Practice Test for Final - Answers

Practice Test for Final Chapter 27 27. The total force on the proton is F = e(E + v × B) = e{(3.0i – 4.2 j) × 10 3 V/m + [(6.0i + 3.0 j – 5.0k) × 103 m/s × (0.45i + 0.20 j)T]} = (1.60 × 10–19 C)[(3.0i – 4.2 j) + (1.0i – 2.25 j – 0.15k )] × 103 V/m = (6.4i – 10.3j – 0.24k )] × 10 –16 N. 28. For the magnetic force we have F = – ev × B; (3.8i – 2.7j ) × 10 –13 N = – (1.60 × 10 –19 C)(v xi + v yj + v zk) × (0.35 T)k ; (3.8i – 2.7j ) × 10 –13 N = – (1.60 × 10 –19 C)(0.35 T)( – vx j + vyi). When we equate each component, we get v x = (– 2.7 × 10–13 N)/(1.60 × 10 –19 C)(0.35 T) = – 4.8 × 106 m/s; v y = – (3.8 × 10–13 N)/(1.60 × 10 –19 C)(0.35 T) = – 6.8 × 106 m/s. Thus the velocity is v = (– 4.8 × 10 6 m/s)i + ( – 6.8 × 106 m/s) j. θ 31. (a) Because the velocity is perpendicular to the magnetic field, the proton will travel in a circular arc. From the symmetry of the motion we see that the upper half is a mirror image of the lower half, so the exit angle is the same as the incident angle: 45°. (b) T he magnetic force provides the radial acceleration, so we have F = evB = m v2 /r, so r = mv/eB = (1.67 × 10–27 kg)(2.0 × 10 5 m/s)/(1.60 × 10–19 C)(0.850 T) = 2.46 × 10–3 m. Thus the distance x is x = rv 2 = 3.5 × 10–3 m. v r B x r 51. The magnetic force must be toward the center of the circular path, so the magnetic field must be up. The magnetic force provides the centripetal acceleration: qvB = m v2 /r, or mv = qBr; 4.8 × 10–16 kg · m/s = (1.60 × 10–19 C)B(1.0 × 103 m), which gives B = 3.0 T up. 64. (a) To make the path straight, the forces from the electric field and the magnetic field balance: eE = evB ; 10,000 V/m = (4.8 × 106 m/s)B, which gives B = 2.1 × 10–3 T . (b) Because the electric force is down, the magnetic force must be up, so the magnetic field is out of the page. (c) If there is only the magnetic field, the radius of the circular orbit is r = mv /qB. The time to complete a circle is T = 2πr/v = 2πm/qB, so the frequency is f = 1/T = qB/2πm = (1.60 × 10 –19 C)(2.08 × 10 –3 T)/2π(9.11 × 10 –31 kg) = Chapter 28 v 45° FB E –e – v B FE 5.8 × 107 H z. Example 28-3, p711 12. Because the currents are in opposite directions, the fields will be in opposite directions. For the net field we have B = B1 – B2 = [(µ0 /4π)2I1 /r 1] – [(µ 0/4π)2I2/r2 ] = [(µ 0/4π)2I]{[1/(L – ! d)] – [1/(L + !d)]} = [(µ 0/4π)2I/L ]{[1/(1 – ! d/L )] – [1/(1 + !d/L)]}. B1 I2 L × I1 d Because d « L, we can use the approximation 1/(1 ± x) ≈ 1 — x: B = [(µ 0/4π)2I/L ][(1 + ! d/L ) – (1 – ! d/L)] = [(µ 0/4π)2I/L ](d/L ) = (µ 0/4π)2Id/L2 = (10–7 T · m/A)2(25 A)(2.8 × 10–3 m)/(0.100 m)2 = 1.4 × 10 –6 T up, with the currents as shown. This is (1.4 × 10–6 T)/(5.0 × 10 –5 T) = 0.028 = 2.8% of the Earth’s field. B2 18. We find the direction of the field for each wire from y the tangent to the circle around the wire, as shown. B θ 1 1 For their magnitudes, we have P B 1 = (µ 0/4π)2I1/L1 = (10–7 T · m/A)2(16.5 A)/(0.120 m) = 2.75 × 10–5 T. θ2 B 2 = (µ 0/4π)2I2/L2 B B2 = (10–7 