THEMAGNETIC FIELD
32.1. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure Ex32.1. Solve: Because the north poles of the magnets point counterclockwise, the magnetic force is counterclockwise. When you point fingers of your right hand counterclockwise, the thumb points up. Thus, the current in the wire is out of the page.
32.2. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure Ex32.2. The magnetic field is into the page on the left of the wire and it is out of the page on the right of the wire. Solve: Grab the wire with your right hand in such a way that your fingers point out of the page to the right of the wire. Since the thumb now points down, the current in the wire is down.
32.3. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure Ex32.3. Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right hand rule. Grab the current carrying wire so that your thumb points in the direction of the current. Because your fingers at point 1 point into the page, gl = (2.0 mT, into the page). At point 2, the magnetic field due to the bottom wire is into the page. The right-hand rule tells us that the magnetic field from the top wire is also into the page. At point 2, = (4.0 mT, into the page).
&
32.4. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure 32.4. Solve: The current in the wire is directed to the right. B2 = 20 mT + 20 mT = 40 mT because the two overlapping wires are carrying current in the same direction and each wire produces a magnetic field having the same direction at point 2. B, = 20 mT - 20 mT = 0 mT, because the two overlapping wires cany currents in opposite directions and each wire produces a field having opposite directions at point 3. The currents at 4 are also in opposite directions, but the point is to the right of one wire and to the left of the other. From the right-hand rule, the field of both currents is out of the page. Thus B,= 20 mT + 20 mT = 40 mT.
32.5. Model: The magnetic field is that of a moving charged particle. Visualize: V
32-1
32-2
Chapter 32
The first point is on the x-axis, with 6, = 90". "he second point is on the y-axis, with 0, = 180", and the third point is on the -y-axis with 0,= 0". Solve: (a) Using Equation 32.1, the Biot-Savart law, the magnetic field strength is
paqvsine B, = 4n r2
(lo-' T m / A ) ( 1 . 6 0 ~ 1 0 -C~)~( 1 . 0 ~ 1 0 ' m/s)sin90" ~ (1.0 x lo-?
m)'
= 1.60 x 10-1~ T
To use the right-hand rule for finding the direction of 13, point your thumb in the direction of $ . The magnetic field vector 13 is perpendicular to the plane of r' and ? and points in the same direction that your fingers point. In the present case, the fmgers point along the i direction. Thus, El = 1.60 x l O - I 5 i 2'. (b) B2= 0 T because sin 0, = sin 180" = 0. (c)B, = 0 T because sin 6, = sin 0" = 0. 32.6. Model: The magnetic field is that of a moving charged particle. Visualize: Y
-1 -1
I
1
The first point is on the y-axis, the second point is on the x-axis, and the third point is in the xy plane. Solve: (a) Using Equation 32.1, the Biot-Savart law, the magnetic field strength is
p o gvsine B, =--= 4z
(lo" T m / A)(1.60 x
C)(l.O x lo7 m / s)sinO"
=OT
( 1 . 0 ~IO-' m)'
r2
i2= 1.60 x l O - " i -( -i)= i .
(b) At point 2, 0, = 90" and the magnetic field is
negative charge, to find that the direction of Bz is
T. We applied the right-hand rule to the
(c)The field has the same direction as in part (b). The magnitude is B3 = and SO
T m / A)(1.60x (1.0 x IO-' m)'
C)(l.O x lo7 m / s)sin45"
+ (1.0 x IO-'
m)'
= 5.66 x
T
= 5.66 x 1O-I6i T.
32.7. Model: The magnetic field is that of a moving charged particle. Visualize: Please refer to Figure Ex32.7. Solve: Using the Biot-Savart law, qvsine B = -pa -4z r2
T m / A)(1.60 ~ 1 O - C)(2.0 l~ x lo7 m / s)sin135" (1.0 x IO-' m)'
The right-hand rule applied to the proton points
+ (1.0 x 10-2 m)*
out of the page. Thus,
= 1.13x lo-'' T
5= 1.13 x l O - " i
T.
32.8. Model: The magnetic field is that of a moving charged particle. Visualize: Please refer to Figure Ex32.8. Solve: The Biot-Savart law is B
p vsine (10" T m / A ) ( 1 . 6 0 ~ 1 0 - 'C)(2.0x1O7 ~ m/s)sin135" A L = = 2.83 x 4n r2 (2.0 x IO-? m)' + (2.0 x lo-' m)'
=
The right-hand rule for the positive charge indicates the field points out of the page. Thus,
T
= 2.83 x 10-l6iT.
The Magnetic Field
32-3
32.9. Model: The magnetic field is that of a moving proton. Visualize:
Z
The magnetic field lies in the xy-plane. Solve: Using the right-hand rule, the charge is moving along the +z-direction. That is, V = v i . Using the Biot-Savart law,
po qvsin8 B=-4~
1.0 x 10-l3T =
(lo-’ T m / A)(1.60 x
C)vsin90°
(1.0 x 10-~m)’
r2
-v=6.25x1O6
m/s
32.10. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 32.3, the magnetic field strength of a long straight wire carrying current I at a distance d from the wire is B,
= Po I 2a d
--
The current needed to produce the earth’s magnetic field is calculated as follows:
Beanhsurfsce = 5 x 10” T = ( 2 x
T m / A)Z 0.01 m
I = 2.5 A
Likewise, the currents needed for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 250 A, 5000-50,000 A, and 500,000 A.
32.11. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 32.3, the magnetic field strength of a long straight wire carrying current I at a distance d from the wire is
B=--Po
*
2z d The distance d at which the magnetic field is equivalent to Earth’s magnetic field is calculated as follows:
T m / A)=
*
d = 4.0 cm d Likewise, the corresponding distances for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 0.4 mm, 20 p to 2 p,and 0.20 pm.
BeanhrMacr = 5 x lo-’ T = (2 x
32.12. Model: The magnetic field is that of a current loop. Solve: From Example 32.5, the on-axis magnetic field of a current loop is
Also, B , , , (: = 0 m) = p01/2R. The ratio of these two fields is z3/’ = 0.354.
32.13. Model: The magnetic field is that of a current loop. Solve: (a) From Equation 32.7, the magnetic field strength at the center of a loop is 2 RBI,, crnler - 2(0.5 x lo-’ m)(2.5 x Pol B,oo,,”~r= -3 I = 4z(10-’ T m / A) 2R PO
-
.
