Physics Solution Manual for 1100 and 2101

As the net emf decreases the current in the rod

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Unformatted text preview: of Fmagnetic = q vB sin θ, where q is the magnitude of the charge, B is the magnitude of the magnetic field, v is the speed, and θ = 90° is the angle of the velocity with respect to the field. Thus, Fmagnetic = q vB. According to Equation 18.2, the electric force has a magnitude of Felectric = q E. Using the Pythagorean theorem, we find the magnitude of the net force to be 2 2 F = Fmagnetic + Felectric = ( = 1.8 ×10−6 C ( q vB ) + ( q E ) 2 2 =q ( vB )2 + E 2 ) (3.1×106 m/s )(1.2 ×10−3 T ) + ( 4.6 ×103 N/C) 2 2 = 1.1× 10−2 N 80. REASONING The two rods attract each other because they each carry a current I in the same direction. The bottom rod floats because it is in equilibrium. The two forces that act on the bottom rod are the downward force of gravity mg and the upward magnetic force of attraction to the upper rod. If the two rods are a distance s apart, the magnetic field generated by the top rod at the location of the bottom rod is (see Equation 21.5) B = µ 0 I / ( 2 π s ) . According to Equation 21.3, the magnetic force exerted on the bottom rod is F = I LB sin θ = µ 0 I 2 L sin θ / ( 2 π s ) , where θ is the angle between the magnetic field at Chapter 21 Problems 1185 the location of the bottom rod and the direction of the current in the bottom rod. Since the rods are parallel, the magnetic field is perpendicular to the direction of the current (RHR-2), and θ = 90. 0 ° , so that sin θ = 1.0 . SOLUTION Taking upward as the positive direction, the net force on the bottom rod is µ 0 I 2 L sin θ − mg = 0 2π s Solving for I, we find I= 2 π mgs = µ0L 2 π (0.073 kg)(9.80 m / s 2 )( 8.2 × 10 –3 m) = 190 A (4 π × 10 –7 T ⋅ m / A)(0.85 m) 81. SSM WWW REASONING The particle travels in a semicircular path of radius r, mv where r is given by Equation 21.2 r = . The time spent by the particle in the q B magnetic field is given by t = s / v , where s is the distance traveled by the particle and v is its speed. The distance s is equal to one-half the circumference of a circle ( s = π r). SOLUTION We find that t= πr π mv πm π (6.0 × 10 –8 kg) s = = = = 8.7 ×10 –3 s = v v v q B qB (7.2 ×10 –6 C)(3.0 T) 82. REASONING Two wires that are parallel and carry current in the same direction exert attractive magnetic forces on one another, as Section 21.7 discusses. This attraction between the wires causes the spring to compress. When compressed, the spring exerts an elastic restoring force on each wire, as Section 10.1 discusses. For each wire, this restoring force acts to push the wires apart and balances the magnetic force, thus keeping the separation between the wires from decreasing to zero. Equation 10.2 (without the minus sign) gives the magnitude of the restoring force as Fx = kx, where k is the spring constant and x is the magnitude of the displacement of the spring from its unstrained length. By setting the magnitude of the magnetic force equal to the magnitude of the restoring force, we will be able to find the separation between the rods when the current is present. SOLUTION 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 is the angle of the wire with respect to the field. Using RHR-2 reveals that the magnetic field produced by either wire is perpendicular to the other wire, so that θ = 90° in Equation 21.3, which becomes F = ILB. According to Equation 21.5 the 1186 MAGNETIC FORCES AND MAGNETIC FIELDS magnitude of the magnetic field produced by a long straight wire is B = µ0I/(2π r). Substituting this expression into Equation 21.3 gives the magnitude of the magnetic force as µ F I I= µ I L Gπ r J 2π r 2 HK 2 F = IL 0 0 Equating this expression to the magnitude of the restoring force from the spring gives µ0I 2L = kx 2π r Solving for the separation r, we find c b h gb g g bg 4 π × 10 −7 T ⋅ m / A 950 A 0.50 m µ I2L r= 0 = = 0.030 m 2π k x 2 π 150 N / m 0.020 m 2 b 83. REASONING The drawing on the left shows the directions of the two magnetic fields, as well as the velocity v of the particle. Each component of the magnetic field is perpendicular to the velocity, so each exerts a magnetic force on the particle. The magnitude of the force is F = q0 vB sinθ (Equation 21.1), and the direction can be determined by using Right-Hand Rule No. 1 (RHR-1). The magnitude and direction of the net force can be found by using trigonometry. +z +z +y v +y By = 0.065 T F1 F θ +x +x F2 Bx = 0.048 T SOLUTION a. The magnitude F1 of the magnetic force due to the 0.048-T magnetic field is ( )( ) F1 = q0 vBx sin 90.0° = 2.0 ×10−5 C 4.2 × 103 m/s ( 0.048 T ) = 4.0 ×10−3 N The magnitude F2 of the magnetic force due to the 0.065-T magnetic field is Chapter 21 Problems ( )( 1187 ) F2 = q0 vBy sin 90.0° = 2.0 ×10−5 C 4.2 ×103 m/s ( 0.065 T ) = 5.5 ×10−3 N The directions of the forces are found using RHR-1, and they are indica...
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