T · m/A)2(16.5 A)/(0.130 m) = 2.54 × 10–5 T. α We use the property of the triangle to find the angles shown: r2 2 = r12 + d2 – 2r 1d cos θ1 ; r1 (13.0 cm)2 = (12.0 cm)2 + (7.0 cm)2 – 2(12.0 cm)(7.0 cm) cos θ1 , which gives cos θ 1 = 0.143, θ 1 = 81.8°; r1 2 = r22 + d2 – 2r 2d cos θ2 ; θ1 (12.0 cm)2 = (13.0 cm)2 + (7.0 cm)2 – 2(7.0 cm)(13.0 cm) cos θ2 , d I1 which gives cos θ 2 = 0.407, θ 2 = 66.0°; From the vector diagram, we have B = B1(– sin θ1 i + cos θ 1 j) + B2(– sin θ 2 i – cos θ2 j) = (2.75 × 10–5 T)(– sin 81.8° i + cos 81.8° j) + (2.54 × 10–5 T)(– sin 66.0° i – cos 66.0° j) = (– 5.04 × 10–5 T) i + (– 6.41 × 10–6 T) j. For the direction of the field, we have tan θ = By/Bx = (– 6.41 × 10–6 T)/(– 5.04 × 10–5 T) = 0.127, θ = 187.2°. We find the magnitude from Bx = B cos θ; – 5.04 × 10 –6 T = B cos 187.2°, which gives B = 5.08 × 10–5 T 187.2° from the x-axis. 47. Because the currents and the separations are the same, we find the force per unit length between any two wires from F/L = I 1(µ 0I2 /2πd) = µ0 I2 /2πd = (4π × 10–7 T · m/A)(8.00 A)2/2π(0.380 m) = 3.37 × 10 –5 N/m. The directions of the forces are shown on the diagram. The symmetry of the force diagrams simplifies the vector addition, so we have F M/L = 2(F/L ) cos 30° = 2(3.37 × 10 –5 N/m) cos 30° = 5.84 × 10–5 N/m up. F N/L = F/L = 3.37 × 10–5 N/m 60° below the line toward P. FP /L = F/L = 3.37 × 10–5 N/m 60° below the line toward N. r2 θ2 I2 y FM F F IM d IN × F 60° F FN IP F × FP x F x y 48. Because the currents and the separations are the same, we find the force on a length L of the top wire from F F either of the two bottom wires from F = I 1(µ 0I 2 /2πd)L = µ0I 1I 2L/2πd I1 = (4π × 10–7 T · m/A)(20.0 A)I 1L /2π (0.380 m) L = (1.05 × 10–5 N/A · m)I1L. The directions of the forces are shown on the diagram. mg 60° The symmetry of the force diagram simplifies the I2 I3 vector addition, so for the net force to be zero, we have × × 2F cos 30° = mg = ρ πr 2Lg; 2(1.05 × 10 –5 N/A · m)I1L cos 30° = (8.9 × 103 kg/m3 )π(1.00 × 10–3 m) 2(9.80 m/s 2)L, which gives I1 = 1.50 × 10 4 A. x 52. (a) For the force produced by the magnetic field of the upper wire to balance the weight, it must be up, i. e., an attractive force. Thus IT the currents must be in the same direction. When we equate the magnitudes of the two forces for a length L, we get I BB TL = mg; FT IB (µ0 /4π)2I BIT L/d = ρ(πr2 L)g; (10–7 T · m/A)2I B(48 A)/(0.15 m) = BT (8.9 × 10 3 kg/m 3)π(1.25 × 10–3 m) 2(9.80 m/s 2), which gives I B = 6.7 × 10 3 A to the right . mg (b) The magnetic force will decrease with increasing separation. If the wire is moved a small distance above or below the equilibrium position, there will be a net force, away from equilibrium, and the wire will be unstable. (c) If the second wire is above the first, there must be a repulsive magnetic force between the two wires to balance the weight, which means the currents must be opposite. Because the separation is the same, the magnitude of the current is the same: I2 = 6.7 × 10 3 