T) = 2 0 A
32-4
Chapter 32
(b) For a long, straight wire that carries a current I , the magnetic field strength is
B,,
4n(10-' T m / A)(20 A)
P I 3 2 . 5 0 ~ 1 0 -T~ = =0 2nd
2nd
= 3 d = 1 . 6 0 ~ 1 0 - ~m
32.14. Model: Assume the wires are infinitely long. Visualize: Please refer to Figure Ex31.14. Solve: The magnetic field strength at point 1 is
Pol
3
Bat, =-
3
E,,
--
27z 2 c m
(4+2)cm
= ( 2 x lo-' T m / A)(10 A)
1 2 x lo-' m 6 x lo-' m
= (6.67 x lo9 T, out of page)
At points 2 and 3,
ia2= i,, =
(z,
= (2.0
into page
(z,
x 10-4 T, into page)
= (6.67 x
into page
T, out of page)
32.15. Model: Assume the wires are infinitely long. @ 10A
Visualize:
@ 10A The field vectors are tangent to circles around the currents. The net magnetic field is the vectorial sum of the fields imp and Bbamm. Points 1 and 3 are at a distance d = f i cm from both wires and point 3 is at a distance d = 1 cm. Solve: The magnetic field at points 1, 2 , and 3 are
-
-
B, = Bmp+
&,
= -(cos45"i^ POI
2nd
- sin45"j) + -(cos45"1+ POI 27ld
sin45"j)
& x IO-' m B- - P0 I1 : '-2nd
+
-POI I
2nd
c
=2
( 2 x 10" T m / A)(10 A) 1x IO-' m
-
i = 4.0x 10";
T
B3 - = AP (Ic 0 s 4 5 ~ +1 s i n 4 5*" j ) P+ ~I( c o s 4 5 "-sin45'j) ~ = 2.OX lo4; T 2 d 2nd
The Magnetic Field
32.16. Solve:
32-5
(a) The magnetic dipole moment of the superconducting ring is
,~=((ltR~)I=n(l.OxlO-~ m)2(100A)=3.14x104 A m 2 (b) From Example 32.5, the on-axis magnetic field of the superconducting ring is
=yo
B,
1R2
2 ( z +~p ) 3 ’ 1
M~
- 2Z(lO-’ T m / A)(100 A)(1.0x10-3 m)’ = 5.02 x [(O.OS mI2 + (0.001 m)‘]yL
T
32.17. Model: Assume that the 10 cm distance is much larger than the size of the small bar magnet. Solve: (a) From Equation 32.9, the on-axis field of a magnetic dipole is B=P o-a 2P 4n z3
4~ Bz3
(5.0 X IO4 T)(0.10 m)3
Po 2
2(10-’ T m A)
p =--=
= 0.025Am’
(b) As the bar magnet is rotated end-over-end by 180°, the magnitude of the magnetic field remains the same 5.0 pT, but the vector points in the opposite direction.
32.18. Model: The radius of the earth is much larger than the size of the current loop. Solve: (a) From Equation 32.9, the magnetic field strength at the surface of the earth at the earth’s north pole is B=bk?= 2K
(2 x 10” T m / A ) ( ~ . ox IO?’ A m 2 )
z3
(6.38 x lo6 m)’
= 6.16 x 10” T
T given in Table 32.1. This value is close to the value of 5 x (b) The current required to produce a dipole moment like that on the earth is
p = A I = ( ~ ~ ) I = $ 8 x 1 O 2 ‘ A r n 2 = l r ( 6 . 3106m)*Z 8x -1=6.26x Assess:
108A
This is an extremely large current to run through a wire around the equator.
32.19. Model: The distance of 50 cm is much larger than the radius of the loop. Solve: The on-axis field of a magnetic dipole (due to a loop of current) is given by Equation 32.9:
- P 2P - P 2AI 4n z 3 4 n 2’
(3.5 x
A = 4n B~~
(z)? =
T)(SOx IO-’ m)’ = 8.75 x T m / A)2(25 A)
m’
Assess: The above area yields a radius of 5.28 mm for the loop. This is certainly much smaller than 50 cm, as assumed.
32.20. Visualize: Please refer to Figure Ex32.20. Solve: The line integral of B between points i and f is
Because B is perpendicular to the integration path from i to f, the dot product is zero at all points and the line integral is zero.
32.21. Visualize: Please refer to Figure Ex32.21. Solve: Because B is in the same direction as the integration path ? from i to f, the dot product of B and 6 is simply Bds. Hence the line integral
jg-a? = j B d s = B!ds I
I
= (,/(OS0
m)’ +(OS0 m)’ =(0.10 T)fi(0.50 m) = 0.0707 T m
32-6
Chapter 32
32.22. Model: The magnetic field is that of a current flowing into the plane of the paper. The current carrying wire is very long. Visualize: Please refer to Figure Ex32.22. Solve: Divide the line integral into three parts: j:i.dS=
j
B.ds+
B.dS+
left h e
I
5.d;
ngbt h e
sermcucle
The magnetic field of the current-carrying wire is tangent to clockwise circles around the wire. is everywhere perpendicular to the left line and to the right line, thus the first and third parts of the line integral are zero. Along the semicircle, is tangent to the path and has the same magnitude B = , ~ d / 2 z dat every point. Thus
P I(& p = (4zx10-7 fi.& = O + BL+O = A =2 ) 2ad 2
m/A)(2.0 A) = 1-26x l O d T 2
where L = z d is the length of the semicircle, which is half the circumference of a circle of radius d.
32.23. Model: The magnetic field is that of three currents. Visualize: Please refer to Figure Ex32.23. Solve: Ampere's law gives the line integral of the magnetic field around the closed path:
$i.ds' = poIthrough = 3.77 x lod T m = po(I,- I2 +I,) = (
4 x~
T m / A)(6 A - 4 A + 13)
( I 3 + 2 A) = 3'77x10" T m *13=1.0A ~ z x X O - ~T m I A Assess: The right-hand rule was used above to assign positive signs to I, and I3 and a negative sign to 12. 3
32.24. Model: Assume that the solenoid is an ideal solenoid. Solve: We can use Equation 32.16 to find the current that will generate a 3.0 mT field inside the solenoid:
Using L = 0.15 m and N = 0.15 m/O.OOl m = 150,
I=
(3.0 x IO" T)(0.15 m) (4z)(10-' T m / A)(150)
= 2.39 A
Assess: This is a reasonable current to pass through a good conducting wire of diameter 1 nun.
32.25. Model: Assume that the solenoid is an ideal solenoid. Solve: We can use Equation 32.16 to find the current that will generate a magnetic field of 1.5 T inside the solenoid
Using L = 1.8 m and N = 1.8 m/0.002 = 900,
Assess: Large currents can be passed through conducting wires. The current density through the superconducting wire is (2390 A)/~(0.001m)' = 7.6 x 10' A / m 2 . This value is reasonable for a superconducting wire.
32.26. Model: A magnetic field exerts a force on a moving charge. Visualize: Please refer to Figure Ex32.26. Solve: (a) The force on a charge moving in a magnetic field is
Po,, = qv x E = (qvBsincr, direction of right-hand rule) The direction of the force on a negative- charge is opposite the direction determined by the right-hand rule. A positive charge moving to the right with B into the page gives a force that is up. (b) A negative charge moving up with B out of the page gives a force to the left.
-
The Magnetic Field
32-7
32.27. Model: A magnetic field exerts a force on a moving charge. Visualize: Please refer to Figure Ex32.27. Solve: (a) The force on a charge moving in a magnetic field is
Enq= qc x B =(qvBsina, direction of right-hand rule) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. A positive charge moving to the right with B down gives a force into the page. (b) A negative charge moving parallel to the field has no force and no deflection.
