CollegePhysics2e-SSM-Ch22

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OpenStax College Physics 2e Student Solutions Manual Chapter 22 CHAPTER 22: MAGNETISM 22.4 MAGNETIC FIELD STRENGTH: FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD 1. What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in Figure 22.47 ? Solution (a) left (West) (b) into the page (c) up (North) (d) no force (e) right (East) (f) down (South) 3. What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in Figure 22.48 , assuming it moves perpendicular to B? Solution (a) right (East) (b) into the page (c) down (South) 5. What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming B is perpendicular to v ? Solution (a) into the page (b) left (West) (c) out of the page 7. What is the maximum force on an aluminum rod with a 0.100 − μC charge that you pass between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s? In what direction is the force? Solution Examining the equation F = qvB sin θ , we see that the maximum force occurs when sin θ = 1 , so that: F max = qvB =( 0.100 × 10 − 6 C )( 5.00 m / s )( 1.50 T )= 7.50 × 10 − 7 N The direction of the force is perpendicular to both the velocity and the magnetic field. 9. (a) A cosmic ray proton moving toward the Earth at 5.00 × 10 7 m / s experiences a magnetic force of 1.70 × 10 − 16 N . What is the strength of the magnetic field if there is a 45 ° angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.

OpenStax College Physics 2e Student Solutions Manual Chapter 22 Solution (a) F = qvB sin θ ⇒ B = F qv sin θ = 1.70 × 10 − 16 N ( 1.60 × 10 − 19 C )( 5.00 × 10 7 m / s )( sin 45 ° ) = 3.01 × 10 − 5 T (b) This value for the magnetic field is on the same order of magnitude as the known strength of the Earth’s magnetic field. Since the actual magnetic field strength of the Earth varies based on where you are around the Earth, the value in part (a) seems consistent with the known value. 11. (a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than 1.00 × 10 − 12 N . What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent. Solution (a) F = qvB ⇒ q = F vB = 1.00 × 10 − 12 N ( 30.0 m / s )( 0.500 × 10 − 4 T ) = 6.67 × 10 − 10 C taking 50.0 μT as the maximum field (the earth’s field). (b) Common static electricity involves charges ranging from nanocoulombs to microcoulombs. Therefore, it would seem to be difficult to limit the charge to less than the value found in part (a) because that charge is smaller than typical static electricity. 22.5 FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD: EXAMPLES AND APPLICATIONS 12. A cosmic ray electron moves at 7.50 × 10 6 m / s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00 × 10 − 5 T . What is the radius of the circular path the electron follows? Solution r = mv qB = ( 9.11 × 10 − 31 kg )( 7.50 × 10 6 m / s ) ( 1.60 × 10 − 19 C )( 1.00 × 10 − 5 T ) = 4.27 m 14. (a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise . One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 × 10 7 m / s in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility? Solution (a) r = mv qB ⇒ B = mv qr = ( 1.67 × 10 − 27 kg )( 5.00 × 10 7 m / s ) ( 1.60 × 10 − 19 C )( 2.00 m ) = 0.261 T (b) This strength is definitely obtainable with today’s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets. 16. What radius circular path does an electron travel if it moves at the same speed and in the same magnetic field as the proton in Exercise 22.13 ?

OpenStax College Physics 2e Student Solutions Manual Chapter 22 Solution r = mv qB = ( 9.11 × 10 − 31 kg )( 7.50 × 10 7 m / s ) ( 1.60 × 10 − 19 C )( 9.785 × 10 − 1 T ) = 4.36 × 10 − 4 m 18. An electron in a TV CRT moves with a speed of 6.00 × 10 7 m / s , in a direction perpendicular to the Earth’s field, which has a strength of 5.00 × 10 − 5 T . (a) What strength electric field must be applied perpendicular to the Earth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.) Solution (a) v = E B ⇒ E = vB =( 6.00 × 10 7 m / s )( 5.00 × 10 − 5 T )= 3.00 kV / m (b) V = Ed =( 3000 V / m )( 0.0100 m )= 30.0 V 20. A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial ice. (The relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) The ratio of the masses of these two ions is 16 to 18, the mass of oxygen- 16 is 2.66 × 10 − 26 kg , and they are singly charged and travel at 5.00 × 10 6 m / s in a 1.20-T magnetic field. What is the separation between their paths when they hit a target after traversing a semicircle? Solution m 16 = 2.66 × 10 − 26 kg; m 18 = ( 18 16 ) 2.66 × 10 − 26 kg = 2.99 × 10 − 26 kg ; Δd = 2 r 18 − 2 r 16 = 2 ( r 18 − r 16 ) r = mv qB ⇒ Δd = 2 ( m 18 − m 16 ) v qB = 2 ( 2.99 × 10 − 26 kg − 2.66 × 10 − 26 kg )( 5.00 × 10 6 m / s ) ( 1.60 × 10 − 19 C )( 1.20 T ) = 0.173 m 22.6 THE HALL EFFECT 22. A large water main is 2.50 m in diameter and the average water velocity is 6.00 m/s. Find the Hall voltage produced if the pipe runs perpendicular to the Earth’s 5.00 × 10 − 5 − T field. Solution E = Blv =( 5.00 × 10 − 5 T )( 2.50 m )( 6.00 m / s )= 7.50 × 10 − 4 V 24. (a) What is the speed of a supersonic aircraft with a 17.0-m wingspan, if it experiences a 1.60-V Hall voltage between its wing tips when in level flight over the north magnetic pole, where the Earth’s field strength is 8.00 × 10 − 5 T ? (b) Explain why very little current flows as a result of this Hall voltage. Solution (a) E = Blv ,v = E Bl = 1.60 V ( 8.00 × 10 − 5 T )( 17.0 m ) = 1176 m / s = 1.18 × 10 3 m / s (b) Once established, the Hall voltage would push charges one direction and the magnetic force acts in the opposite direction resulting in no net force on the charges. Therefore, there will be no current.

