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PHYSICS 172 WQ 2010 Solutions to Homework #7 1. Giancoli Chapter 27, Problem 6 The magnetic force must be equal in magnitude to the force of gravity on the wire. The maximum magnetic force is applicable since the wire is perpendicular to the magnetic field. The mass of the wire is the density of copper times the volume of the wire. FB mg I l B 12 d l g 2 3 3 3 2 d 2 g 8.9 10 kg m 1.00 10 m 9.80 m s I 1400 A 4B 4 5.0 105 T 2 This answer does not seem feasible. The current is very large, and the resistive heating in the thin copper wire would probably melt it. 2. Giancoli Chapter 27, Problem 8 We find the force per unit length from Eq. 27-3. Note that while the length is not known, the direction is given, and so l l ˆi. F I l B I l ˆi B B FB l I ˆi B 3.0 A ˆi ˆj kˆ 1 0 0 0.20 T 1m 0.75ˆj 1.08kˆ N m 100 cm 0.36 T 0.25T 7.5 ˆj 11 kˆ 103 N cm 3. Giancoli Chapter 27, Problem 20 The velocity of each charged particle can be found using energy conservation. The electrical potential energy of the particle becomes kinetic energy as it is accelerated. Then, since the particle is moving perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will cause the ion to move in a circular path, and the radius can be determined in terms of the mass and charge of the particle. Einitial Efinal qV 12 mv 2 v Fmax qvB m v 2 r r mv qB m 2qV m 2qV 1 m qB B 2mV q rd rp r rp 1 B 2mdV qd 1 B 2mpV 1 2mV B q 1 2mpV B qp md mp qd qp qp 2 1 m mp q qp 4 2 2 rd 2 rp 2 r 2 rp 4. Giancoli Chapter 27, Problem 25 The total force on the proton is given by the Lorentz equation, Eq. 27-7. ˆi ˆj kˆ FB q E v B e 3.0ˆi 4.2ˆj 103 V m 6.0 103 m s 3.0 103 m s 5.0 103 m 0.45T 0.38T 0 1.60 10 C 4.9ˆi 6.45ˆj 0.93kˆ 10 N C 7.84 10 ˆi 1.03 10 ˆj 1.49 10 kˆ N C 0.78ˆi 1.0ˆj 0.15kˆ 10 N s 1.60 1019 C 3.0ˆi 4.2ˆj 1.9ˆi 2.25ˆj 0.93kˆ 103 N C 19 16 3 15 16 15 5. Giancoli Chapter 27, Problem 34 (a) Since the velocity is perpendicular to the magnetic field, the particle will follow a circular trajectory in the x-y plane of radius r. The radius is found using the centripetal acceleration. mv 2 mv qvB r r qB From the figure we see that the distance is the chord distance, which is twice the distance r cos . Since the velocity is perpendicular to the radial vector, the initial direction and the angle are complementary angles. The angles and are also complementary angles, so 30. 2mv0 mv 2r cos cos30 3 0 qB0 qB0 (b) From the diagram, we see that the particle travels a circular path, that is 2 short of a complete circle. Since the angles and are complementary angles, so 60. The trajectory distance is equal to the circumference of the circular path times the fraction of the complete circle. Dividing the distance by the particle speed gives t. l 2 r 360 2 60 2 mv0 2 4 m t v0 qB0 3 3qB0 v0 v0 360 6. Giancoli Chapter 27, Problem 36 With the plane of the loop parallel to the magnetic field, the torque will be a maximum. We use Eq. 27-9. 0.185m N NIAB sin B 3.32 T 2 NIAB sin 1 4.20 A 0.0650 m sin 90 7. Giancoli Chapter 27, Problem 44 Use Eq. 27-13. 260V m q E 2 1.5 105 C kg m B r 0.46T 2 0.0080m 8. Giancoli Chapter 28, Problem 8 At the location of the compass, the magnetic field caused by the wire will point to the west, and the Earth’s magnetic field points due North. The compass needle will point in the direction of the NET magnetic field. 7 I 4 10 T m A 43A Bwire 0 4.78 105 T B net 2 r 2 0.18 m tan 1 BEarth Bwire tan 1 4.5 105 T 4.78 105 T 43 N of W B Earth B wire 9. Giancoli Chapter 28, Problem 18 The magnetic field at the loop due to the long wire is into the page, and can be calculated by Eq. 28-1. The force on the segment of the loop closest to the wire is towards the wire, since the currents are in the same direction. The force on the segment of the loop farthest from the wire is away from the wire, since the currents are in the opposite direction. Because the magnetic field varies with distance, it is more difficult to calculate the total force on the left and right segments of the loop. Using the right hand rule, the force on each small piece of the left segment of wire is to the left, and the force on each small piece of the right segment of wire is to the right. If left and right small pieces are chosen that are equidistant from the long wire, the net force on those two small pieces is zero. Thus the total force on the left and right segments of wire is zero, and so only the parallel segments need to be considered in the calculation. Use Eq. 28-2. Fnet Fnear Ffar 1 0 I 1 I 2 II 1 l near 0 1 2 l far 0 I1 I 2 l 2 d near 2 d far 2 d near d far 4 107 T m A 2 6 5.1 10 N, towards wire 0.030 m 0.080 m 3.5A 2 0.100 m 1 