32.28. Model: A magnetic field exerts a force on a moving charge. Visualize: Please refer to Figure Ex32.28. Solve: (a) The force on a charge moving in a magnetic field is
-
-
Fonq= qv x B =(qvBsina, direction of right-hand rule) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. The magnetic field must be in a plane perpendicular to both the v’ and F vectors. Using the right-hand rule for a psirive charge moving to the right, the b field must be out of *e page. (b) The force F on the negative charge is into the page. Since the velocity is to the right, the magnetic field must be up.
32.29. Model: A magnetic field exerts a force on a moving charge. Visualize: Please refer to Figure Ex32.29. Solve: (a) The force on a charge moving in a magnetic field is
E,,= qV x 1?.=(qvBsina, direction of right-hand rule) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. Since the force F. is our of h e page and the velocity of the negative charge is to the left and up in the plane of the paper, the magnetic field must be in the plane of the page, 45” clockwise from straight up. (b)The magnetic field on the positive charge is in the plane of the page, 45” counterclockwise from straight down.
3230. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure Ex32.30. Solve: (a) The force on the charge is
= (1.60 x A
C)(0.50 T)
( 1 . 0 ~ 1 0m~/ s )
Jz
(i x ; + k x ~ ) = + 5 . 6 6 x l O - ” f N
1
(b) Because the cross product i x i in the equation for the force is zero,
En,= 8 N .
3231. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure Ex32.31. Solve: (a) The force is
c,,
= qV x b = ( - 1 . 6 0 ~lo-’’ e)(-l.0x 1073 m / s ) x (0.50; T) = - 8 . 0 ~ lO-”iN
(b) The force is
,F,
=(-1.60~10-’~ C)(l.Ox IO’ m/s)(-cos45of+sin45ok^)x(0.50~T)=5.66xlO~13(-~-~) N
3232. Solve: The cyclotron frequency of a charged particle in a magnetic field is
= qBI2nm. An ion has charge 4 = -e. Its mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example. the mass of N I is
m=m,+m,-m,,,,=2(14.0031
u)(1.661 x lO-“kg/u)-9.11
x
kg=4.65174x 10-’6kg
32-8
Chapter 32
Note that we're given the atomic masses very accurately, and we need to retain that accuracy to tell the difference between N i and CO'. Calculating each mass and frequency: Ion
f (MHz)
Mass (kg) 4.65174 x 5.31341 x 4.64986 x
N: 0: CO'
1.6423 1.4378 1.6429
Assess: The difference between N; and CO' is not large but is easily detectable.
32.33. Model: A charged particle moving perpendicular to a uniform magnetic field moves in a circle. Solve: The frequency of revolution of a charge moving at right angles to the magnetic field is 21Rnf,, f q c =4B G j B = - =
21~(9.11x
kg)(45 x lo6 Hz)
1.60 x 10-l9 c
4
32.34. Solve: (a) From Equation 32.19, the cyclotron radius is mv (9.11 x 10"' kg)(l.O x lo6 m / s) re,= = -= qB (1.60 x
C)(5.0x lo-' T)
= 1 . 6 1 ~ 1 0 -T~
= 0.114m = 11.4cm
(b) For the proton, (1.67 x rprotoln
kg)(5.0 x lo4 m / s)
= ( 1 . 6 0 ~ 1 0 - 'C)(5.0~10-~ ~ T)
= 10.4 m
32.35. Model: Assume the magnetic field is uniform over the Hall probe. m. Visualize: Please refer to Figure 32.41(a). The thickness is t = 4.0 x Solve: The Hall voltage is given by Equation 32.24: (15 A)(I.O T) = 2.9 x 10'' m-3 (1.0 x low3m)(1.60 x Cl(3.2 xlOd V) m-3 (Table 28.1). The value Assess: The conduction electron density in metals is of the order of = 5 x IB IB AVH=-jn=-= tne teAV,
obtained for the charge carrier density is reasonable.
32.36. Model: Assume that the field is uniform. The wire will float in the magnetic field if the magnetic force on the wire points upward and has a magnitude mg, allowing it to balance the downward gravitational force.
Visualize: Please refer to Figure Ex32.36. Solve: We can use the right-hand rule to determine which current direction experiences an upward force. The current being from right to left, the force will be up if the magnetic field points out of the page. The forces will balance when mg (2.0 x kg)(9.8 m / s') F=ILB=mg* B = - = = 0.131 T IL (1.5 A)(0.10 m) Thus
= (0.131 T, out of page).
32.37. Model: Assume that the magnetic field is uniform over the 10 cm length of the wire. Force on top and bottom pieces will cancel. Visualize: Please refer to Figure 32.37. The figure shows a 10-cm-segment of a circuit in a region where the magnetic field is directed into the page. Solve: The current through the 10-cm-segment is
and is flowing down. The force on this wire, given by the right-hand rule, is to the right and perpendicular to the current and the magnetic field. The magnitude of the force is F = ILB = (5 A)(0.10 m)(50 mT) = 0.025 N Thus
F = (0.025 N, right).
The Magnetic Field
32-9
3238. Model: Two parallel wires carrying currents in opposite directions exert repulsive magnetic forces on each other. Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure Ex32.38. Solve: The magnitudes of the various forces between the parallel wires are &on1
T m / A)(0.50 m)(lO A)(10 A) =5.0x104N = & m 3 = & o n 2 = F ; , z 0.02 m
- POLZlZ2 = ( 2 x -2nd
( 2 x lo-' T m / A)(0.50 m)(lO A)(10 A)
u113 =o-
2nd 0.04 m Now we can find the net force each wire exerts on the other as follows:
F,, =
eonl +Eonl
= ( ~ . o x 1 0 4 j ) ~ + ( - 2 . 5 ~ 1 0 4N j ) = ~ s x I o ~ ~ N = ( ~ . ~up) x ~ o ~ N ,
En2=
En3= E
=2.5x104 N = q m 3
on3
+
= (-5.Ox1O4j) N+(+5.0x1045) N = 0 N
+ KOn3 = ( 2 . 5 104j) ~ N +(-5.Ox
104j) N = - 2 . 5 ~1043 N = (2.5 x lo4 N, down)
32.39. Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other.
Visualize: Please refer to Figure Ex32.39. The current in the circuit on the left is 1, and has a clockwise direction. The current in the circuit on the right is l2and has a counterclockwise direction. Solve: Since Z, = 9 V/2 C2 = 4.5 A , the force between the two wires is F = 5 . 4 ~ 1 0 - ~= N
21cd P O W 2
( 2 x 1 0 - ~T m / A)(O.IO ml(4.5 A)Z,
af2=3.0A*R=--
0.005 m
9v 3.0 A
- 3.0R
32.40. Model: The torque on a current loop is due to the magnetic field. Loop 1 Loop 2 visualize: Loop 1 -
Âś
.
++-I
side View
-
3if-
-
I
w
,
,
B ; ; -4I
--
,
No net force, no torque
Torque restores position
(a)
Torque causes flip (b)
Solve: (a) We can use the right-hand rule to find the force direction on the currents at the top, bottom, left, and right segments of loop 1 and loop 2 in Figure Ex32.40. We see that
-
-
-
-
and
-
F,,
are equal and opposite to each other.
