Physics Solution Manual for 1100 and 2101

# 1 i q t where q is the amount of charge that passes by

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Unformatted text preview: tic force that acts on a wire of length L that is directed at an angle θ with respect to a magnetic field of magnitude B and carries a current I. When θ = 0° or 180°, F = 0 N. Therefore, we need only apply Equation 21.3 to the horizontal component of the earth’s magnetic field in this problem. The direction of the magnetic force can be determined with the aid of RHR-1 (fingers point in direction of the field, thumb points in the direction of the current, palm faces in the direction of the magnetic force). SOLUTION According to Equation 21.3, the magnitude of the magnetic force exerted on the wire by the horizontal component of the earth’s field is ( ) F = ILB sin θ = ( 28 A )( 6.0 m ) 1.8 × 10−5 T sin 90.0° = 3.0 ×10−3 N Note that θ = 90.0° because the field component points toward the geographic north and the current is directed perpendicularly into the ground. The application of RHR-1 (fingers point due north, thumb points perpendicularly into the ground, palm faces due east) reveals that the direction of the magnetic force is due east 31. SSM REASONING The magnitude F of the magnetic force experienced by the wire is given by F = ILB sin θ (Equation 21.3), where I is the current, L is the length of the wire, B is the magnitude of the earth’s magnetic field, and θ is the angle between the direction of the current and the magnetic field. Since all the variables are known except B, we can use this relation to find its value. SOLUTION Solving F = ILB sin θ for the magnitude of the magnetic field, we have B= F 0.15 N = = 5.1 × 10−5 T I L sin θ ( 75 A )( 45 m ) sin 60.0° 32. REASONING The magnitude B of the external magnetic field is proportional to the magnitude F of the magnetic force exerted on the wire, according to F = ILB sin θ (Equation 21.3), where L is the length of the wire, I is the current it carries, and θ is the angle between the directions of the current and the magnetic field. When the wire is horizontal, the magnetic force is zero, indicating that sin θ = 0 . The only angles θ for which this holds are θ = 0° and θ = 180°. Therefore, the external magnetic field must be horizontal, and when the wire is tilted upwards at an angle of 19°, the angle between the directions of the current and the magnetic field must be θ = 19°. Chapter 21 Problems 1157 SOLUTION Solving F = ILB sin θ (Equation 21.3) for B yields B= F 4.4 × 10 −3 N = = 3.4 × 10 −3 T IL sin θ ( 7.5 A )( 0.53 m ) sin19o 33. REASONING AND SOLUTION The force on each side can be found from F = ILB sin θ. For the top side, θ = 90.0°, so F = (12 A)(0.32 m)(0.25 T) sin 90.0° = 0.96 N The force on the bottom side (θ = 90.0°) is the same as that on the top side, F = 0.96 N . For each of the other two sides θ = 0°, so that the force is F = 0 N . 34. REASONING We begin by noting that segments AB and BC are both perpendicular to the magnetic field. Therefore, they experience magnetic forces. However, segment CD is parallel to the field. As a result no magnetic force acts on it. According to Equation 21.3, the magnitude F of the magnetic force on a current I is F = ILB sin θ, where L is the length of the wire segment and θ is the angle that the current makes with respect to the magnetic field. For both segments AB and BC the value of the current is the same and the value of θ is 90°. The length of segment AB is greater, however. Because of its greater length, segment AB experiences the greater force. SOLUTION Using F = ILB sin θ (Equation 21.3), we find that the magnitudes of the magnetic forces acting on the segments are: Segment AB Segment BC Segment CD bgg g bb F = ILB sin θ = b.8 A g .55 m g .26 T g 90 ° = 0.40 N 2 0 0 b b sin F = ILB sin θ = b.8 A g .55 m g .26 T g 0 ° = 0 N 2 0 0 b b sin F = ILB sin θ = 2 .8 A 1.1 m 0.26 T sin 90 ° = 0.80 N 35. SSM REASONING According to Equation 21.3, the magnetic force has a magnitude of F = ILB sin θ, where I is the current, B is the magnitude of the magnetic field, L is the length of the wire, and θ = 90° is the angle of the wire with respect to the field. 1158 MAGNETIC FORCES AND MAGNETIC FIELDS SOLUTION Using Equation 21.3, we find that 7 .1 × 10 −5 N F L= = = 2 .7 m IB sin θ 0.66 A 4 .7 × 10 −5 T sin 58 ° b g c h 36. REASONING Each wire experiences a force due to the magnetic field. The magnitude of the force is given by F = ILB sin θ (Equation 21.3), where I is the current, L is the length of the wire, B is the magnitude of the magnetic field, and θ is the angle between the direction of the current and the magnetic field. Since the currents in the two wires are in opposite directions, the magnetic force acting on one wire is opposite to that acting on the other. Thus, the net force acting on the two-wire unit is the difference between the magnitudes of the forces acting on each wire. SOLUTION The length L of each wire, the magnetic field B, and the angle θ are the same for both wires. Denoting the current in one of the wires as I1 = 7.00 A and the current in the other as I, the magnitude Fnet of the net magnetic force acting on the two-wire unit is Fnet = I1...
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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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