b mg (b) This is about one-sixth the magnitude of the Earth s field. It will affect the compass reading.

Transcription

1 Chapter 9 (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by i = μ p r With r = ft = 6 m, we have = c4p TmA hbag p6 b mg 6 = 33 T= 33 μ T (b) This is about one-sixth the magnitude of the Earth s field It will affect the compass reading Equation 9- is maximized (with respect to angle) by setting θ = 9º ( = π/ rad) Its value in this case is μi ds dmax = 4π R From Fig 9-34(b), we have max = 6 T We can relate this max to our d max by setting ds equal to 6 m and R = 5 m This allows us to solve for the current: i = 375 A Plugging this into Eq 9-4 (for the infinite wire) gives = 3 μt 3 (a) The field due to the wire, at a point 8 cm from the wire, must be 39 μt and must be directed due south Since = μ i p r, 6 b gc Th pr p 8 m 39 i = = μ 4p TmA = 6A (b) The current must be from west to east to produce a field that is directed southward at points below it 4 The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem Magnetic field at the center of a circular arc of current ) Also, the fields from the two semicircular loops cancel at C (by symmetry) Therefore, C = 5 (a) We find the field by superposing the results of two semi-infinite wires (Eq 9) and a semicircular arc (Eq 9-9 with φ = π rad) The direction of is out of the page, as can be checked by referring to Fig 9-6(c) The magnitude of at point a is therefore 7

2 8 CHAPTER 9 a μ i μ iπ μ i π = 4 R + = + = + = π 4πR R π (5 m) π (4 T m/a)( A) 3 T upon substituting i = A and R = 5 m (b) The direction of this field is out of the page, as Fig 9-6(c) makes clear (c) The last remark in the problem statement implies that treating b as a point midway between two infinite wires is a good approximation Thus, using Eq 9-4, b μ i μ i π = R = = = π πr π(5 m) (4 T m/a)( A) 4 8 T (d) This field, too, points out of the page 6 With the usual x and y coordinates used in Fig 9-37, then the vector r pointing from a current element to P is r = sˆi+ Rˆ j Since ds = ds î, then ds r = Rds Therefore, with r = s + R, Eq 9-3 gives μ ir ds d = 4 π ( s + R ) 3/ (a) Clearly, considered as a function of s (but thinking of ds as some finite-sized constant value), the above expression is maximum for s = Its value in this case is d = μ i ds π R max /4 (b) We want to find the s value such that d = d / max This is a nontrivial algebra exercise, but is nonetheless straightforward The result is s = /3 R If we set R = cm, then we obtain s = 38 cm 7 (a) Recalling the straight sections discussion in Sample Problem Magnetic field at the center of a circular arc of current, we see that the current in the straight segments collinear with P do not contribute to the field at that point Using Eq 9-9 (with φ = θ) and the right-hand rule, we find that the current in the semicircular arc of radius b contributes μ iθ 4p b (out of the page) to the field at P Also, the current in the large radius arc contributes μ iθ 4p a (into the page) to the field there Thus, the net field at P is μiθ (4p T m A)(4A)(74 π /8 ) = = 4 b a 4π 7m 35m = T

3 9 (b) The direction is out of the page 8 (a) Recalling the straight sections discussion in Sample Problem Magnetic field at the center of a circular arc of current, we see that the current in segments AH and JD do not contribute to the field at point C Using Eq 9-9 (with φ = π) and the right-hand rule, we find that the current in the semicircular arc H J contributes μ i 4R (into the page) to the field at C Also, arc D A contributes μ i 4R (out of the page) to the field there Thus, the net field at C is μ i (4p T m A)(8A) = = = 4 R R 4 35m 78m (b) The direction of the field is into the page T 9 (a) The currents must be opposite or antiparallel, so that the resulting fields are in the same direction in the region between the wires If the currents are parallel, then the two fields are in opposite directions in the region between the wires Since the currents are the same, the total field is zero along the line that runs halfway between the wires (b) At a point halfway between they have the same magnitude, μ i/πr Thus the total field at the midpoint has magnitude = μ i/πr and ( m)( 3 6 T) pr p 4 i = = = 3A μ 4p T m A (a) Recalling the straight sections discussion in Sample Problem Magnetic field at the center of a circular arc of current, we see that the current in the straight segments collinear with C do not contribute to the field at that point Equation 9-9 (with φ = π) indicates that the current in the semicircular arc contributes μ i 4R to the field at C Thus, the magnitude of the magnetic field is μ i (4p T m A)(348A) = = = 8 T 4R 4(96m) (b) The right-hand rule shows that this field is into the page (a) P = μ i /πr where i = 65 A and r = d + d = 75 cm + 5 cm = 5 cm, and = μ i /πr where r = d = 5 cm From P = P we get P r 5 cm ( 65A) 43A = i = = r 5 cm i