- cefi
-
are equal and opposite. Thus, Fmp + Fbottom = 0 and + F,,, = 0. Since the top-bottom or Similarly qchand FnSht left-right forces act along the same line, they cause no torque. Thus, both the loops are in static equilibrium. (b) Now let us rotate each loop slightly and re-examine the forces. The two forces on loop 1 still give = 0, but now there is a torque that tends to rotate the left loop back to its upright position. This is a restoring torque, so this loop position is stable. But for loop 2 , the torque causes the loop to rotate even further. Any small angular displacement gets magnified into a large displacement until the loop gets flipped over. So, the position of loop 2 is unstable.
c,,
32.41. Solve: From Equation 32.28, the torque on the loop exerted by the magnetic field is
f = @ x B a I = @sin8
= IABsinO = (0.500 A)(0.05 m x 0.05 m)(1.2 T)sin30° = 7.5 x lo4 N m
32-10
32.42.
Chapter 32
Solve: From Equation 32.28, the torque on the bar magnet exerted by the magnetic field is
32.43.
Model: The torque on the current loop is due to the magnetic field produced by the current-carrying wire. Assume that the wire is very long. Visualize: Please refer to Figure Ex32.43. Solve: (a) From Equation 32.27, the magnitude of the torque on the current loop is z = @sin 8 , where p = I,w# and Bis the magnetic field produced by the current Iwi,in the wire. The magnetic field of the wire is tangent to a circle around the wire. At the position of the loop, 5 points up and is 8= 90" from the axis of the loop. Thus, 7 = (I,-A)-sinB= POIW
(0.20 A)rc(0.001 m)'(2 x
2nd
T m / A)(2.0 A)sin90째
2.0x1O-~m
=1.26xlO-" N m
Note that the magnetic field produced by the wire on the current loop is up so that the angle 8 between normal to the loop is 90". (b) The loop is in equilibrium when e= 0" or 180". That is, when the coil is rotated by 90".
2 and the
32.44. Model: Magnetic and electric fields exert forces on a moving charge. Solve: Consider a top view of the cathode ray tube (CRT), as seen from the ceiling. If an electric field is causing the deflection, the electric field must point to the left, as seen when viewing the screen. If a magnetic field is causing the deflection, the magnetic field must point up toward the ceiling. These conclusions both depend on the fact that the electron is negative. The figure below illustrates these two possibilities.
+@g
Electric field pointing to the left
Magnetic field pointing up
The first step is to turn the CRT around 180" to face the opposite wall. If the deflection is caused by an electric field, the deflection will reverse and appear as a deflection to the left. However, if the deflection is caused by a magnetic field or if the CRT is broken, the deflection will still be to the right. The figure below illustrates these two possibilities.
@$+e
The next step is to point the CRT at the ceiling. If there is a magnetic field, the electrons are now moving parallel to the field and won't be deflected at all. So a centered spot after the second rotation means that the cause is a magnetic field. If the spot is still to the right, the CRT is broken.
32.45. Model: An electric current produces a magnetic field. Solve: (a) The field of a household wire is I ( 2 x io-' T m / A)(IO A) B=--= = 1 . 0 ~ 1 0T~= ].OPT 271 d 2m (b) The earth's field is Be- = 5 x lo-' T = 50 pT, so B,,,/B,, = 1.0 pT/50 pT = 0.02 = 2.0%. (c) The field of a transmission line is (2 x lo-' T m / A)(200 A) B=--= I = 2.0 x lod T = 2.0 pT 271 d 20 m This is twice the field of part (a), which would probably not be significantly worse. (d) Let's estimate that a fetus is IO cm (= 4 inches) from a 1 A current. Here the field is 1 (2 x IO-' T m / A)(1 A) B=--= =2.0x1O6 T = 2.0pT 2rc d 0.1 m This is twice the field of part (a).
The Magnetic Field
32-1 1
32.46. Model: The magnetic field is the superposition of the magnetic fields of three wire segments. Visualize: Please refer to Figure P32.46 Solve: The magnetic field of the horizontal wire, with current Z,encircles the wire. Because the dot is on the axis of the wire, the input current creates no magnetic field at this point. The current divides at the junction, with 112 traveling upward and I12 traveling downward. The right-hand rule tells us that the upward current creates a field at the dot that is into the page; the downward current creates a field that is out of the page. Although we could calculate the strength of each field, the symmetry of the situation (the dot is the same distance and direction from the base of each wire) tells us that the fields of the upward and downward current must have the same strength. Since they are in opposite directions, their sum is 6. Altogether, then, the field at the dot is = 6 T.
32.47. Model: Assume that the wires are infinitely long and that the magnetic field is due to currents in both the wires.
Visualize: Point 1 is a distance d, away from the two wires and point 2 is a distant d, away from the two wires. A right triangle with a 75" degree angle is formed by a straight line from point 1 to the intersection and a line from point 1 that is perpendicular to the wire. Likewise, point 2 makes a 15" right triangle.
O
d2 = (4.0 cm)sinl5"
d, = (4.0 cm) sin75O
1
Solve: First we determine the distances d, and 4 of the points from the two wires: d, = (4.0 cm)sin75" = 3.86 cm = 0.0386 m d, = (4.0 cm)sin15" = 1.04 cm = 0.0104 m
At point 1, the fields from both the wires point up and hence add. The total field is B, = B,,,
p,, I, p I. Po (5.0 A) (4x , + BwKe2= -+ 2-= -- = 27r d,
In vector form,
21c dl
7r
T m / A ) ( ~ . oA) = 5.2 x 10" T 0.0386 m
d,
g, = (5.2 x
T, out of page). Using the right-hand rule at point 2, the fields are in opposite directions but equal in magnitude. So, E, = 6 T.
32.48. Model: The magnetic field is that of two long wires that carry current. Visualize: 3.0A
5.0 A
@-- - I
-7
0
1
5.0 A
3.0 A
x (cm)
8
2
-2-1
0
1
2
1
3
I
,
1
,
?
4
5
6
7
8
x (cm)
x(a)
(b)
Solve: (a) For x > +2 cm and for x < -2 cm, the magnetic fields due to the currents in the two wires add. The point where the two magnetic fields cancel lies on the x-axis in between the two wires. Let that point be a distance x away from the origin. Because the magnetic field of a long wire is B = p 0 1 / 2 m ,we have
Pn (5.0A) _- pn (3.0A) 3 5(0.02 m - x ) = 3(0.02 m + x ) 2rr (0.02 m + x ) #27r(0.02 m - x )
x = 0.005 m = 0.50 cm
(b) The magnetic fields due to the currents in the two wires add in the region -2.0 cm < x < 2.0 cm. For x < -2.0 cm, the magnetic fields subtract, but the field due to the 5.0 A current is always larger than the field due to the 3.0 A current. However, for x > 2.0 m, the two fields will cancel at a point on the x-axis. Let that point be a distance x away from the origin, so p,, 5.0A p o 3.0A -3 5(x - 0.02 m) = 3(x + 0.02 m) 3 x = 8.0 cm 27r x + 0.02 m 27r x - 0.02 m
32-12
Chapter 32
32.49. Model: The resistor and capacitor form an RC circuit. Assume that the length of the wire next to the dot is much larger than 1.O cm. visualize: B (TI 2x
-
I x
I
I
I
I
I
tbs)
Please refer to Figure P32.49
Solve: As the capacitor discharges through the resistor, the current through the circuit will decrease as
B = & = h ( 1 0 A)e-‘/io* = 2nr 2m A graph of t h i s equation is shown above.