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OpenStax College Physics 2e Student Solutions Manual Chapter 22 26. Calculate the Hall voltage induced on a patient’s heart while being scanned by an MRI unit. Approximate the conducting path on the heart wall by a wire 7.50 cm long that moves at 10.0 cm/s perpendicular to a 1.50-T magnetic field. Solution E = Blv =( 1.50 T )( 0.0750 m )( 0.100 m / s )= 1.13 × 10 − 2 V = 11.3 mV 28. What would the Hall voltage be if a 2.00-T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.0-A current? Solution I = nqAv d ∧ E = Bl v d ⇒ E = IB ned = 20.0 A× 2.00 T ( 8.34 × 10 28 / m 3 )( 1.60 × 10 − 19 C )( 2.588 × 10 − 3 m 3 ) = 1.16 × 10 − 6 V = 1.16 μV 30. A patient with a pacemaker is mistakenly being scanned for an MRI image. A 10.0-cm- long section of pacemaker wire moves at a speed of 10.0 cm/s perpendicular to the MRI unit’s magnetic field and a 20.0-mV Hall voltage is induced. What is the magnetic field strength? Solution E = Blv ⇒ B = E lv = 20.0 × 10 − 3 V ( 0.100 m )( 0.100 m / s ) = 2.00 T 22.7 MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR 31. What is the direction of the magnetic force on the current in each of the six cases in Figure 22.50 ? Solution (a) left (West) (b) into the page (c) up (North) (d) no force (e) right (East) (f) down (South) 33. What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases in Figure 22.52 , assuming B is perpendicular to I ? Solution (a) into the page (b) left (West) (c) out of the page 35. (a) A DC power line for a light-rail system carries 1000 A at an angle of 30.0 ° to the Earth’s 5.00 × 10 − 5 − T field. What is the force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any. Solution (a) F = IlB sin θ =( 1000 A )( 100 m )( 5.00 × 10 − 5 T ) sin 30 ° = 2.50 N (b) This means that the right-rail power lines must be attached in order not to be moved by the force caused by the earth’s magnetic field. 37. A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength? Solution F = IlB ⇒ B = F Il = 2.16 N ( 30.0 A )( 0.0400 m ) = 1.80 T

OpenStax College Physics 2e Student Solutions Manual Chapter 22 39. (a) What is the angle between a wire carrying an 8.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N? (b) What is the force on the wire if it is rotated to make an angle of 90 ° with the field? Solution (a) F = IlB sin θ ⇒ θ = sin − 1 [ F IlB ] = si n − 1 [ 2.40 N ( 8.00 A )( 0.500 m )( 1.20 T ) ] = 30.0 ° (b) F = IlB =( 8.00 A )( 0.500 m )( 1.20 T )= 4.80 N 22.8 TORQUE ON A CURRENT LOOP: MOTORS AND METERS 41. (a) By how many percent is the torque of a motor decreased if its permanent magnets lose 5.0% of their strength? (b) How many percent would the current need to be increased to return the torque to original values? Solution (a) τ = NIAB. So τ decreases by 5.00% if B decreases by 5.00%. (b) 100 95.0 = 1.0526 ⇒ 5.26% increase 43. Find the current through a loop needed to create a maximum torque of 9.00 N ⋅ m. The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-T magnetic field. Solution 0.150 m ¿ 2 ( 0.800 T ) ¿ ( 50 ) ¿ τ = NIAB ⇒ I = τ NAB = 9.00 N ⋅ m ¿ 45. Since the equation for torque on a current-carrying loop is τ = NIABsinθ , the units of N ⋅ m must equal units of . Verify this. Solution 47. A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop in radius with a current of (no kidding). Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.) Solution 49. Repeat Exercise 22.48 but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where the Earth’s field is north, but at an angle below the horizontal and with a strength of .