4 3 CHAPTER 9 (b) Using the right-hand rule, we see that the current i carried by wire must be out of the page (a) Since they carry current in the same direction, then (by the right-hand rule) the only region in which their fields might cancel is between them Thus, if the point at which we are evaluating their field is r away from the wire carrying current i and is d r away from the wire carrying current 3i, then the canceling of their fields leads to μ (3 ) 6 cm i μ i d = r = = = 4 cm πr π( d r) 4 4 (b) Doubling the currents does not change the location where the magnetic field is zero 3 Our x axis is along the wire with the origin at the midpoint The current flows in the positive x direction All segments of the wire produce magnetic fields at P that are out of the page According to the iot-savart law, the magnitude of the field any (infinitesimal) segment produces at P is given by i d = μ θ sin dx 4p r where θ (the angle between the segment and a line drawn from the segment to P ) and r (the length of that line) are functions of x Replacing r with x + R and sin θ with Rr= R x + R, we integrate from x = L/ to x = L/ The total field is μ ir dx μ ir x μ i L = = = 4p 4p R pr L + 4R L L L 3 L ( x + R ) ( x + R ) ( p )( ) p( 3 ) 4 T m A 58 A 8m = = m (8m) + 4(3m) 8 53 T 4 We consider Eq 9-6 but with a finite upper limit (L/ instead of ) This leads to μi L / = πr ( L/) + R In terms of this expression, the problem asks us to see how large L must be (compared with R) such that the infinite wire expression (Eq 9-4) can be used with no more than a % error Thus we must solve =

5 3 This is a nontrivial algebra exercise, but is nonetheless straightforward The result is R L L= 4 R 4 R 5 (a) As discussed in Sample Problem Magnetic field at the center of a circular arc of current, the radial segments do not contribute to P and the arc segments contribute according to Eq 9-9 (with angle in radians) If k designates the direction out of the page then μ ( 4 A)( π rad) ˆ ( 8 A)( / 3rad) ˆ 6 k μ π = k = (7 T)kˆ 4π 5 m 4π 4 m 6 or = 7 T (b) The direction is ( ) ˆk, or into the page (c) If the direction of i is reversed, we then have ( ) or ( )( ) π ( ) ( )( ) π ( ) μ 4 A π rad ˆ μ 8 A π / 3rad ˆ 6 = k k = (67 T)kˆ 4 5 m 4 4 m 6 = 67 T (d) The direction is ˆk, or into the page 6 Using the law of cosines and the requirement that = nt, we have θ = cos + = 44, where Eq 9- has been used to determine (68 nt) and (5 nt) 7 Our x axis is along the wire with the origin at the right endpoint, and the current is in the positive x direction All segments of the wire produce magnetic fields at P that are out of the page According to the iot-savart law, the magnitude of the field any (infinitesimal) segment produces at P is given by i d = μ θ sin dx 4p r

