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28 HMK #7 Chp 28-29a

# 28 HMK #7 Chp 28-29a - 28-1Chapter 28 19 26 31 36 38 42 45...

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Unformatted text preview: 28-1Chapter 28 Problems 11, 15, 19, 26, 31, 36, 38, 42, 45, 47, 49, 52, 59, 62, 67, 70, 74, 80, 81, 83. Chapter 29 Problems 1, 4, 6, 14, 15. Note: appears a few times in this problem set as a typesetting mistake… this means 28.11. IDENTIFY:A current segment creates a magnetic field. SET UP:The law of Biot and Savart gives . EXECUTE:Applying the law of Biot and Savart gives (a)= 4.40 ×10–7T, out of the paper. (b)The same as above, except and φ= arctan(5/14) = 19.65°, giving dB=1.67 ×10–8T, out of the page. (c)dB= 0 since φ= 0°. EVALUATE:This is a very small field, but it comes from a very small segment of current. 28.15. IDENTIFY:A current segment creates a magnetic field. SET UP:The law of Biot and Savart gives Both fields are into the page, so their magnitudes add. EXECUTE:Applying the Biot and Savart law, where= 2.121 cm, we have = 1.76 ×10–5T, into the paper. EVALUATE:Even though the two wire segments are at right angles, the magnetic fields they create are in the same direction. 28.19. IDENTIFY:We can model the current in the heart as that of a long straight wire. It produces a magnetic field around it. SET UP: For a long straight wire, B=µI2πr.µ=4π×10−7T⋅m/A.1 gauss=10−4T.SOURCES OF MAGNETIC FIELDANDELECTROMAGNETIC INDUCTION2928 28-2Chapter 28EXECUTE:Solving for the current gives I=2πrBµ=2π.050 m( )1.×10−10T( )4π×10−7T⋅m/A=2.5×10−5A=25 µA.EVALUATE: By household standards, this is a very small current. But the magnetic field around the heart ( ≈1 µG) is also very small. 28.26. IDENTIFY:Each segment of the rectangular loop creates a magnetic field at the center of the loop, and all these fields are in the same direction. SET UP:The field due to each segment is 2224IaBxxaμπ=+. Bis into paper so Iis clockwise around loop. EXECUTE:Long sides:4.75 cma=. 2.10 cmx=. For the two long sides, B=2(1.00×10−7T⋅m/A)I2(4.75×10−2m)(2.10×10−2m)(0.0210 m)2+(0.0475 m)2=(1.742×10−5T/A)I.Short sides:2.10 cma=. 4.75 cmx=. For the two short sides, B=2(1.00×10−7T⋅m/A)I2(2.10×10−2m)(4.75×10−2m)(0.0475 m)2+(0.0210 m)2=(3.405×10−6T/A)I.Using the known field, we have 55(2.08210T/A)5.5010TBI−−=×=×, which gives 2.64 AI=. EVALUATE:This is a typical household current, yet it produces a magnetic field which is about the same as the earth’s magnetic field. 28.31. IDENTIFY:Apply Eq.(28.11). SET UP:Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. EXECUTE:(a) and the force is repulsive since the currents are in opposite directions. (b) Doubling the currents makes the force increase by a factor of four to EVALUATE:Doubling the current in a wire doubles the magnetic field of that wire. For fixed magnetic field, doubling the current in a wire doubles the force that the magnetic field exerts on the wire....
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28 HMK #7 Chp 28-29a - 28-1Chapter 28 19 26 31 36 38 42 45...

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