Physics Solution Manual for 1100 and 2101

06 10 3 a882 10 21 m 2 9 3 10 24 a m 2 chapter 22

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Unformatted text preview: .5) to determine BI, where r = 0.280 m is 2π r the radial distance between the wire and the center of the compass needle and µ 0 = 4π × 10 −7 T ⋅ m/A . Then we will use Equation 1.3 to find the magnitude BE of the horizontal component of earth’s magnetic field. SOLUTION a. Because the magnetic field BI due to the current in the wire points east at the location of the compass, the magnetic field lines must circulate clockwise around the wire, as viewed from above. Right-Hand Rule No. 2, then, indicates that the current I flows into the page. Since the drawing shows a top view of the situation, the current flows toward the earth’s surface. b. Solving tan θ = BI (Equation 1.3) for BE, we obtain BE BE = Substituting BI = BI (1) tan θ µ0 I (Equation 21.5) into Equation (1), we find that 2π r µ0 I 2π r 4 π ×10−7 T ⋅ m/A ( 25.0 A ) BI = µ0 I BE = = = = 4.21×10−5 T tan θ tan θ 2π r tan θ 2 π ( 0.280 m ) tan 23.0o ( ) 65. SSM REASONING According to Equation 21.6 the magnetic field at the center of a circular, current-carrying loop of N turns and radius r is B = Nµ 0 I / ( 2 r ) . The number of turns N in the coil can be found by dividing the total length L of the wire by the circumference after it has been wound into a circle. The current in the wire can be found by using Ohm's law, I = V / R . 1176 MAGNETIC FORCES AND MAGNETIC FIELDS SOLUTION The number of turns in the wire is N= L 2π r The current in the wire is I= V 12 .0 V 2 .03 × 10 3 = = A R ( 5.90 × 10 –3 Ω / m) L L Therefore, the magnetic field at the center of the coil is µ µ F I I = FL IF I I = µ LI Gr J Gπ r J 2 r J 4π r GK 2 2 HKHK H 2.03 4 LG cπ × 10 T ⋅ m / A hF × 10 A I H L J= K = B=N 0 0 0 2 3 –7 4 π ( 0.140 m ) 2 1.04 × 10 –2 T 66. REASONING The turns of the coil are as close together as possible without overlapping, meaning that there are no gaps between the turns. Therefore, the width of a single turn is equal to the diameter D = 2r of the wire, where r is the wire’s radius. The number N of turns, therefore, is N = L/(2r), where L is the length of the solenoid. The number of turns per unit length is n, where N L / ( 2r ) 1 (1) n= = = L L 2r The magnitude B of the magnetic field inside a solenoid with n turns per meter and a current I is given by B = µ0nI (Equation 21.7), where µ 0 = 4π × 10 −7 T ⋅ m/A is the permeability of V (Equation 20.2), to determine the current I in the I solenoid when the battery (Emf = V) is connected to it. The resistance R of the silver wire L will be found from R = ρ (Equation 20.3), where ρ = 1.59×10−8 Ω·m is the resistivity of A silver (see Table 20.1), L is the length of the wire, and A is its cross-sectional area. free space. We will use Ohm’s law, R = SOLUTION Substituting Equation (1) into B = µ0nI (Equation 21.7) and solving for the radius r yields µI µI (2) B = µ0 nI = 0 or r= 0 2r 2B Chapter 21 Problems 1177 V V (Equation 20.2) for I, we obtain I = . Substituting this result into I R Equation (2), we find that µ I µV (3) r= 0 = 0 2 B 2 BR Solving R = Substituting R = ρ L (Equation 20.3) into Equation (3), we obtain A r= µ0V 2 BR = µ0V µ VA =0 ρ L 2Bρ L (4) 2B A The wire has a circular cross-section, so we may express the area A in Equation (4) as A = πr2. Substituting this expression into Equation (4) and simplifying the resulting expression, we find that r= µ0V π r 2 2Bρ L or 1= µ0V π r 2B ρ L (5) Solving Equation (5) for r yields ( )( ) −3 −8 2 B ρ L 2 6.48 ×10 T 1.59 × 10 Ω ⋅ m ( 25.0 m ) r= = = 4.35 ×10−4 m −7 µ0V π 4π × 10 T ⋅ m/A ( 3.00 V ) π ( ) 67. REASONING AND SOLUTION The currents in wires 1 and 2 produce the magnetic fields B1 and B2 at the empty corner, as shown in the following drawing. The directions of these fields can be obtained using RHR-2. Since there are equal currents in wires 1 and 2 and since these wires are each the same distance r from the empty corner, B1 and B2 have equal magnitudes. Using Equation 21.5, we can write the field magnitude as B1 = B2 = µ 0 I / 2 π r . bg B3 Wire 1 X B2 r B1 B1+2 X Wire 2 Since the fields B1 and B2 are perpendicular, it follows Wire 3 from the Pythagorean theorem that they combine to produce a net magnetic field that has the direction shown in the drawing at the right and has a magnitude B1+2 given by 1178 MAGNETIC FORCES AND MAGNETIC FIELDS µ µ F I I +F I I = Gπ r J Gπ r J 2 2 H KH K 2 B1+ 2 = B +B 2 1 = 2 2 2 0 0 2 µ0I 2π r The current in wire 3 produces a field B3 at the empty corner. Since B3 and B1+2 combine to give a zero net field, B3 must have a direction opposite to that of B1+2. Thus, B3 must point upward and to the left, and RHR-2 indicates that the current in wire 3 must be directed out of the plane of the paper . Moreover, the magnitudes of B3 and B1+2 must be the same. Recognizing that wire 3 is a distance of d = r2 +r2 = B3 = B1+ 2 or 2 r from the empty corner, we have µ0I3 2π di 2r = 2 µ0I 2π r so that I3 I =2 68. R...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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