Solutions to Homework Questions 5 Chapt19, Problem-2: (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat part (a), assuming the moving particle is an electron. (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are: (a ) in plane of page and to left (b ) into the page (c ) out of the page (d ) in plane of page and toward the top (e ) into the page (f ) out of the page (b) For a negatively charged particle, the direction of the force is exactly opposite what the right hand rule predicts for positive charges. Thus, the answers for part (b) are reversed from those given in part (a). Chapt19, Problem-5: At the Equator near Earth s surface, the magnetic field is approximately 50.0 µt northward and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron with an instantaneous velocity of 6.00x10 6 m/s directed to the east in this environment. Gravitational force: F g = mg = ( 9.11! 10 "31 kg) ( 9.80 m s 2 )= 8.93! 10 "30 N downward Electric force: F e = qe=!1.60" 10!19 C ( )= 1.60! 10 "17 N upward ( )!100 N C Magnetic force: F m = qvbsin! = "1.60 #10 "19 C ( ) 6.00 # 10 6 ms ( ) 50.0 #10 "6 T ( )sin 90.0 ( ) = 4.80!10 "17 N in direction opposite right hand rule prediction F m = 4.80! 10"17 N downward Chapt19, Problem-6: A proton travels with a speed of 3.0x10 6 m/s at an angle of 37 with the direction of a magnetic field of 0.30 T in the +y direction. What are (a) the magnitude of the magnetic force on the proton and (b) the proton s acceleration? ( a ) F = qvbsin! ( b ) ( )( 3.0!10 6 ms) 0.30 T = 1.60!10 "19 C a = F m = 8.7! 10"14 N 1.67!10-27 kg = 5.2! 1013 ms 2 ( )sin( 37 )= 8.7! 10"14 N 1
Chapt19, Problem-10: Sodium ions (Na + ) move at 0.851 m/s through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of 0.254 T and makes an angle of 51.0 with the motion of the sodium ions. The arm contains 100 cm 3 of blood with 3.00x10 20 Na + ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm? The force on a single ion is F 1 = qvbsin! ( )( 0.851 m s) 0.254 T = 1.60 "10 #19 C The total number of ions present is " N = 3.00! 10 20 ions% $ # cm 3 ' ( 100 cm 3 )= 3.00! 10 22 & ( )sin( 51.0 )= 2.69"10 #20 N Thus, assuming all ions move in the same direction through the field, the total force is F = N!F 1 = ( 3.00" 10 22 )( 2.69" 10 #20 N)= 806 N Chapt19, Problem-11: A current I = 15 A is directed along the positive x axis and perpendicularly to a magnetic field. The conductor experiences a magnetic force per unit length of 0.12 N/m in the negative y direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. From F = BIL sin!, the magnetic field is B = FL Isin! = 0.12 N m ( 15 A)sin 90 = 8.0!10 "3 T The direction of B must be the + z direction to have F in the y direction when I is in the +x direction. Chapt19, Problem-15: A wire carries a current of 10.0 A in a direction that makes an angle of 30.0 with the direction of a magnetic field of strength 0.300 T. Find the magnetic force on a 5.00-m length of the wire F = BILsin! = ( 0.300 T) ( 10.0 A) ( 5.00 m)sin( 30.0 )= 7.50 N Chapt19, Problem-23: An 8-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm (Fig. P19.23). The coil lies in the plane of the page and has a 6.00-A current flowing clockwise around it. If the coil is in a uniform magnetic field of 2.00x10 4 T, directed toward the left of the page, what is the magnitude of the torque on the coil? (Hint: The area of an ellipse is A =!ab, where a and b are the semi-major and semi-minor axes of the ellipse.) The area is A =!ab =!( 0.200 m )( 0.150 m )= 0.0942 m 2 (Note we have to use half the lengths of the major & minor axes). Since the field is parallel to the plane of the loop,! = 90.0 and the magnitude of the torque is! = NBIAsin" = 8( 2.00!10 "4 T) ( 6.00 A) ( 0.0942 m 2 )sin 90.0 = 9.05! 10"4 N# m The torque is directed to make the left-hand side of the loop move toward you and the right-hand side move away. 2