6 3 CHAPTER 9 where θ (the angle between the segment and a line drawn from the segment to P ) and r (the length of that line) are functions of x Replacing r with x + R and sin θ with Rr= R x + R, we integrate from x = L to x = The total field is μ ir dx μ ir x μ i L = = = 4p 4p R 4pR L + R L 3 L ( x + R ) ( x + R ) ( p )( ) 4p( 5 ) 4 T m A 693 A 36m = = m (36m) + (5m) 3 T 8 In the one case we have small + big = 475 μt, and the other case gives small big = 575 μt (cautionary note about our notation: small refers to the field at the center of the small-radius arc, which is actually a bigger field than big!) Dividing one of these equations by the other and canceling out common factors (see Eq 9-9) we obtain (/ rsmall ) + (/ rbig ) + ( rsmall / rbig ) = = 3 (/ r ) (/ r ) ( r / r ) small big small big The solution of this is straightforward: r small = r big / Using the given fact that the r = 4 cm, then we conclude that the small radius is r small = cm big 9 The contribution to net from the first wire is (using Eq 9-4) μ i ˆ (4π T m/a)(3 A) ˆ 6 ˆ = k = k = (3 T)k πr π( m) The distance from the second wire to the point where we are evaluating net m = m Thus, is r = 4 m μ i ˆ (4π T m/a)(4 A) ˆ 6 = i = i = (4 T)i ˆ πr π( m) and consequently is perpendicular to The magnitude of net is therefore net = (3 T) + (4 T) = 5 T (a) The contribution to C from the (infinite) straight segment of the wire is C i = μ π R

7 33 The contribution from the circular loop is C i = μ Thus, R 3 ( 4π T m A)( 578 A) μ i R π m π C = C+ C = + = + = 53 T ( 89 ) C points out of the page, or in the +z direction In unit-vector notation, = (53 T)kˆ C (b) Now, so C C 3 ( 4p T m A)( 578 A) μi C = C+ C = + = + = T R π m π ( 89 ) and C points at an angle (relative to the plane of the paper) equal to In unit-vector notation, C = = C π C tan tan 766 ˆ ˆ ˆ 8 = T(cos766 i + sin766 k) = (9 T)i + (6 T)kˆ Using the right-hand rule (and symmetry), we see that net points along what we will refer to as the y axis (passing through P), consisting of two equal magnetic field y- components Using Eq 9-7, μi net = sinθ π r where i = 4 A, r = r= d + d / 4 = 5 m, and d 4 m 4 θ = tan = tan = tan = 53 d / 6 m / 3 Therefore, μi (4π T m A)(4 A) net = sinθ = sin 53 = 56 T πr π( 5 m) The fact that y = at x = cm implies the currents are in opposite directions Thus,

8 34 CHAPTER 9 y μ i μ i μ i 4 = = π( L+ x) πx π L+ x x using Eq 9-4 and the fact that i = 4i To get the maximum, we take the derivative with respect to x and set equal to zero This leads to 3x Lx L =, which factors and becomes (3x + L)(x L) =, which has the physically acceptable solution: x = L This produces the maximum y : μ o i /πl To proceed further, we must determine L Examination of the datum at x = cm in Fig 9-49(b) leads (using our expression above for y and setting that to zero) to L = 3 cm (a) The maximum value of y occurs at x = L = 3 cm (b) With i = 3 A we find μ o i /πl = nt (c) and (d) Figure 9-49(b) shows that as we get very close to wire (where its field strongly dominates over that of the more distant wire ) y points along the y direction The right-hand rule leads us to conclude that wire s current is consequently is into the page We previously observed that the currents were in opposite directions, so wire s current is out of the page 3 We assume the current flows in the +x direction and the particle is at some distance d in the +y direction (away from the wire) Then, the magnetic field at the location of a proton with charge q is = ( μ ˆ i/ πd)k Thus, In this situation, v (( ) ) F = iq qv = μ pd v k e = ve jj (where v is the speed and is a positive value), and q > Thus, 9 μiqv μ iqv (4p T m A)(35A)(6 C)(m/s) F = ˆj kˆ = ˆi= ˆi pd pd π (89 m) -3 = ( 775 N)i ˆ 4 Initially, we have net,y = and net,x = + 4 = (μ o i /πd) using Eq 9-4, where d = 5 m To obtain the 3º condition described in the problem, we must have net, y = net, xtan(3 ) 3 = tan(3 ) j μ i π d where 3 = μ o i /πd and = μ i/ π d Since tan(3º) = / 3, this leads to