(2.0 x 10-~T m A)QO A) e-,/,oM -
j
0.010 m
- (2.0 x IO4 T)e-r’ioM
32.50. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P32.50. Solve: (a) The Biot-Savart law (Equation 32.6) for the magnetic field of a current segment & is - p, I G x i g=-4n r2 where the unit vector i points from current segment As to the point, a distance r away, at which we want to evaluate the field. For the two linear segments of the wire, is in the same direction as i , so G x i = 0. For X i = A s . Thus the curved segment, A.? and i are always perpendicular, so g=--Po IAS 4n r z Now we are ready to sum the magnetic field of all the segments at point P. For all segments on the arc, the distance to point P is r = R. The superposition of the fields is
where L = R e is the length of the arc. (b) Substituting 8 = 2n in the above expression,
This is Equation 32.7, which is the magnetic field at the center of a 1-turn coil.
32.51. Model: Assume that the superconducting niobium wire is very long. Solve: The magnetic field of a long wire carrying current I is
We’re interested in the magnetic field of the current right at the surface of the wire, where d = 1.5 mm. The maximum field is 0.10 T, so the maximum current is I=
m)(O.lO T) (215d)~,,, = 2n(1.5x = 750A PO 4axIO-’Tm/A
Assess: The current density in this superconducting wire is o f the order of 1 x lo* Nm’. This is a typical value for conventional superconducting materials.
The Magnetic Field
32-13
32.52. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P32.52. The distance from P to the inner arc is r, and the distance from P to the outer arc is r,. Solve: As given in Equation 32.6, the Biot-Savart law for a current carrying small segment 6 is - Po Ih-x; B=-4 n r2 For the linear segments of the loop, B , = 0 T because ,G x i = 0. Consider a segment Ai on length on the inner arc. Because 6 is perpendicular to the r^ vector, we have
7
p IAB P Id6 --POI IC=- POI B = - pco - - - I- A s - p,, & A 0 = 0 B,, = 047r 52 4 K r,' 4n r, -np 4m; 4% 45 A similar expression applies for B,, . The right-hand rule indicates an out-of-page direction for B , page direction for B , ,. Thus,
(2:
B = -,
-
into page
The field strength is
B= Thus
)[P:(K).1
I (K +
and an into-
' J , out of page = 0- - - , into page
= 7.85 x lo-'
4
T
= (7.85 x lo-' T, into page).
-*
32.53. Model: A 1000-km-diameter ring makes a loop of diameter 3000 km. Visualize:
iron
Solve: (a) The current loop has a diameter of 3000 km, so its nominal area, ignoring curvature effects, is A,,, = m'= lljl500 x lo3 m)' = 7.07 x 10" m' Because the magnetic dipole moment of the earth is modeled to be due to a current flowing in such a loop, p = I A,,,. The current in the loop is
(b) The current density J in the above loop is
(c) The current density in the wire is
I 1.0A =-= =1.3x106 A / m 2 A K ( + X I . O X ~ O - m) ~ You can see that J,,, << Jw,,e.The current in the earth's core is large, but the current density is actually quite small. J W. E
32.54. Model: Assume that the wire is infinitely long. Visualize: Please refer to Figure P32.54. The wire, looped as it is, consists of a circular part and a linear part. Solve: Using Equation 32.7 and Example 32.3, the magnetic field at P is
4n(10-' T m / A)(5.0 A) 4z(10-' T m / A)(5.0 A) + = 4.1 x I O J T 2(0.01 m)
2740.01 m)
32-14
Chapter 32
32.55. Model: Assume that the solenoid is an ideal solenoid. Solve: The magnetic field of a solenoid is B,, = p o N Z / L ,where L is the length and N the total number of turns of wire. If the wires are wound as closely as possible, the spacing between one turn and the next is simply the diameter d, of the wire. The number of turns that will fit into length L is N = L/d,, . For #18 wire, N = (20 cm)/(O.l02 cm) = 196 turns. The current required is
LB,, - (0.20 m)(5 x 10-~T) = 4.06 A I*,8 = -p o N ( 4 x~lo-’ T m / A)(196) For #26 wire, dwi,= 0.41 mm, leading to N = 488 turns and I,,, = 1.63 A. The current that would be needed with #26 wire exceeds the current limit of 1 A, but the current needed with #18 wire is within the current limit of 6 A. So use #18 wire with a current of 4.06 A.
32.56. Model: Assume that the solenoid is an ideal solenoid. Solve: A solenoid field is B,, = p,$II/L, so the necessary number of turns is N = -u-s 0 1 -
pol
(0.08 m)(O. 10 T) = 3 180 ( 4 x~ T m / A)(2 A)
Problem 32.55 states that the rating is 6 A for a wire with d,, = 1.02 mm and the rating is 1 A for a wire having d,, = 0.41 mm. This means the rating of a wire can be increased 1 A by increasing the diameter by approximately 0.12 mm. We can assume that a wire that can carry a 2 A current will probably have a diameter larger than 0.41 mm + 0.12 mm. Let us take the wire’s diameter to be 0.6 mm. The number of turns that will fit in a length of 8 cm is
8 cm N, layer = -= 133 0.06 cm Thus 3180 turns would require 3180/133 = 24 layers. Assess: With this many layers, the thickness of the layers of wires becomes larger than the radius of the solenoid. Maybe not impossible, but it is probably not feasible.
32.57. Model: The coils are identical, parallel, and carry equal currents in the same direction. The magnetic field is that of the currents in these coils. Solve: (a) The on-axis magnetic field of an N-turn current loop at a distance z from the loop center is
If the spacing between the loops is R, then the midpoint between them is z = R/2. Since the currents are in the same direction, the field of each loop is in the same direction and the net field is simply twice the field due to a single loop. Thus
B=
PONIR’ 3,2 =(1.25)- 312 PONI (R2/4+ R’) R
(b) The magnetic field is B = (0.716)
4n(10-’ T m / A)10(1.0 A)
0.05 m
= 1.80 x lo4 T
32.58. Solve: From Equation 32.7, the magnetic field at the center ( z = 0 m) of an N-turn coil is B,,
center
=
poNI
2 R
-
p 1 0 mI 2
221rRR where we used the requirement that the entire length of wire be used. That is, 1.0 m = (2nR)N. Solving for R,
241.0 x ~ o - ~ T )
= 0.010m = 1.0cm
The Magnetic Field
32-15
32.59. Model: The magnetic field is that of a current in the wire. Visualize: Please refer to Figure P32.59. Solve: As given in Equation 32.6 for a current carrying small segment &-, the Biot-Savart law is
and i point along the same line. That is not the case with the For the straight sections, & X i = 0 because both curved section over which & and ? are perpendicular. Thus,
where we used As = RA8 = RdB for the small arc length As. Integrating to obtain the total magnetic field at the center of the semicircle,
32.60. Model: The toroid may be viewed as a solenoid that has been bent into a circle. Visualize: Please refer to Figure P 32.60. Solve: (a) A long solenoid has a uniform magnetic field inside and it is roughly parallel to the axis. If we bend the solenoid to make it circular, we will have circular magnetic field lines around the inside of the toroid. However, as explained in uart (c) the field is not uniform.
