution

# 826_PartUniversity Physics Solution - 28-4 28.9 28.10...

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28-4 Chapter 28 28.9. I DENTIFY : A current segment creates a magnetic field. S ET U P : The law of Biot and Savart gives 0 2 sin 4 Idl dB r μ φ π = . E XECUTE : Applying the law of Biot and Savart gives (a) 7 2 4 π 10 T m/A (10.0 A)(0.00110 m) sin90° 4 π (0.0500 m) dB × = = 4.40 × 10 ±7 T, out of the paper. (b) The same as above, except 2 2 (5.00 cm) (14.0 cm) r = + and φ = arctan(5/14) = 19.65°, giving dB = 1.67 × 10 ±8 T, out of the page. (c) dB = 0 since φ = 0°. E VALUATE : This is a very small field, but it comes from a very small segment of current. 28.10. I DENTIFY : Apply 0 0 2 3 ² 4 4 μ Id μ Id d π r π r × × l r l r B = = ! ! ! ! . S ET U P : The magnitude of the field due to the current element is 0 2 sin 4 μ Idl dB π r φ = , where φ is the angle between r ! and the current direction. E XECUTE : The magnetic field at the given points is: 6 0 0 2 2 sin (200 A)(0.000100 m) 2.00 10 T. 4 4 (0.100 m) a μ Idl μ dB π r π φ = = = × 6 0 0 2 2 sin (200 A)(0.000100 m)sin45 0.705 10 T. 4 4 2(0.100 m) b μ Idl μ dB π r π φ ° = = = × 6 0 0 2 2 sin (200 A)(0.000100 m) 2.00 10 T. 4 4 (0.100 m) c μ Idl μ dB π r π φ = = = × 0 0 2 2 sin sin(0 ) 0 4 4 d μ Idl μ Idl dB π r π r φ ° = = = . 6 0 0 2 2 sin (200 A)(0.00100 m) 2 0.545 10 T 4 4 3(0.100 m) 3 e μ Idl dB π r φ μ π = = = × The field vectors at each point are shown in Figure 28.10. E VALUATE : In each case d B ! is perpendicular to the current direction. Figure 28.10 28.11. I DENTIFY and S ET U P : The magnetic field produced by an infinitesimal current element is given by Eq.(28.6). 0 2 ² 4 I d r μ π l r B ! ! × = As in Example 28.2 use this equation for the finite 0.500-mm segment of wire since the 0.500 mm l Δ = length is much smaller than the distances to the field points. 0 0 2 3 ² 4 4 I I r r μ μ π π Δ Δ = l r l r B ! ! ! ! × × = I is in the ( ) 3 ² -direction, so 0.500 10 m z + Δ × l k ! = E XECUTE : (a) Field point is at x = 2.00 m, y = 0, z = 0 so the vector r ! from the source point (at the origin) to the field point is ( ) ² 2.00 m . r i ! = ( ) ( ) ( ) 3 3 2 ² ² ² 0.500 10 m 2.00 m 1.00 10 m Δ × + × l r k i j ! ! × = × = ( ) ( ) ( ) ( ) ( ) 7 3 2 11 3 1 10 T m/A 4.00 A 1.00 10 m ² ² 5.00 10 T 2.00 m × × × B j j ! = =

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Sources of Magnetic Field 28-5 (b) ( ) ² 2.00 m , 2.00 m. r = r j ! = ( ) ( ) ( ) 3 3 2 ² ² ² 0.500 10 m 2.00 m 1.00 10 m Δ × × l r k j i ! ! × = × = ( ) ( ) ( ) ( ) ( ) 7 3 2 11 3 1 10 T m/A 4.00 A 1.00 10 m ² ² 5.00 10 T 2.00 m × × = − × B = i i ! (c) ( ) ( ) ( ) ² ² 2.00 m , 2 2.00 m . r = r = i + j ! ( ) ( ) ( ) ( ) ( ) 3 3 2 ² ² ² ² ² 0.500 10 m 2.00 m 1.00 10 m Δ × × + = ° l r = k i j j i ! ! × × ( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 3 2 11 3 1 10 T m/A 4.00 A 1.00 10 m ² ² ² ² 1.77 10 T 2 2.00 m × × × ° = ° B = j i i j ! (d) ² (2.00 m) , 2.00 m. r = r = k ! 3 ² ² (0.500 10 m)(2.00 m) 0; 0. Δ × l r = k k = B = ! ! ! × × E VALUATE : At each point B ! is perpendicular to both and . Δ r l ! ! B = 0 along the length of the wire. 28.12. I DENTIFY : A current segment creates a magnetic field. S ET U P : The law of Biot and Savart gives 0 2 sin 4 Idl dB r μ φ π = .
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826_PartUniversity Physics Solution - 28-4 28.9 28.10...

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