826_PartUniversity Physics Solution

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

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28-4 Chapter 28 28.9. IDENTIFY: A current segment creates a magnetic field. SET UP: The law of Biot and Savart gives 0 2 sin 4 Idl dB r μ φ π = . EXECUTE: 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 22 (5.00 cm) (14.0 cm) r =+ and = arctan(5/14) = 19.65&, giving dB = 1.67 × 10 ±8 T, out of the page. (c) dB = 0 since = 0&. EVALUATE: This is a very small field, but it comes from a very small segment of current. 28.10. IDENTIFY: Apply 00 23 ² 44 μ Id μ Id d π r π r ×× lr B= = !! ! ! . SET UP: The magnitude of the field due to the current element is 0 2 sin 4 μ Idl dB π r = , where is the angle between r ! and the current direction. EXECUTE: The magnetic field at the given points is: 6 sin (200 A)(0.000100 m) 2.00 10 T. 4 4 (0.100 m) a μ Idl μ dB π r π == = × 6 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 ° = × 6 sin (200 A)(0.000100 m) 4 4 (0.100 m) c μ Idl μ dB π r π = × sin sin(0 ) 0 d μ Idl μ Idl dB π r π r ° = . 6 sin (200 A)(0.00100 m) 2 0.545 10 T 4 4 3(0.100 m) 3 e μ Idl dB π r φμ = × The field vectors at each point are shown in Figure 28.10. EVALUATE: In each case d B ! is perpendicular to the current direction. Figure 28.10 28.11. IDENTIFY and SET UP: The magnetic field produced by an infinitesimal current element is given by Eq.(28.6). 0 2 ² 4 I d 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. ² II rr μμ ππ ΔΔ = B ! ! = I is in the ( ) 3 ² -direction, so 0.500 10 m z × lk ! = EXECUTE: (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 . ri ! = ( ) ( ) 33 2 ²² ² 0.500 10 m 2.00 m 1.00 10 m −− Δ× + × ki j ! ! ×= ( ) ( ) 73 2 11 3 11 0 Tm /A 4 . 0 0 A 1 . 0 01 0 m 5.00 10 T 2.00 m × × Bj j !
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Sources of Magnetic Field 28-5 (b) () & 2.00 m , 2.00 m. r = rj ! = ( ) ( ) 33 2 && & 0.500 10 m 2.00 m 1.00 10 m −− Δ× × lr k j i ! ! ×= ( ) ( ) 73 2 11 3 11 0 Tm /A 4 . 0 0 A1 . 0 01 0 m 5.00 10 T 2.00 m ×⋅ × −= × B =i i ! (c) 2.00 m 2 2.00 m . r = r= i+j ! ( ) ( ) ( )( ) 2 &&& & & × += & = k i jj i ! ! ×× 2 11 3 1 10 T m/A 4.00 A 1.00 10 m 1.77 10 T 22 . 0 0 m × −× ⎡⎤ ⎣⎦ &= & B = j ii j ! (d) & (2.00 m) , 2.00 m. r = k ! 3 (0.500 10 m)(2.00 m) 0; 0. = kk = B = ! ! ! EVALUATE: At each point B ! is perpendicular to both and . Δ rl ! ! B = 0 along the length of the wire. 28.12. IDENTIFY: A current segment creates a magnetic field. SET UP: The law of Biot and Savart gives 0 2 sin 4 Idl dB r μ φ π = . Both fields are into the page, so their magnitudes add. EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives 7 2 2.50 cm (12.0 A)(0.00150 m) 4 π 10 T m/A 8.00 cm 4 π (0.0800 m) dB ⎛⎞ ⎜⎟ ⎝⎠ = = 8.79 × 10 ±8 T The field from the 24.0-A segment is twice this value, so the total field is 2.64 × 10 ±7 T, into the page.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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