A to the left . The magnetic force will decrease with increasing separation. If the wire is moved a small distance above or below the equilibrium position, there will be a net force back toward equilibrium, and the wire will be stable for vertical displacements d Chapter 29 14. As the loop is pulled from the field, the flux through the loop will decrease. We find the induced emf from å = – ∆Φ B/∆t = – B ∆A/∆t = – B¬ ∆x/∆t = – B¬ (– v) = B¬v. FB Because the inward flux is decreasing, the induced flux will be into the page, so the induced current is clockwise, given by I = å/R. B Because this current in the left hand side of the loop is in a downward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, which we find from F = I¬B = (å/R)¬ B = B 2¬ 2v/R = (0.450 T)2(0.350 m) 2(3.40 m/s)/(0.230 Ω) = ¬ x 0.367 N. 15. For the resistance of the loop, we have R = ρL/A = ρπD/( πd2 = 4ρD/d2 . The induced emf is å = – ∆Φ B/∆t = – ( πD 2 ∆B/∆t; so the induced current is I = å/R = – (πDd2/16ρ) ∆B/∆t. In the time ∆t the amount of charge that will pass a point is Q = I ∆t = – (πD d2/16ρ) ∆B = – [π(0.156 m)(2.05 × 10 –3 m )2/16(1.68 × 10 –8 Ω · m)](0 – 0.550 T) = 19. The flux through the loop is F 4.21 C . Φ B = BA. The induced emf is å = – dΦ B/dt = – B dA/dt = – (0.48 T)(– 3.50 × 10 –2 m 2/s) = 1.7 × 10–2 V. Because the area changes at a constant rate, this is the induced emf for both times. 21. At a distance r from the wire, the magnetic field is directed into the paper with magnitude B = µ 0I/2πr. Because the field is not constant over the square, we find the magnetic flux by integration. We choose a differential element parallel to the wire at position r with area a dr. The magnetic field through this element is constant, so the flux is B a a I a ΦΒ = a+ a B · dA = B dA = a µ 0I µ Ia a dr = 0 l n 2a = 2πr 2π a r µ 0Ia l n 2. 2π 22. Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from å = B¬v = (0.750 T)(0.190 m)(0.250 m/s) = 3.56 × 10 –2 V = 35.6 mV. B dr v ¬ 40. We find the ratio of the number of turns from N S/N P = VS/VP = (12 × 10 3 V)/(220 V) = 55. If the transformer is connected backward, the role of the turns will be reversed: V S/VP = N S/NP ; V S/(220 V) = 1/55, which gives VS = 4.0 V. 41. (a) We assume 100% efficiency, so we have IPVP = ISVS ; (0.65 A)(120 V) = (15 A)V S , which gives VS = 5.2 V. (b) Because V S < VP , this is a step-down transformer. 42. (a) We assume 100% efficiency, so we find the input voltage from P= IP VP ; 100 W = (26 A)VP , which gives VP = 3.8 V. Because V S > VP , this is a step-up transformer. (b) For the voltage ratio we have VS/VP = (12 V)/(3.84 V) = 3.1 . Chapter 30 13. If D represents the diameter of the solenoid, the length of the wire is N(πD). Because this is constant, we have N 1 πD 1 = N 2πD 2 , or N 2 /N1 = D 1/D 2 = @ . The solenoid is tightly wound, so the length of the solenoid is ¬ = Nd, where d is the diameter of the wire. Thus we have ¬ 2/¬ 1 = N 2 /N1 = @. We use the inductance of a solenoid: L = µ 0AN 2/¬, and form the ratio of inductances for the two conditions, so we have L2 /L 1 = (D2 /D 1)2(N 2 /N1 )2 /(¬ 2 /¬ 1) = (3)2(@ )2/(@ ) = 3. 