Physics Solution Manual for 1100 and 2101

1 thus the bar in circuit 1 produces twice the emf

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Unformatted text preview: π × 10 −7 T ⋅ m / A 48 A µ0I r= = = 0.12 m 2π B 2 π 8.0 × 10 −5 T c (21.5) 74. REASONING The angle θ between the electron’s velocity and the magnetic field can be found from Equation 21.1, F sin θ = q vB According to Newton’s second law, the magnitude F of the force is equal to the product of the electron’s mass m and the magnitude a of its acceleration, F = ma. SOLUTION The angle θ is ( )( ) 9.11 × 10−31 kg 3.50 × 1014 m/s 2 ma −1 = 19.7° θ = sin = sin q vB 1.60 × 10−19 C 6.80 × 106 m/s 8.70 × 10−4 T −1 ( )( )( ) 75. SSM REASONING The magnitude τ of the torque that acts on a current-carrying coil placed in a magnetic field is specified by τ = NIAB sinφ (Equation 21.4), where N is the number of loops in the coil, I is the current, A is the area of one loop, B is the magnitude of 1182 MAGNETIC FORCES AND MAGNETIC FIELDS the magnetic field, and φ is the angle between the normal to the coil and the magnetic field. All the variables in this relation are known except for the current, which can, therefore, be obtained. SOLUTION Solving the equation τ = NIAB sinφ for the current I and noting that φ = 90.0° since τ is specified to be the maximum torque, we have I= τ NAB sin φ = (1200 ) (1.1 × 5.8 N ⋅ m ) 10−2 m 2 ( 0.20 T ) sin 90.0° = 2.2 A 76. REASONING The magnitude of the magnetic force acting on the particle is F = q0 vB sin θ (Equation 21.1), where q0 and v are the charge magnitude and speed of the particle, respectively, B is the magnitude of the magnetic field, and θ is the angle between the particle’s velocity and the magnetic field. The magnetic field is produced by a very long, straight wire, so its value is given by Equation 21.5 as B = µ0 I / ( 2π r ) . By combining these two relations, we can determine the magnitude of the magnetic force. SOLUTION The direction of the magnetic field B produced by the current-carrying wire can be found by using Right-Hand Rule No. 2. At the location of the charge, this field points perpendicularly into the page, as shown in the drawing. Since the direction of the particle’s velocity is perpendicular to the magnetic field, θ = 90.0°. Substituting B = µ0 I / ( 2π r ) into F = I v B (into page) q× F q0 vB sin θ gives µ I F = q0 vB sin θ = q0 v 0 sin θ 2π r ( 6.00 × 10−6 C )( 7.50 × 104 m/s )( 4π ×10−7 T ⋅ m/A ) ( 67.0 A ) sin 90.0° = 1.21 × 10−4 N = 2π ( 5.00 × 10−2 m ) The direction of the magnetic force F exerted on the particle can be determined by using Right-Hand Rule No. 1. This direction, which is shown in the drawing, is perpendicular to the wire and is directed away from it . Chapter 21 Problems 1183 77. SSM REASONING AND SOLUTION The current associated with the lightning bolt is I= ∆q 15 C = = 1.0 × 104 A −3 ∆t 1.5 × 10 s The magnetic field near this current is given by ( )( ) 4π × 10−7 T ⋅ m/A 1.0 × 104 A µ0 I B= = = 8.0 × 10−5 T 2π r 2π ( 25 m ) 78. REASONING The magnitude of the magnetic force exerted on a long straight wire is given by Equation 21. 3 as F = ILB sin θ. The direction of the magnetic force is predicted by Right-Hand Rule No. 1. The net force on the triangular loop is the vector sum of the forces on the three sides. SOLUTION a. The direction of the current in side AB is opposite to the direction of the magnetic field, so the angle θ between them is θ = 180°. The magnitude of the magnetic force is FAB = ILB sin θ = ILB sin 180° = 0N For the side BC, the angle is θ = 55.0°, and the length of the side is L= 2 .00 m = 3.49 m cos 55.0 ° The magnetic force is b g b gg b FBC = ILB sin θ = 4 .70 A 3.49 m 1.80 T sin 55.0 ° = 24 .2 N An application of Right-Hand No. 1 shows that the magnetic force on side BC is directed perpendicularly out of the paper , toward the reader. For the side AC, the angle is θ = 90.0°. We see that the length of the side is L = (2.00 m) tan 55.0° = 2.86 m The magnetic force is b g b gg b F AC = ILB sin θ = 4 .70 A 2 .86 m 1.80 T sin 90.0 ° = 24 .2 N 1184 MAGNETIC FORCES AND MAGNETIC FIELDS An application of Right-Hand No. 1 shows that the magnetic force on side AC is directed perpendicularly into the paper , away from the reader. b. The net force is the vector sum of the forces on the three sides. Taking the positive direction as being out of the paper, the net force is b g ∑ F = 0 N + 24.2 N + −24.2 N = 0 N 79. REASONING The magnetic field applies the maximum magnetic force to the moving charge, because the motion is perpendicular to the field. This force is perpendicular to both the field and the velocity. The electric field applies an electric force to the charge that is in the same direction as the field, since the charge is positive. These two forces are shown in the drawing, and they are perpendicular to one another. Therefore, the magnitude of the net field can be obtained using the Pythagorean theorem. Fmagnetic E Felectric B v SOLUTION According to Equation 21.1, the magnetic force has a magnitude...
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