Chapter 24: Magnetic iels an orces Solutions Questions: 4, 13, 16, 18, 31 Exercises & Problems: 3, 6, 7, 15, 21, 23, 31, 47, 60 Q24.4: Green turtles use the earth s magnetic fiel to navigate. They seem to use the fiel to tell them their latitue how far north or south of the equator they are. Explain how knowing the irection of the earth s fiel coul give this information. Q24.4. Reason: The earth s magnetic fiel vectors ten to point below the horizon in the northern hemisphere. The angle below the horizon is calle the ip angle. ut the ip angle changes with latitue. As illustrate in the figure, near the equator, the ip angle is small an near the north pole, the ip angle is nearly 90. Each ifferent latitue has a certain ip angle an the turtle is able to etermine its latitue from the ip angle it senses. Assess: The way the turtle fins its latitue is similar to the way sailors can fin their latitue by looking at the position of stars. or example, a sailor can use the ip angle of the sun s rays at noon to fin his or her latitue. Q24.13: The figure shows a solenoi as seen in cross section. Compasses are place at points 1 an 2. In which irection will each compass point when there is a large current in the irection shown? Explain. Q24.13. Reason: If a current is present in a long solenoi, the magnetic fiel that is create is uniform along its central axis. The irection of the magnetic fiel is in the irection the thumb on your right han points when your fingers are curle in the irection of the current in accorance with the right-han rule for fiels. The magnetic fiel points to the right. The magnetic fiel lines shown in igure 24.17 in the textbook show that the fiel curves aroun the outsie of the solenoi in the other irection. This means the compass neele shoul point left. Assess: The right-han rule can be use to etermine the irection of the magnetic fiel ue to any current, even a curve current.
Q24.16: What is the initial irection of eflection for the charge particles entering the magnetic fiels shown? Q24.16. Reason: The irection of the magnetic force etermines the irection that the particle will be eflecte when it enters the magnetic fiel. Use the right-han rule for forces to etermine the irection of the force. See igure 24.27 in the textbook. (a) The velocity points to the right an the magnetic fiel points towar the bottom of the page, so the force points into the page. (b) In this case the velocity an the magnetic fiel are pointing in the same irection, so there is no force. Assess: The reason there is no force for part (b) can be shown using the = qvsin α, equation where α is the angle between the velocity an the magnetic fiel. So if the velocity is parallel to the magnetic fiel, this angle is zero, which makes the sine an the force equal zero as well. Q24.18: Determine the magnetic fiel irection that causes the charges particles shown in the figure below to experience the inicate magnetic forces. Q24.18. Reason: Use the right-han rule for forces to etermine the irection of the magnetic fiel. or (a), the component of the magnetic fiel affecting the charge particle points 90 clockwise from the irection of the velocity, v is parallel to the plane of the page (note the charge is negative). or (b), the component of the magnetic fiel affecting, the charge particle points 90 counterclockwise from the irection of the force vector, an is parallel to the plane of the page. Assess: The irection of the magnetic fiel for a particle is governe by the right-han rule regarless of what irection the velocity an force are pointing. Q24.31: The moon oes not have a molten iron core like the earth, but the moon oes have a small magnetic fiel. What might be the source of this fiel? Q24.31. Reason: The moon contains rocks that were magnetize, an in fact some parts of the moon s surface are magnetize. One theory suggests that the moon at one time ha a molten core, an when the outer surface of the Moon coole, the magnetic fiel of the molten core aligne the atoms of the metallic elements on the moon s surface.
Since the surface of the moon has not been isturbe or heate, the ferromagnetic material continue to keep the alignment of the atoms, an therefore staye magnetize. Assess: One theory why the earth is magnetize suggests that the earth s molten core cycles the metal core through convection an moves the charge particles with it. This cyclic motion, like any circular electric current, creates a ipole magnetic fiel. This earth s magnetic fiel over time magnetizes any ferromagnetic material on earth (i.e., by aligning the ipole moments in the atoms). P24.3: The magnetic fiel at the center of a 1.0-cm-iameter loop is 2.5 mt. a) What is the current in the loop? b) A long, straight wire carries the same current you foun in part a. At what istance from the wire is the magnetic fiel 2.5 mt? P24.3. Prepare: The equations for the magnetic fiel at the center of a loop an a wire are = I R an = I/(2 πr) wire 0 loop center 0 /(2 ) Solve: (a) The raius of the loop is 0.5 cm an the magnetic fiel is 2.5 mt so the current is: loop center 0I 2R = I = 2R 0 loop center 2 3 2(0.5 10 m)(2.5 10 T) = = 19.89 A 4 π (10 T m/a) which will be reporte as 20 A. (b) or a long, straight wire that carries a current I, the magnetic fiel strength is 0I 3 4 π(10 T m/a)(19.89 A) 3 wire = 2.50 10 T = = 1.6 10 m 2π 2π Assess: The result for obtaine is consistent with the equations for wire an loop center. P24.6: Although the evience is weak, there has been concern in recent years over possible health effects from the magnetic fiels generate by transmission lines. A typical high-voltage transmission is 20 m off the groun an carries a current of 200 A. Estimate the magnetic fiel strength on the groun unerneath such a line. What percentage of the earth s magnetic fiel oes this represent?? P24.6. Prepare: Use the equation for the magnetic fiel for a long straight wire. Solve: (a) The fiel of a transmission line is aroun 0 I (2 10 T m/a)(200 A) 6 = = = 2.0 10 T = 2.0 T 2π 20 m 5 (b) The earth s fiel is earth = 5 10 T = 50 T, so wire / earth = 2.0 T/(50 T) = 0.04 = 4.0%. Assess: The fiel prouce by a transmission line on the groun is much smaller than the earth s magnetic fiel. P24.7: Some consumer groups urge pregnant women not to use electric blankets, in case there is a health risk from the magnetic fiels from the approximately 1 A current in the heater wires.