Chapt19, Problem-29: Figure P19.29a is a diagram of a device called a velocity selector, in which particles of a specific velocity pass through undeflected but those with greater or lesser velocities are deflected either upward or downward. An electric field is directed perpendicularly to a magnetic field. This produces on the charged particle an electric force and a magnetic force that can be equal in magnitude and opposite in direction (Fig. P19.29b), and hence cancel. Show that particles with a speed of v = E/B will pass through undeflected. For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives F m = F e, or qvb = qe which reduces to v = EB Chapt19, Problem-37: At what distance from a long, straight wire carrying a current of 5.0 A is the magnetic field due to the wire equal to the strength of Earth s field, approximately 5.0x10 5 T? From B = µ 0 I 2!r, the required distance is r = µ 0 I 2!B ( ) 5.0 A ( ) 4! "10-7 T# ma = 2! 5.0 "10 $5 T ( ) = 2.0" 10 $2 m = 2.0 cm Chapt19, Problem-45: A wire with a weight per unit length of 0.080 N/m is suspended directly above a second wire. The top wire carries a current of 30.0 A, and the bottom wire carries a current of 60.0 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. In order for the system to be in equilibrium, the repulsive magnetic force per unit length on the top wire must equal the weight per unit length of this wire. Thus, F L = µ 0 I 1 I 2 2!d d = = 0.080 N m, and the distance between the wires will be µ 0 I 1 I 2 2! 0.080 N m ( ) ( ) 60.0 A 2! ( 0.080 N m) 4! "10 #7 T$m A = ( )( 30.0 A) = 4.5! 10 "3 m = 4.5 mm 3
Chapt19, Problem-49: A single-turn square loop of wire, 2.00 cm on a side, carries a counterclockwise current of 0.200 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns per centimeter and carries a counterclockwise current of 15.0 A. Find the force on each side of the loop and the torque acting on it. The magnetic field inside the solenoid is + B = µ 0 ni 1 = 4! " 10 #7 % ( T$ ma) 30 turns ( % 100 cm (. -' * ' *,& cm ) & 1 m ) 0 ( 15.0 A)= 5.65 "10 #2 T / Therefore, the magnitude of the magnetic force on any one of the sides of the square loop is F = BI 2 Lsin 90.0 = ( 5.65! 10 "2 T) ( 0.200 A) ( 2.00! 10 "2 m )= 2.26!10 "4 N The forces acting on the sides of the loop lie in the plane of the loop, are perpendicular to the sides, and are directed away from the interior of the loop. Thus, they tend to stretch the loop but do not tend to rotate it. The torque acting on the loop is! = 0. Chapt19, Conceptual-3: A proton moving horizontally enters a region where a uniform magnetic field is directed perpendicular to the proton s velocity, as shown in Figure Q19.3. Describe the subsequent motion of the proton. How would an electron behave under the same circumstances? The proton moves in a circular path upward on the page. After completing half the circle, it exits the field and moves in a straight line back in the direct whence it came. An electron behaves simialarly, but the direction of traversal of the circle is downward, and the radius of the circular path is smaller. Chapt19, Conceptual-4: A current-carrying conductor experiences no magnetic force when placed in a certain manner in a uniform magnetic field. Explain If the current is in a direction parallel (or at 180 o ) to the magnetic field, then there is no force. Chapt19, Conceptual-8: A magnet attracts a piece of iron. The iron can then attract another piece of iron. On the basis of domain alignment, explain what happens in each piece of iron. The magnet causes domain alignment in the iron such that the iron becomes magnetic and is attracted to the original magnet. Now that the iron is magnetic, it can produce an identical effect in another piece of iron. 4
Chapt19, Conceptual-11: Suppose you move along a wire at the same speed as the drift speed of the electrons in the wire. Do you now measure a magnetic field of zero? If you are moving along with the electrons, you will measure zero current for the electrons, so the electrons would not produce a magnetic field according to your observations. However, the fixed position charges in the metal are now moving backwards relative to you and creating a current equivalent to the forward motion of the electrons when you were stationary. Thus, you will measure the same magnetic field as when you were stationary, but it will be due to the positive charges moving in your reference frame. Chapt19, Conceptual-14: It is found that charged particles from outerspace, called cosmic rays strike the Earrth more frequently at the poles than at the Equator. Why? Near the poles the magnetic field of the Earth points almost straight downward (or straight upward), in the direction (or opposite to the direction) the charges are moving. As a result, there is little or no magnetic force exerted on the charged particles at the pole to deflect them away from the Earth. 5