Physics Solution Manual for 1100 and 2101

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Unformatted text preview: ted in the drawing on the right. Also shown is the net force F, as well as the angle θ that it makes with respect to the +x axis. Since the forces are at right angles to each other, we can use the Pythagorean theorem to find the magnitude F of the net force: F = F12 + F22 = ( 4.0 ×10−3 N ) + (5.5 ×10−3 N ) 2 2 = 6.8 ×10−3 N b. The angle θ can be determined by using the inverse tangent function: −3 F1 N −1 4.0 × 10 = tan = 36° −3 F2 5.5 ×10 N θ = tan −1 84. REASONING AND SOLUTION The forces acting on each wire are the magnetic force F, the gravitational force mg, and the tension T in the strings. Each string makes an angle of 7.5° with respect to the vertical. From the drawing below at the right we can relate the magnetic force to the gravitational force. Since the wire is in equilibrium, Newton’s second law requires that ΣFx = 0 and ΣFy = 0. These equations become − T sin2.44F = 0 14 7 5° +3 4 ∑ Fx and T 442 ° − mg 1cos 7.544 = 0 3 ∑ Fy +y T o 7.5 F +x Wir e mg Solving the first equation for T, and then substituting the result into the second equation gives (after some simplification) tan 7.5° = F mg The magnetic force F exerted on one wire by the other is F = (1) µ0I 2 L , where d is the 2π d distance between the wires [d/2 = (1.2 m) sin 7.5°, so that d = 0.31 m], I is the current (which is the same for each wire), and L is the length of each wire. Substituting this relation for F into Equation (1) and then solving for the current, gives 1188 MAGNETIC FORCES AND MAGNETIC FIELDS I= 2 m F I g tan 7.5° Fπ d I GJ GJ HK L µ HK 0 = 2 O 0 tan b.050 kg / mg9.80 m / s h 7.5° Lπ b.31 m g = c M 0µ P N Q 2 320 A 0 85. SSM REASONING The magnetic moment of the orbiting electron can be found from the expression Magnetic moment = NIA . For this situation, N = 1. Thus, we need to find the current and the area for the orbiting charge. SOLUTION The current for the orbiting charge is, by definition (see Equation 20.1), I = ∆q / ∆t , where ∆q is the amount of charge that passes a given point during a time interval ∆t. Since the charge (∆q = e) passes by a given point once per revolution, we can find the current by dividing the total orbiting charge by the period T of revolution. I= ∆q ∆q (1.6 × 10 –19 C)(2.2 × 10 6 m / s) = = = 1.06 × 10 –3 A T 2π r / v 2 π (5.3 × 10 –11 m) The area of the orbiting charge is A = π r 2 = π ( 5.3 × 10 –11 m ) 2 = 8.82 × 10 –21 m 2 Therefore, the magnetic moment is Magnetic moment = NIA = (1)( 1.06 × 10 –3 A)(8.82 × 10 –21 m 2 ) = 9 .3 × 10 –24 A ⋅ m 2 CHAPTER 22 ELECTROMAGNETIC INDUCTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. 3.5 m/s 2. (e) The work done by the hand equals the energy dissipated in the bulb. The energy dissipated in the bulb equals the power used by the bulb times the time. Since the time is the same in each case, more work is done when the power used is greater. The power, however, is the voltage squared divided by the resistance of the bulb, according to Equation 20.6c, so that a smaller resistance corresponds to a greater power. Thus, more work is done when the resistance of the bulb is smaller. 3. (c) The magnetic flux Ф that passes through a surface is Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the field and the normal to the surface. Knowing Ф and A, we can calculate B cos φ = Φ / A , which is the component of the field parallel to the normal or perpendicular to the surface. 4. (b) The magnetic flux Ф that passes through a surface is Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the field and the normal to the surface. It has the greatest value when the field strikes the surface perpendicularly (φ = 0° ) and a value of zero when the field is parallel to a surface (φ = 90° ) . The field is more nearly perpendicular to face 1 (φ = 20° ) than to face 3 (φ = 70° ) and is parallel to face 2. 5. (d) Faraday’s law of electromagnetic induction states that the average emf ξ induced in a coil of N loops is ξ = − N ( ∆Φ / ∆t ) (Equation 22.3), where ∆Ф is the change in magnetic flux through one loop and ∆t is the time interval during which the change occurs. Reducing the time interval ∆t during which the field magnitude increases means that the rate of change of the flux will increase, which will increase (not reduce) the induced emf. 6. 3.2 V 7. (c) According to Faraday’s law, the magnitude of the induced emf is the magnitude of the change in magnetic flux divided by the time interval over which the change occurs (see Equation 22.3). In each case the field is perpendicular to the coil, and the initial flux is zero since the coil is outside the field region. Therefore, the changes in flux are as follows: ∆Φ A = BL2 , ∆Φ B = ∆Φ C = B 2 L2 (see Equation 22.2). The corresponding time intervals ar...
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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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