a) Estimate, stating any assumptions you make, the magnetic strength a fetus might experience. What percentage of the earth s magnetic fiel is this? b) It is becoming stanar practice to make electric blankets with minimal external magnetic fiel. Each wire is paire with another wire that carries current in the opposite irection. How oes this reuce the external magnetic fiel? P24.7. Prepare: We nee the formula for the magnetic fiel of a straight wire. Solve: (a) Let s estimate that the baby is 10 cm ( 4 inches ) from a 1 A current. Here the magnetic fiel is 0 I (2 10 T m/a)(1 A) 6 = = = 2.0 10 T = 2.0 T 2π 0.1 m (b) The two magnetic fiels prouce will be in opposite irections so they will cancel each other out. Assess: The magnetic fiel prouce by an electric blanket on a baby, even if the wires are not oppositely paire, is much smaller than the earth s magnetic fiel. P24.15: What is the magnetic fiel at the center of the loop in the figure below? P24.15. Prepare: This problem can be envisione as a superposition of a straight current an a current loop. The magnetic fiel at the center of a loop is the combination of both the magnetic fiel of the loop plus the magnetic fiel of the straight part of the wire. = = π loop 0I/(2 R) an wire 0I/(2 r) = + total loop wire Note that for the magnetic fiel of the loop, R is the raius of the loop, whereas, for the magnetic fiel of the wire, r is the istance from the straight section of the wire. In this situation, r is the istance from the point where the wire begins to ben into a loop. y using the right-han rule, we see the irections of both magnetic fiels point in the same irection, own. Therefore, the two magnetic fiels a together. Solve: The magnetic fiel at the center of the loop is: 0I 0I 0I 1 1 total = + = 2R 2πr 2 + R πr (4π 10 T m/a)(5.0 A) 1 1 = + 2 0.01m π (0.01m) = Assess: 4 4.1 10 T Since magnetic fiels are vector fiels, we can use superposition. P24.21: An electron moves with a spee of 1.0 10 7 m/s in the irections shown in the figure below. A 0.50 T magnetic fiel points in the positive x-irection. or each, what is the magnetic force on the electron? Give your answers as a magnitue an a irection.
P24.21. Prepare: A magnetic fiel exerts a magnetic force on a moving charge. The = q vsin α. magnitue of the magnetic force is calculate using the equation The righthan rule for forces is use to fin the irection on a positive charge. Solve: (a) The magnitue of the magnetic force on the charge is = q v α = = 19 7 13 sin (1.60 10 C)(1.0 10 m/s)(0.50 T)(sin90 ) 8.0 10 N Using the right-han rule for forces, you fin that the irection of the force is in the +zirection or up. So, on the electron the force is in the z irection, or own. (b) The magnitue of the magnetic force on the charge is q v α 19 7 13 = sin = (1.60 10 C)(1.0 10 m/s)(0.50 T)(sin90 ) = 8.0 10 N Using the right-han rule for forces, you fin that the irection of the force is in the +yzirection, or 45 up an into the page. So, on the electron the force is 45 own an out of the page. Assess: The velocity, magnetic fiel, an magnetic force are always perpenicular to each other. Using the right-han rule for forces, you can etermine the irection of the magnetic force without calculating all the iniviual components of velocity (v) an magnetic fiel (). P24.23: The aurora is cause when electrons an protons, moving in the earth s magnetic fiel (5.0 10-5 T), collie with molecules of the atmosphere an cause them to glow. What is the raius of the circular orbit for a. An electron with spee 1.0 10 6 m/s? b. A proton with spee 5.0 10 4 m/s? P24.23. Prepare: Charge particles moving perpenicular to a uniform magnetic 2 ( a v /) r fiel unergo uniform circular motion at a constant spee (v): r = vm/( q). or Solve: The calculations for the raius of the circular orbit is 6 31 vm (1.0 10 m/s)(9.11 10 kg) electron 19 5 r 4 27 vm (5.0 10 m/s)(1.67 10 kg) proton 19 5 r = = = = = q (1.60 10 C)(5 10 T) = = = 10.4 m q (1.60 10 C)(5 10 T) 1 1.14 10 m 11.4 cm = mv r = qv 2 net / Assess: In a magnetron, the charge particles have an angular velocity relate to the linear velocity by v= ωr = 2 π fr, where f is the frequency an r is the raius of the circular motion. These charge particles create light that can be seen at the North Pole an South Pole, an the patterns are from these orbits.