OpenStax College Physics 2e Student Solutions Manual Chapter 22 Solution points to the west (the loops will rotate about an east/west axis clockwise as viewed from the east) 22.10 MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS 50. (a) The hot and neutral wires supplying DC power to a light-rail commuter train carry 800 A and are separated by 75.0 cm. What is the magnitude and direction of the force between 50.0 m of these wires? (b) Discuss the practical consequences of this force, if any. Solution (a) The force is repulsive because the currents are in opposite directions. (b) This force is repulsive and therefore there is never a risk that the two wires will touch and short circuit. 52. A 2.50-m segment of wire supplying current to the motor of a submerged submarine carries 1000 A and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is the direction and magnitude of the current in the other wire? Solution in the opposite direction. 54. An AC appliance cord has its hot and neutral wires separated by 3.00 mm and carries a 5.00-A current. (a) What is the average force per meter between the wires in the cord? (b) What is the maximum force per meter between the wires? (c) Are the forces attractive or repulsive? (d) Do appliance cords need any special design features to compensate for these forces? Solution (a) (b) (c) The force is repulsive because the currents are in opposite directions. (d) These forces are rather small, so the casing around the wires is all that is needed to keep the wires from pushing apart. 56. Find the direction and magnitude of the force that each wire experiences in Figure 28.55 (a), using vector addition.

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OpenStax College Physics 2e Student Solutions Manual Chapter 22 Solution Opposites repel, likes attract, so we need to consider each wire’s relationship with the other two wires. Let f denote force per unit length, then by Look at each wire separately: Wire A Wire B Wire C For wire A:

OpenStax College Physics 2e Student Solutions Manual Chapter 22 For Wire B: For Wire C: 22.11 MORE APPLICATIONS OF MAGNETISM 58. Indicate whether the magnetic field created in each of the three situations shown in Figure 22.56 is into or out of the page on the left and right of the current. Solution (a) Right: into the page, Left: out of the page (b) Right: out of the page, Left: into the page (c) Right: out of the page, Left: into the page 60. What are the directions of the currents in the loop and coils shown in Figure 22.58 ? Solution (a) Clockwise (b) Clockwise as seen from the left

OpenStax College Physics 2e Student Solutions Manual Chapter 22 (c) Clockwise as seen from the right 61. To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop in radius carrying . What is the field at the center of such a loop? Solution Inside the MRI solenoid, For the proton “loop”: 63. Nonnuclear submarines use batteries for power when submerged. (a) Find the magnetic field 50.0 cm from a straight wire carrying 1200 A from the batteries to the drive mechanism of a submarine. (b) What is the field if the wires to and from the drive mechanism are side by side? (c) Discuss the effects this could have for a compass on the submarine that is not shielded. Solution (a) (b) There is no field since the net current is zero. (c) If the wires to and from the drive mechanism are not side-by-side, a compass on the submarine would read the wrong magnetic field, and therefore the submarine’s navigational system would not work properly. If the wires are side-by- side, then there would be no effect on the compasses, and the system would be just fine. 65. What current is needed in the solenoid described in Exercise 22.64 to produce a magnetic field times the Earth’s magnetic field of ? Solution 67. Measurements affect the system being measured, such as the current loop in Figure 22.53. (a) Estimate the field the loop creates by calculating the field at the center of a circular loop 20.0 cm in diameter carrying 5.00 A. (b) What is the smallest field strength this loop can be used to measure, if its field must alter the measured field by less than 0.0100%? Solution (a) (b)

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OpenStax College Physics 2e Student Solutions Manual Chapter 22 69. Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.55 (a), using the rules of vector addition to sum the contributions from each wire. Solution 71. What current is needed in the top wire in Figure 22.55 (a) to produce a field of zero at the point equidistant from the wires, if the currents in the bottom two wires are both 10.0 A into the page? Solution First, find the field from the bottom two wires. The field in the y direction is zero, by symmetry.

