Physics Solution Manual for 1100 and 2101

# 140 m 2 104 10 2 t 66 reasoning the turns of the coil

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Unformatted text preview: we have that ( ) τ = N I AB sinφ = N I π r 2 B sinφ (1) Since the length of the wire is the circumference of the circle, or L = 2π r , it follows that the L radius of the circle is r = . Substituting this result into Equation (1) gives 2π L 2 N I L2 B τ = N I π r B sinφ = N I π B sinφ = sinφ 4π 2π ( 2 ) The maximum torque τmax occurs when φ = 90.0°, so that τ max ( (1)( 4.30 A ) 7.00 ×10−2 m N I L2 B = sin 90.0° = 4π 4π ) 2 ( 2.50 T ) sin 90.0° = 4.19 ×10−3 N ⋅ m 46. REASONING AND SOLUTION The torque is given by τ = NIAB sin φ. a. The maximum torque occurs when φ = 90.0° (sin φ = 1). In this case we want the torque to be 80.0% of the maximum value, so b NIAB sin φ = 0.800 NIAB sin 90.0 ° g so that φ = sin −1 e .800 j = 53.1° 0 Chapter 21 Problems b. The edge-on view of the coil at the right shows the normal to the plane of the coil, the magnetic field B, and the angle φ . 1165 B φ = 53.1° Coil Normal 47. SSM WWW REASONING The torque on the loop is given by Equation 21.4, τ = NIABsin φ . From the drawing in the text, we see that the angle φ between the normal to the plane of the loop and the magnetic field is 90° − 35° = 55° . The area of the loop is 0.70 m × 0.50 m = 0.35 m2. SOLUTION a. The magnitude of the net torque exerted on the loop is τ = NIAB sin φ = (75)(4.4 A)(0.35 m 2 )(1.8 T) sin 55° = 170 N ⋅ m b. As discussed in the text, when a current-carrying loop is placed in a magnetic field, the loop tends to rotate such that its normal becomes aligned with the magnetic field. The normal to the loop makes an angle of 55° with respect to the magnetic field. Since this angle decreases as the loop rotates, the 35° angle increases . 48. REASONING When the wire is used to make a single-turn 1 square coil, each side of the square has a length of 4 L . When the wire is used to make a two-turn coil, each side of the square 1 has a length of 8 L . The drawing shows these two options and indicates that the total effective area of NA is greater for the single-turn option. Hence, more torque is obtained by using the single-turn option. SOLUTION According to Equation 21.4 the maximum torque experienced by the coil is τmax = (NIAB) sin 90º = NIAB, where N is the number of turns, I is the current, A is the area of each turn, and B is the magnitude of the magnetic field. Applying this expression to each option gives L/4 L/8 1166 MAGNETIC FORCES AND MAGNETIC FIELDS Single - turn τ max = NIAB = NI ( 1 L) 4 2 B 4 = (1)(1.7 A ) 1 (1.00 m ) Two - turn 2 ( 0.34 T ) = 0.036 N ⋅ m ( 0.34 T ) = 0.018 N ⋅ m τ max = NIAB = NI ( 1 L ) B 8 2 = ( 2 )(1.7 A ) 1 (1.00 m ) 8 2 49. REASONING The magnitude τ of the torque that acts on a current-carrying coil placed in a magnetic field is given by τ = NIAB sinφ (Equation 21.4), where N is the number of loops in the coil (N = 1 in this problem), I is the current, A is the area of one loop, B is the magnitude of the magnetic field (the same for each coil), and φ is the angle (the same for each coil) between the normal to the coil and the magnetic field. Since we are given that the torque for the square coil is the same as that for the circular coil, we can write (1) Isquare Asquare B sin φ 144424444 4 3 τ square = (1) I circle Acircle B sin φ 1444 2444 3 τ circle This relation can be used directly to find the ratio of the currents. SOLUTION Solving the equation above for the ratio of the currents yields Isquare I circle = Acircle Asquare If the length of each wire is L, the length of each side of the square is the square coil is Asquare = 1 4 L , and the area of 1 ( 1 L ) ( 1 L ) = 16 L2 . The area of the circular coil is 4 4 Acircle = π r 2 , where r is the radius of the coil. Since the circumference (2π r) of the circular coil is equal to the length L of the wire, we have 2π r = L, or r = L / ( 2π ) . Substituting this value for r into the expression for the area of the circular coil gives Acircle = π [ L / ( 2π )] . Thus, the ratio of the currents is 2 Isquare I circle = Acircle Asquare = 2 L 2π = 4 = 1.27 1 L2 π 16 π Chapter 21 Problems 1167 50. REASONING AND SOLUTION According to Equation 21.4, the maximum torque for a single turn is τmax = IAB. When the length L of the wire is used to make the square, each side of the square has a length L/4. The area of the square is Asquare = (L/4)2. For the rectangle, since two sides have a length d, while the other two sides have a length 2d, it follows that L = 6d, or d = L/6. The area is Arectangle = 2d 2 = 2(L/6)2. Using Equation 21.4 for the square and the rectangle, we obtain τ square τ rectangle = IAsquare B IArectangle B Asquare = Arectangle L b / 4g = 2b / 6g L 2 = 2 1.13 51. REASONING The coil in the drawing is oriented such that the normal to the surface of the coil is perpendicular to the magnetic field (φ = 90°). The magnetic torque is a maximum, and Equation 21.4 gives its magnitude as τ = NIAB sin φ. In this expression N is the number of loops in the coil, I is the current, A is the area of one loop, and B is th...
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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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