P24.31: What is the net force (magnitue an irection) on each wire in the figure below? P24.31. Prepare: Two parallel wires carrying currents in opposite irections exert repulsive magnetic forces on each other. Two parallel wires carrying currents in the same irection exert attractive magnetic forces on each other. Solve: The magnitues of the various forces between the parallel wires are 0LI1I2 (2 10 T m/a)(0.50 m)(10 A)(10 A) 4 2 on 1 = = = 5.0 10 N = 2 on 3 = 3 on 2 = 1 on 2 π 2 0.02 m LI I 2π 0.04 m 0 1 3 (2 10 T m/a)(0.50 m)(10 A)(10 A) 4 3 on 1 = = = 2.5 10 N = 1 on 3 Now we can fin the net force each wire exerts on the other as follows: on 1 = 2 on 1 + 3 on 1 = + = 4 4 on 2 = 1 on 2 + 3 on 2 = ( 5.0 10 N, up) + ( + 5.0 10 N, own) = 0 N = + = + = on 3 1 on 3 2 on 3 4 4 4 (5.0 10 N, up) ( 2.5 10 N, own) (2.5 10 N, up) 4 4 (2.5 10 N, up) ( 5.0 10 N, own) (2.5 10 4 N, own) Assess: The forces on a wire are cumulative, so the contribution ue to the both wires is ae. P24.47: An electron travels with spee 1.0 10 7 m/s between the two parallel charges shown in the figure. The plates are separate by 1.0 cm an are charge by a 200 V battery. What magnetic fiel strength an irection will allow the electron to pass between the planes without being eflecte? Hint: E = an E = ΔV/ P24.47. Prepare: An electric an magnetic fiel exerts two inepenent forces on the moving electron. The magnitue of the electric force is E = qe irection of the electric fiel. The magnitue of the magnetic force is irection is given by the right-han rule for forces. Solve: The electric fiel is V 200 V E = =, own = (20,000 V/m, own) 0.01m 15 E = qe = ee = (3.2 10 N, up). an the irection is the = qv an the The force this fiel exerts on the electron is The electron will pass through without eflection if the magnetic fiel also exerts a force on the electron such that net = E + = 0N or = : E i.e., the electric an magnetic forces cancel each other. So,
the magnetic force is 15 = (3.2 10 N, own). or a negative charge with v to the right to have own requires, from the right-han rule for forces, that the irection of the magnetic fiel point into the page. The magnitue of the magnetic force on a moving charge is = qv, so the neee fiel strength is 15 3.2 10 N 3 = = = = eν Thus, the require magnetic fiel is = 19 7 (1.60 10 C)(1.0 10 m/s) (2.0 mt, into page). 2.0 10 T 2.0 mt P24.60: A mass spectrometer is esigne to separate atoms of carbon to etermine the fraction of ifferent isotopes. There are three main isotopes of carbon, with the following atomic masses: The atoms of carbon are singly ionize an enter a mass spectrometer with magnetic fiel strength = 0.200 T at a spee of 1.50 10 5 m/s. The ions move along a semicircular path an exit through an exit slit. How far from the entrance will the beams of the ifferent isotope ions en up? P24.60. Prepare: Picture the setup of a mass spectrometer. A particle enters a magnetic fiel an is move in a semicircular path. The raius of the circle is given r = mv/ q, by but the net istance travele by the ion is the iameter of the circle. q = 19 1.60 10 C. Solve: The isotopes are each singly ionize so the charge on each is or the mass, we use the atomic masses liste in the table given in the problem text. The total istance travele is therefore: 26 5 2(1.99 10 kg)(1.5 10 m/s) 12 r12 19 = 2 = = 18.7 cm (1.60 10 C)0.200 T 26 5 2(2.16 10 kg)(1.5 10 m/s) 13 r13 19 = 2 = = 20.2 cm (1.60 10 C)0.200 T 26 5 2(2.33 10 kg)(1.5 10 m/s) 14 r14 19 = 2 = = 21.8 cm (1.60 10 C)0.200 T Assess: The more massive the particle, the further it travels from its entrance into the magnetic fiel. This is how the mass spectrometer separates particles of ifferent masses.