(b)
Ampere's of integrat Current out
A top view of the toroid is shown. Current is into the page for the inside windings and out of the page for the outside windings. The closed Ampere's path of integration thus contains a current NI where I is the current flowing through the wire and N is the number of turns. Applied to the closed line path, the Ampere's law is
=polthrough =poNZ Because B and
ds are along the same direction and B is the same along the line integral, the above simplifies to
(c) The magnetic field for a toroid depends inversely on r which is the distance from the center of the toroid. As r increases from the inside of the toroid to the outside, Btamiddecreases. Thus the field is not uniform.
32-16
Chapter 32
32.61. Model: The magnetic field - - -is-that _ of the current which is distributed uniformly in the hollow wire. Vialize:
...
,,
< -
Ampere’s path r > R , Ampere’s path R , < r < R, ere’s path 0 < r < R,
..- - - _ _ - . -
I
Ampere’s integration paths are shown in the figure for the regions 0 m < r < R,, R, < r <R2, and R2 < r. Solve: For the region 0 m < r < R,, 4B-C= p,,Z,,,,, . Because the current inside the integration path is zero, B = 0 T . To find Ithroughin the region R, < r < R,, we multiply the current density by the area inside the integration path that carries the current. Thus,
where the current density is the first term. Because the magnetic field has the same magnitude at every point on the circular path of integration, Ampere’s law simplifies to
For the region R2 < r, IMugh is simply I because the loop encompasses the entire current. Thus,
fs.
= Bjds = B 2 w =
P I *B =0 2m
Assess: The results obtained for the regions r > R, and R, < r < R2 yield the same result at r = R,. Also note that a hollow wire and a regular wire have the same magnetic field outside the wire.
32.62. Visualize: Please refer to Figure P32.62. Solve: The electric field is
I? = [E, down 1 cm
The force this field exerts on the electron is without deflection is,
Fmag = (3.2 x
= (20,000 V / m, down)
E,== q E = -eE
if the magnetic field also exerts
= (3.2 x
N, up). The electron will pass through
a force on the electron such that
Fne,= E, + Fmag = 0 N . That
N, down). In this case, the electric and magnetic forces cancel each other. For a negative
point into the page. The down requires, from the right-hand rule, that charge with to the right to have Fmg = qvB, so the needed field strength is magnitude of the magnetic force on a moving charge is Fmag ’mag
B=-ev
-
(1.60 x
3 . 2 ~ 1 0 - lN~ =2.0~10T - ~ = 2.0mT C)(l.O x lo7 m / s )
Thus, the required magnetic field is B = (2.0 mT, into page).
The Magnetic Field
32-17
32.63. Model: Energy is conserved as the electron moves between the two electrodes. Assume the electron starts from rest. Once in the magnetic field, the electron moves along a circular arc. :ov ; 10,Ooo v
visualize:
The electron is deflected by 10" after moving along a circular arc of angular width IO". Solve: Energy is conserved as the electron moves from the 0 V electrode to the 10,000 V electrode. The potential energy is U = qV with q =-e, so
K , +V, = K , + V i3 +mv2 -eV = 0 + 0
The radius of cyclotron motion in a magnetic field is r = mv/eB. From the figure we see that the radius of the circular arc is r = (2.0 cm)/sinlO". Thus (9.11 x kg)(5.93 x IO' m/s) = 2.9 x (1.60 x lo-'' C)(0.02 m) I sinl0"
mv B=-= er
T
32.64. Model: Charged particles moving perpendicular to a uniform magnetic field undergo uniform circular motion at a constant speed. Solve: (a) From Equation 32.19, the magnetic field is
27@ 2n(2.4 x lo9 Hz)(9.11 x e 1.6 x lo-'' C
kg)
B=--
= 0.086T = 86mT
(b) The maximum kinetic energy is for an orbit with radius 1.25 cm.
K = +mv' = +m(2lnf)' = t(9.11 x
kg)[2*(0.0125 m)(2.4 x IO9 Hz)]~= 1.6 x
J
32.65. Model: Electric and magnetic fields exert forces on a moving charge. The fields are uniform throughout the region.
Visualize: Please refer to Figure P32.65. Solve: (a) We will first find the net force on the antiproton, and then find the net acceleration using Newton's second law. The magnitudes of the electric and magnetic forces are
Ffi = eâ&#x201A;Ź = (1.60 x
C)(1000 V I m) = 1.60 x
FB = cvB = (1.60 x
C)(500m I s)(2.5 T) = 2.0 x
The directions of these two forces on the antiproton are opposite.
F,
N
F'
N
points up whereas, using the right-hand rule,
points down. Hence,
F,,
N - 1.60 x
= (2.0 x
*7i= (b) If
were reversed, both
Fnel= (1.6 x
F, N
N, down) 3 F,,, = 0.40 x
N = ma
0.40 x N = 2.4 x 10" m I s', down 1.67 x lo-" kg
and
F,
+
2.0 x
will point up. Thus,
N, up)
=$
Z
3.6~
N
= 1.67 x IO-" kg =2.2x101' m / s ' , up
I
32-18
Chapter 32
32.66. Model: A magnetic field exerts a magnetic force on a moving charge given by = 4.' x 5.Assume the magnetic field is uniform. Visualize: Please refer to Figure P32.66. The magnetic field points in the +z-direction. If the charged particle is moving along
B,
F = 0 N . If 3 is perpendicular to g, the motion of the charged particle is a circle. However,
when v' makes an angle with 5,the motion of the charged particle is like a helix or a spiral. The perpendicular component of the velocity is responsible for the circular motion, and the parallel component is responsible for the linear motion along the magnetic field direction. Solve: From the figure we see that vy = vcos30째 and vz = vsin30". For the circular motion, the magnetic force causes a centripetal acceleration. That is, (9.11 x kg)(5.0 x lo6 m I s)cos30째 mv: =$ r = mvY evyB = = 0.82 mm r eB (1.60 X C)(0.030 T) The time for one revolution is 2m T=-= vy
21c(8.2 x lo4 m)
= 1.2 x 1 0 - ~s
(5.0 x lo6 m / s)cos30~
The pitch p is the vertical distance covered in time T. We have
p = vLT= (5.0 x IO6 m / s)sin30"(1.2 x
s ) = 3.0 x
m = 3 mm
32.67. Model: Charged particles moving perpendicular to a uniform magnetic field undergo uniform circular motion at constant speed. Solve: (a) The magnetic force on a proton causes a centripetal acceleration: mv' eBr evB = - v = r m Maximum kinetic energy is achieved when the diameter of the proton's orbit matches the diameter of the cyclotron: e 2 B 2 r 2 (1.60 X K = 1mv2 = -2m
C)'(0.75 T)*(0.325 m)'
2( 1.67 x 1O-*' kg)
= 4.6 x 10-13 J
(b) The proton accelerates through a potential difference of 500 V twice during one revolution. The energy gained per cycle is 2 qAV= 'e (500 V) = 1.60 x
J
Using the maximum kinetic energy of the proton from part (a), the number of cycles before the proton attains this energy is 4.6 x 1.60 x
J = 2850 J
32.68. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Solve: The magnetic force on a proton causes the centripetal acceleration of mv' evB=-a r
mv B=er The maximum radius r is 0.25 m, and the desired speed v is 0. IC. So, the required field B is B=
( 1 . 6 7 ~ 1 0 - ~ ~ k g ) ( O . 1 ) ( 3m~/1s0) ~ (1.6 x
C)(O.25 m)
= 1.25 T
The Magnetic Field
32-19
