CollegePhysics2e-SSM-Ch22

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.

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