1 10. Giancoli Chapter 28, Problem 24 We break the current loop into the three branches of the triangle and add the forces from each of the three branches. The current in the parallel branch flows in the same direction as the long straight wire, so the force is attractive with magnitude given by Eq. 28-2. II F1 0 a 2 d By symmetry the magnetic force for the other two segments will be equal. These two wires can be broken down into infinitesimal segments, each with horizontal length dx. The net force is found by integrating Eq. 28-2 over the side of the triangle. We set x=0 at the left end of the left leg. The distance of a line segment to the wire is then given by r d 3x . Since the current in these segments flows opposite the direction of the current in the long wire, the force will be repulsive. a/2 a/2 0 II II II 3a F2 dx 0 ln d 3x 0 ln 1 0 0 2d 2 3 2 3 2 d 3x We calculate the net force by summing the forces from the three segments. F F1 2 F2 0 II II 3a 0 II a 3 3a a 2 0 ln 1 ln 1 2 d 2d 2d 3 2d 2 3 11. Giancoli Chapter 28, Problem 31 Because of the cylindrical symmetry, the magnetic fields will be circular. In each case, we can determine the magnetic field using Ampere’s law with concentric loops. The current densities in the wires are given by the total current divided by the cross-sectional area. I I0 J inner 0 2 J outer 2 R1 R3 R22 (a) Inside the inner wire the enclosed current is determined by the current density of the inner wire. 2 B ds 0 I encl 0 J inner R B 2 R 0 I 0 R 2 IR B 0 02 2 R1 2 R1 (b) Between the wires the current enclosed is the current on the inner wire. I B ds 0 I encl B 2 R 0 I 0 B 20 R0 (c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the current from the outer wire. B ds I 0 encl 0 I 0 J outer R 2 R22 2 2 R 2 R22 0 I 0 R3 R B 2 r 0 I 0 I 0 B 2 R R32 R22 R32 R22 (d) (e) Outside the outer wire the net current enclosed is zero. B ds 0 I encl 0 B 2 R 0 B 0 3.0 See the adjacent graph. 2.5 1.5 -5 B (10 T) 2.0 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 2.5 R (cm) 12. Giancoli Chapter 28, Problem 37 (a) The magnetic field at point C can be obtained using the BiotSavart law (Eq. 28-5, integrated over the current). First break the loop into four sections: 1) the upper semi-circle, 2) the lower semi-circle, 3) the right straight segment, and 4) the left straight segment. The two straight segments do not contribute to the magnetic field as the point C is in the same direction that the current is flowing. Therefore, along these segments r̂ and d ˆ are parallel and d ˆ rˆ 0 . For the upper segment, each infinitesimal line segment is perpendicular to the constant magnitude radial vector, so the magnetic field points downward with constant magnitude. I d ˆ rˆ 0 I kˆ I B upper 0 R1 0 kˆ . 2 2 4 r 4 R1 4 R1 Along the lower segment, each infinitesimal line segment is also perpendicular to the constant radial vector. I d ˆ rˆ 0 I kˆ I Blower 0 R2 0 kˆ R1 4 r 2 4 R2 2 4 R2 C Adding the two contributions yields the total magnetic field. I I I I 1 1 B B upper Blower 0 kˆ 0 kˆ 0 kˆ R2 4 R1 4 R2 4 R1 R2 I (b) The magnetic moment is the product of the area and the current. The area is the sum of the two half circles. By the right-hand-rule, curling your fingers in the direction of the current, the thumb points into the page, so the magnetic moment is in the k̂ direction. R2 R2 I 2 μ 1 2 Ikˆ R1 R22 kˆ 2 2 2 3.0 13. Giancoli Chapter 28, Problem 42 We treat the loop as consisting of 5 segments, The first has length d, is located a distance d to the left of point P, and has current flowing toward the right. The second has length d, is located a distance 2d to left of point P, and has current flowing upward. The third has length d, is located a distance d to the left of point P, and has current flowing downward. The fourth has length 2d, is located a distance d below point P, and has current flowing toward the left. Note that the fourth segment is twice as long as the actual fourth current. We therefore add a fifth line segment of length d, located a distance d below point P with current flowing to the right. This fifth current segment cancels the added portion, but allows us to use the results of Problem 41 in solving this problem. Note that the first line points radially toward point P, and therefore by Problem 41(a) does not contribute to the net magnetic field. We add the contributions from the other four segments, with the contribution in the positive z-direction if the current in the segment appears to flows counterclockwise around the point P. B B 2 B3 B 4 B5 0 I I I I d d 2d d kˆ 0 kˆ 0 kˆ 0 kˆ 1/ 2 1/ 2 1/ 2 4 2d 4d 2 d 2 4 d d 2 d 2 4 d d 2 4d 2 4 d d 2 d 2 1/ 2 0 I 5ˆ 2 k 4 d 2