32.69. Model: Assume that the magnetic field is uniform over the Hall probe. Solve: Equation 32.24 gives the Hall voltage and Equation 30.20 gives the cyclotron frequency in terms of the magnetic field. We have
2lnnf,,I =me=-= eAV,
2n(1.67 X lo-” kg)(lO.OX lo6 Hz)(O.150 x
A)
(1.6 x 10-j~c)(0.543 x 1 0 - ~v)
= 0.1812 T A / V
With this value of me, we can once again use the Hall voltage equation to find the magnetic field:
B = (?)me
(
1.735 x = 0.150 x
V 0.1812 TA / V) = 2.10 T A
32.70. Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Visualize: Please refer to Figure P32.70. Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v. Upon entering the magnetic field, the ion follows a circular trajectory with cyclotron radius r = mv/eB. To be detected, an ion’s trajectory must have radius d = 2r = 8 cm. This means the ion needs the speed eBd v = -eBr - -m 2m This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy equation to find the voltage that will accelerate the ion:
K , + U,= K 2+ U,
=j
0 J+eAV= jmv‘+O J
3 AV
mv2 =2e
Using the above expression for v, the voltage that causes an ion to be detected is
An ion’s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of NS is rn = mN+ m, - mrleL = 2( 14.0031 u)( 1.661 x lo-’’ kgh) - 9.1 1 x lo-’‘ kg = 4.65 174 x
kg Note that we’re given the atomic masses very accurately in Exercise 32. We need to retain this accuracy to tell the difference between N; and CO’. The voltage for N i is
AV = Ion Nf 0; CO
(1.6 x
C)(0.200 T)’(0.08 m)* 8(4.65174 x
kg)
= 110.07 V
Accelerating voltage (V)
Mass (kg) 4.65174 x 5.31341 x 4.64986 x
110.07 96.36 110.11 ~~~
Assess: The difference between N f and CO’ is not large but is easily detectable.
32.71. Model: A magnetic field exerts a magnetic force on a length of current carrying wire. We ignore gravitational effects, and focus on the B effects. Visualize: Please refer to Figure P32.71. The figure shows a wire in a magnetic field that is directed out of the page. The magnetic force on the wire is therefore to the right and will stretch the springs. Solve: In static equilibrium, the sum of the forces on the wire is zero: FB + F,,,
+ F , , , = O N * ILB + (-khr) + ( - k h )
3
2 h - k 2(10 N / rn)(0.01 m) = 2,0 A I = -= LB (0.20 m)(0.5 T)
32.72. Model: The wire will float in the magnetic field if the magnetic force on the wire points upward and has a magnitude mg, allowing it to balance the downward gravitational force.
32-20
Chapter 32
-
Visualize:
Fmag
30" I I
\
-
W
I I
\\
4cm
'
\
I
\
\
I I
\
@j __-___ - -- - - ---$ Solve: Each lower wire exerts a repulsive force on the upper wire because the currents are in opposite directions. The currents are of equal magnitude and the distances are equal, so F; = 4. Consider segments of the wires of length L. Then the forces are
The horizontal components of these two forces cancel, so the net magnetic force is upward and of magnitude
Fmg = 2 4 C O S ~ O O =
p 0 U 2cos30째 7rd
In equilibrium, this force must exactly balance the downward weight of the wire. The wire's linear mass density is p = 0.050 kg/m, so the mass of this segment is m = p5 and its weight is w = mg = pLg. Equating these gives
pow2 cos30O
(0.050 kg / m)(9.8 m / s2)n(0.040m)
7rd
(4n x lo-' T m / A)cos30째
= 238 A
32.73. Model: A magnetic field exerts a magnetic force on a length of current carrying wire. Visualize:
Please refer to Figure P32.73. Solve: The above figure shows a side view of the wire, with the current moving into the page. From the righthand rule, the magnetic field B points down to give a leftward force on the current. The wire is hanging in static - - - equilibrium, so F,,, = F,,, + W + T = 0 N. Consider a segment of wire of length L. The wire's linear mass density is p = 0.050 kg/m, so the mass of this segment is m = pL and its weight is W = mg = p5g. The magnetic force on this length of wire is Fmg= ILB. In component form, Newton's first law is
(Fn,I)x=TsinO-F,,
=Tsin@-ILB=ON*TsinO=
ILB
Dividing the first equation by the second.
I
TcosO The magnetic field is
= (0.00864T, down)
(0.050 kg/ m)(9.8 m /s')tanlO" = 0.00864 T 10 A
The Magnetic Field
32-21
32.74. Model: Assume that the 20 cm is large compared with the effective radius of the magnetic dipole (bar magnet).
Solve: From Equation 32.9, the on-axis field of a magnetic dipole at a distance z from the dipole is
The magnetic dipole moment can be obtained from the magnitude of the torque. We have
Thus, the magnetic field produced by the magnet is B
~= ~
T , m~i A)
2(0.15 Am2)
(0.20 m)3
= 3.75 x IO4 T
32.75. Model: A current loop produces a magnetic field. Visualize: B'
Solve: The field at the center of a current loop is B,, = pJi2R. The electron orbiting an atomic nucleus is, on average, a small current loop. Current is defined as I = AqiAt. During one orbital period T , the charge Aq = e goes around the loop one time. Thus the average current is Iavg = e/T. For circular motion, the period is
Thus, the magnetic field at the center of the atom is
p,~,,, (4nx IO-' T m / A)(1.057 x
Bcenle1
=2R =
A)
2(5.3 x lo-'' m)
= 12.5 T
32.76. Model: A magnetic field exerts a force on a segment of current. Visualize: The figure shows the forces on two small segments of current, one in which the current enters the plane of the page and one in which the current leaves the plane of the page.
Current / segmentof segment of length As B
- /"w
2R
Solve: (a) Consider a small segment of the loop of length As. The magnetic force on this segment is perpendicular both to the current and to the magnetic field. The figure shows two segments on opposite sides of the loop. The horizontal components of the forces cancel but the vertical components combine to give a force toward the bar magnet. The net force is the sum of the vertical components of the force on all segments around the loop. For a segment of length As, the magnetic force is F = ZBAs and the vertical component is F, = (ZB&)sin6. Thus the net force on the current loop is F,,, = C F v = C(IBAs)sine = ZBsin6CA.s
32-22
Chapter 32
We could take sin9 outside the summation because 8 is the same for all segments. The sum of all the As is simply the circumference 27rR of the loop, so
F,, = 2nRIBsin9
(b) The net force is F,, = 2N0.02 m)(sin 20”)(0.50 A)(200 x 10” T) = 4.3 x
N
32.77. Model: A ferromagnetic material is divided into small regions called magnetic domains. Visualize:
sizedxd
Solve: A diskette is measured with a ruler and found to have an outer radius r, = 4.5 cm and an inner radius r, 1.0 cm. The area of one surface is Asdace = x(q2- rZZ)= 51 cm2 = 5.1 x m2
a
Each side stores =. 500,000 bytes, each of 8 bits. This is N = 4 x lo6 bits of data. Each bit needs one magnetic domain, so the area of each domain is
If each domain is square and of size d x d, then A , , , = d x d * d = ~ = 3 . 6 ~ 1 0 - ~ r=n 0.036mm
Assess: Your answer may differ somewhat, depending on the assumptions you made, and it will depend on the distance between the domains. This result is consistent with the estimate of 0.1 mm for the size of domains given in Section 31.10.