OpenStax College Physics 2e Student Solutions Manual Chapter 22 and must be to the left and of magnitude (into page) 73. Integrated Concepts (a) A pendulum is set up so that its bob (a thin copper disk) swings between the poles of a permanent magnet as shown in Figure 22.60 . What is the magnitude and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive charge and is released from a height of 30.0 cm above its lowest point? The magnetic field strength is 1.50 T. (b) What is the acceleration of the bob at the bottom of its swing if its mass is 30.0 grams and it is hung from a flexible string? Be certain to include a free-body diagram as part of your analysis. Solution (a) so that The force acts up . (b) At the bottom of the swing, the acceleration due to the pendulum motion is zero. Just before entering the magnetic field, . To calculate the acceleration just after entering the magnetic field, , so that the acceleration is: , or 75. Integrated Concepts Find the radius of curvature of the path of a 25.0-MeV proton moving perpendicularly to the 1.20-T field of a cyclotron. Solution 77. Integrated Concepts (a) Using the values given for an MHD drive in Exercise 22.36 , and assuming the force is uniformly applied to the fluid, calculate the pressure created in (b) Is this a significant fraction of an atmosphere?

OpenStax College Physics 2e Student Solutions Manual Chapter 22 Solution (a) (b) No , this is not a significant fraction of an atmosphere. 79. Integrated Concepts A current balance used to define the ampere is designed so that the current through it is constant, as is the distance between wires. Even so, if the wires change length with temperature, the force between them will change. What percent change in force per degree will occur if the wires are copper? Solution Force is proportional to the length of the wires. Copper has 81. Integrated Concepts A cyclotron accelerates charged particles as shown in Figure 22.61 . Using the results of the previous problem, calculate the frequency of the accelerating voltage needed for a proton in a 1.20-T field. Solution 83. Integrated Concepts (a) What is the direction of the force on a wire carrying a current due east in a location where the Earth’s field is due north? Both are parallel to the ground. (b) Calculate the force per meter if the wire carries 20.0 A and the field strength is . (c) What diameter copper wire would have its weight supported by this force? (d) Calculate the resistance per meter and the voltage per meter needed. Solution (a) Use the right hand rule-1. The direction of the force is up from the ground (out of the page). (b) , or (c) We want the force of the magnetic field to balance the weight force, so . Now, to calculate the mass, recall , where the volume is , so and , or

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OpenStax College Physics 2e Student Solutions Manual Chapter 22 (d) Also 85. Unreasonable Results (a) Find the charge on a baseball, thrown at 35.0 m/s perpendicular to the Earth’s field, that experiences a 1.00-N magnetic force. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Solution (a) (b) It is impossible to have such a large separated charge on such a small object. (c) The 1.00 N force is much too great to be realistic in the earth’s field. 87. Unreasonable Results An inventor wants to generate 120-V power by moving a 1.00- m-long wire perpendicular to the Earth’s field. (a) Find the speed with which the wire must move. (b) What is unreasonable about this result? (c) Which assumption is responsible? Solution (a) (b) The speed is too high to be practical at ≤1% the speed of light. (c) The assumption that you could generate such a voltage with a single wire in the earth’s field is unreasonable. 89. Unreasonable Results A surveyor 100 m from a long straight 200-kV DC power line suspects that its magnetic field may equal that of the Earth and affect compass readings. (a) Calculate the current in the wire needed to create a field at this distance. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Solution (a) (b) This current is unreasonably high. It implies a total power delivery in the line of which is much too high for standard transmission lines. (c) 100 meters is a long distance to obtain the required field strength. Also coaxial cables are used for transmission lines so that there is virtually no field for DC power lines, because of cancellation from opposing currents. The surveyor’s concerns are not a problem for his magnetic field measurements. 92. Critical Thinking A magnetic field of 2.00 × 10 − 7 T in the positive x-direction exists.

OpenStax College Physics 2e Student Solutions Manual Chapter 22 Three electrons pass through the origin, one at a time, each with a velocity of 5.00 × 10 7 m / s . The first electron passes the origin along the x-axis, the second electron along the y-axis, and the third at an angle of 45 ° with respect to both the y- and z-axes in the yz-plane. (a) What is the force acting on each electron? (b) What is the shape of the path of each electron? (c) When, if ever, will each electron return to the origin? Solution (a) F = qvB sin θ Since sin θ = 0, F = 0 for the first electron. For the second and third electrons, sin θ = 1 . So, F =( 1.60 × 10 − 19 C )( 5.00 × 10 7 m / s )( 2.00 × 10 − 7 T )= 1.60 × 10 − 18 N . (b) The first electron moves in a straight line. The other two electrons take a circular path. (c) The first electron will never return to the origin. The other two will return. r = ( 9.11 × 10 − 11 )( 5.00 × 10 7 ) ( 1.60 × 10 − 19 )( 2.00 × 10 − 7 ) m v = r t t = ( 9.11 × 10 − 11 ) ( 1.60 × 10 − 19 )( 2.00 × 10 − 7 ) s = 2.85 × 10 15 s This file is copyright 2022, Rice University. All Rights Reserved.

CollegePhysics2e-SSM-Ch22

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