5. We find the mutual inductance of the system by finding the mutual inductance of the loop. The magnetic field of the long wire depends only on the distance from the wire. To find the magnetic flux through the loop, we choose a strip a distance x from the wire with width dx: ΦΒ = 2 B · dA = 1 µ 0I µ Iw w dx = 0 ln 2πx 2π 2 . 1 The mutual inductance is M = Φ B/I = (µ0w/2π) ln(¬ 2/¬ 1). x dx w I B ¬ 1 ¬ 2 Chapter 31 15. (a) The reactance of the capacitor is XC = 1/2πfC = 1/2π(60 Hz)(0.80 × 10–6 F) = 3.32 × 10 3 Ω = 3.32 kΩ. The impedance of the circuit is Z = (R 2 + XC 2)1/2 = [(6.0 kΩ )2 + (3.32 kΩ)2]1/2 = 6.86 kΩ. The rms current is Irms = Vr m s/Z = (120 V)/(6.86 kΩ) = 18 mA . (b) We find the phase angle from cos φ = R/Z = (6.0 kΩ )/(6.86 kΩ) = 0.875. In an RC circuit, the current leads the voltage, so φ = – 29°. (c) The power dissipated is P = Ir m s2 R = (17.5 × 10–3 A) 2(6.0 × 103 Ω) = 1.8 W. (d) The rms readings across the elements are VR = Irms R = (17.5 mA)(6.0 kΩ ) = 105 V; VC = Irms XC = (17.5 mA)(3.32 kΩ ) = 58 V. Note that, because the maximal voltages occur at different times, the two readings do not add to the applied voltage of 120 V. 16. (a) The reactance of the inductor is XL = 2πfL = 2π(60 Hz)(0.250 H) = 94.2 Ω. The impedance of the circuit is Z = (R 2 + XL 2) 1/2 = [(765 Ω) 2 + (94.2 Ω)2] 1/2 = 770.8 Ω . The rms current is Irms = Vr m s/Z = (120 V)/(770.8 Ω) = 0.156 A. (b) We find the p hase angle from cos φ = R/Z = (765 Ω )/(770.8 Ω) = 0.993. In an RL circuit, the current lags the voltage, so φ = + 7.02°. (c) The power dissipated is P = Ir m s2 R = (0.156 A)2 (765 Ω ) = 18.5 W. (d) The rms readings across the elements are VR = Irms R = (0.156 A)(765 Ω) = 119 V; VL = I rms X L = (0.156 A)(94.2 Ω) = 15 V. Note that, because the maximal voltages occur at different times, the two readings do not add to the applied voltage of 120 V. 17. The reactances and impedance in the circuit are X L = 2πfL = 2π(60 Hz)(35 × 10 –3 H) = 13.2 Ω. X C = 1/2πfC = 1/2π(60 Hz)(20 × 10 –6 F) = 133 Ω. Z = [R2 + (XL – X C)2 ]1/2 = [(2.0 Ω)2 + (13.2 Ω – 133 Ω) 2]1/2 = 119 Ω . (a) The rms current is Irms = Vr m s/Z = (45 V)/(119 Ω) = 0.38 A. (b) We find the phase angle from cos φ = R/Z = (2.0 Ω)/(119 Ω) = 0.0168. Because X C > XL , the current leads the voltage, so φ = – 89°. (c) The power dissipated is P = Ir m s2 R = (0.38 A) 2(2.0 Ω) = 0.29 W. 27. We find the capacitance from the new resonant frequency: f0 = (1/2π)(1/LC) 1/2 ; 2(33.0 × 10 3 Hz) = (1/2π)[1/(4.15 × 10 –3 H)C] 1/2, which gives C = 1.40 × 10–9 F. At the applied frequency the reactances are X L = 2πfL = 2π(33.0 × 10 3 Hz)(4.15 × 10–3 H) = 860 Ω = 0.860 kΩ. X C = 1/2πfC = 1/2π(33.0 × 103 Hz)(1.40 × 10–9 F) = 3.44 × 103 Ω = 3.44 kΩ. The impedance is Z = [R2 + (XL – X C)2 ]1/2 = [(0.220 kΩ)2 + (0.860 kΩ – 3.44 kΩ )2] 1/2 = 2.59 kΩ. The peak current is I0 = V0 /Z = (136 V)/(2.59 kΩ ) = 52.5 mA .