32.78. Visualize:
a aParallel currents attract each other
Solve: A permanent magnet results when the “spinning” electrons in a metal are aligned with each other. Each spinning electron is, in effect, a current loop and it generates a smaU magnetic field. The field of the magnet is the sum of the fields of all the spinning electrons. A powerful microscope would show that the spin axis of each electron tends to be parallel to the axis of the magnet, as shown in the figure. When the north pole of one magnet approaches the south pole of another magnet, the electrons in both magnets are spinning parallel to each other. Parallel currents exert attractive magnetic forces on each other, so the macroscopic phenomenon of attraction between north and south poles is actually the net result of the attractive forces betwen the vast number of microscopic current loops with the currents flowing in the same direction. When two north poles approach, the electrons in the two magnets are spinning in opposite directions. Opposite currents exert repulsive magnetic forces on each other, so the macroscopic phenomenon of repulsion between two north poles is actually the net result of the repulsive forces between current loops with currents flowing in opposite directions.
The Magnetic Field
32-23
32.79. Visualize: Electrons
Domains experience a torque
Electrons
S I
I
s N
I
Induced magnetic poles feel force from magnet
N Aligned domains create induced poles
Solve: A permanent magnetic creates a magnetic field. This is due to the motion of the charge of the “spinning” electrons. The unmagnetized piece of iron also has spinning electrons, but these are organized into magnetic domains that are randomly oriented and give no net magnetic moment. However, each individual domain-due to the aligned atomic spins-is a magnetic moment, so it experiences a torque in the field of the permanent magnet. This torque tends to align the domains with the external field E. Now there are many aligned domains, so the piece of iron has an induced net magnetic moment. That is, the piece of iron has an induced north and south pole. The induced south pole of the iron is facing the permanent magnet’s north pole, so there is an attractive force that lifts the iron. Although the lifting is a macroscopic phenomenon, it is due to magnetic attraction between the spinning electrons in the permanent magnet and the spinning electrons in the piece of iron.
32.80. Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop balances the torque of the weight. Visualize: Please refer to Figure CP32.80. Solve: The rotational equilibrium condition Z?, = 0 N m is about the axle and means that the torque from the weight is equal and opposite to the torque from the magnetic force. We have
(50 x lO-’kg)g(0.025 m) = pBsin90” = (NIA)B *B=
(50 x
kg)(9.8 m / s2)(0.025m)
(10)(2.0 A)(0.05 m)(0.10 m)
= 0.123 T
Assess: The current in the loop must be clockwise for the two torques to be equal.
32.81. Model: The magnetic field is that of a finite length of current carrying wire. Solve: (a) Let the wire be on the x-axis and the point of interest be at distance d on the y-axis. The wire extends from x = -W2 to x = +LE. As explained in Example 32.3 and Figure 32.13, each segment dx of the wire generates a field at the observation point of strength
dB =
pnId dx 472(x2+ d’)”’
These infinitesimal fields all point the same direction, so we can add them (superposition) to find the total field. The sum becomes an integral, giving
B - PnId 472
4
dx Po Id (x’+ d ’ ) 312 = 47r d’(x’
PnIL
X
+ &)“’
4d .-/, -L/?
32-24
(b)
Chapter 32
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A square is formed of four straight-wire segments, with each segment contributing the same field at the center. The length is L = 2R, and the observation point is distance d = R from the center of each segment. Using the formula from part (a>,
(c)The ratio of the field strengths of a square loop and a circular loop, each with â&#x20AC;&#x153;radiusâ&#x20AC;? R, is
The formula for the magnetic field at the center of a circular loop is from Equation 32.7.
32.82. Model: The magnetic field is that of a conducting wire that has a nonuniform current density. Visualize:
Integration Path
through area dA = 2rrrdr
Solve: (a) Consider a small circular disk of width drat a distance r from the center. The current through this disk is ( 2 m ) d r=
~
21dor2dr R
Integrating this expression, we get
(b) Applying & = poImugb to the circular path of integration, we note that the wire has perfect cylindrical symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r. Thus,
(c)At r = R ,
This is the same result as obtained in Example 32.3 for the magnetic field of a long straight wire.
The Magnetic Field
32-25
32.83. Model: The magnetic field is that of a coaxial cable consisting of a solid inner conductor surrounded by a hollow outer conductor of essentially zero thickness.
Visualize: Please refer to Figure CP32.83. The solid inner conductor and the hollow outer conductor cany equal currents but in opposite directions. The coaxial cable has perfect cylindrical symmetry. So the magnetic field must be tangent to circles that are concentric with the wire. Solve: (a) Ampereâ&#x20AC;&#x2122;s law is
For r < R,,
For R , < r < R2,I-uzh =I.Hence, B = p01/2m.For r > R,, ILhmugh = 0 A and B = 0 T. (b) In the region r < R,,B is linearly proportional to r; in the region R , < r < R,, B is inversely proportional to r; and in the region for r > 2R,, B = 0 T. A B versus r graph in the range r = 0 m to r = 2R, is shown below. B
32.84. Model: The current sheet extends infinitely far in both directions. Visualize: (a) The shape of the magnetic field must be consistent with the symmetry of the current sheet. Consider two current-carrying wires. The magnetic field above the midpoint of the wires is horizontal and to the left. The magnetic field below the midpoint of the wires is horizontal and to the right. If we imagine the current sheet to consist of many such pairs of wires, closely spaced, then we see that the magnetic field above the sheet is everywhere horizontal and to the left while the magnetic field below the sheet is everywhere horizontal and to the right. The figure shows a closed path for using Ampereâ&#x20AC;&#x2122;s law.
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32-26
Chapter 32
Solve: (b) The closed path of width L shown in the figure is parallel to the field along the top and bottom edges, perpendicular to the field along the left and right edges. Thus the line integral in Ampereâ&#x20AC;&#x2122;s law is fs.d3=
I-
B.d?+
top
I-
B.ds+
left
I
bomm
5.ds-k
I
B.ds=BL+O+BL+O=2BL
nghl
Because the current per unit width is Js,the amount of current in a length L of the current sheet (the current through the closed path) is Ithrough = JA. Thus Ampereâ&#x20AC;&#x2122;s law gives
$ B . d i = p O I m u g h* 2 B L = p o J , L * B = + p o J , Assess: The magnetic field strength is independent of the distance from the sheet. This is not unexpected since the electric field of an infinite plane of charge is independent of the distance from the plane.