9 37 (a) We now make the assumption that wire # must be at π/ rad ( 9º, the bottom of the cylinder) since it would pose an obstacle for the motion of wire # (which is needed to make these graphs) if it were anywhere in the top semicircle (b) Looking at the θ = 9º datum in Fig 9-57(b)), where there is a maximum in net x (equal to +6 μt), we are led to conclude that x = 6 μt μt = 4 μt in that situation Using Eq 9-4, we obtain 6 π R x π ( m)(4 T) μ 4π T m/a i = = = 4 A (c) The fact that Fig 9-57(b) increases as θ progresses from to 9º implies that wire s current is out of the page, and this is consistent with the cancellation of net y at θ = 9, noted earlier (with regard to Fig 9-57(c)) (d) Referring now to Fig 9-57(b) we note that there is no x-component of magnetic field from wire when θ =, so that plot tells us that x = + μt Using Eq 9-4, we find the magnitudes of the current to be 6 π R x π ( m)( T) μ 4π T m/a i = = = A (e) We can conclude (by the right-hand rule) that wire s current is into the page 3 (a) Recalling the straight sections discussion in Sample Problem Magnetic field at the center of a circular arc of current, we see that the current in the straight segments collinear with P do not contribute to the field at that point We use the result of Problem 9- to evaluate the contributions to the field at P, noting that the nearest wire segments (each of length a) produce magnetism into the page at P and the further wire segments (each of length a) produce magnetism pointing out of the page at P Thus, we find (into the page) μ ( )( ) i μ 4 T m A 3A i μi p P = 8 a 8 ( a) = = p p 8pa 8π ( 47 m) -5-5 = 96 T T (b) The direction of the field is into the page 3 Initially we have i μiφ μiφ = + 4π R 4π r using Eq 9-9 In the final situation we use Pythagorean theorem and write

10 39 cosines of some angle A little trig (and the use of the right-hand rule) leads us to conclude that when wire is at angle θ (shown in Fig 9-6) then its components are = sin θ, = cos θ x y The magnitude-squared of their net field is then (by Pythagoras theorem) the sum of the square of their net x-component and the square of their net y-component: = ( sin θ ) + ( cos θ ) = + cos θ (since sin θ + cos θ =), which we could also have gotten directly by using the law of cosines We have μ i μ i = = 6 nt, = = 4 nt πr πr With the requirement that the net field have magnitude = 8 nt, we find + θ = cos = cos ( / 4) = 4, where the positive value has been chosen 35 Equation 9-3 gives the magnitude of the force between the wires, and finding the x- d component of it amounts to multiplying that magnitude by cosφ = d Therefore, + d the x-component of the force per unit length is μ ii d π = = L π( d d ) [(4 m) (5 m) ] = 884 N/m Fx 3 3 (4 T m/a)(4 A)(68 A)(5 m) + π + 36 We label these wires through 5, left to right, and use Eq 9-3 Then, (a) The magnetic force on wire is F ( π )( ) μ il T m A 3A (m) ˆ μ il j ˆj ˆj π d d 3d 4d 4πd 4π 8 m = = = = 4 (469 N) j (b) Similarly, for wire, we have ˆ ( )

11 4 CHAPTER 9 μil ˆ 5μil ˆ 4 F ˆ = + j= j = (88 N) j π d 3d πd (c) F 3 = (because of symmetry) (d) F (e) F = F = ( 88 N)j = F = 4 (469 N)j ˆ, and 37 We use Eq 9-3 and the superposition of forces: the situation is as shown on the right ˆ F4 = F4 + F4 + F34 With θ = 45, The components of F 4 are given by μ i μ i cos 45 3μ i F4x = F43 F4cosθ = = pa pa 4pa and μi μi sin 45 μi F4y = F4 F4sin θ = = pa pa 4pa Thus, ( x y) ( π )( ) π ( ) 3μi μ 4 T m A 75A i μi 4 = = + = = F F F 4 3 N/m = 4πa 4πa 4πa 4 35m and F 4 makes an angle φ with the positive x axis, where In unit-vector notation, we have F H G I = KJ F HG I K J = F 4 y φ = tan tan 6 F 3 4 x F = (3 N/m)[cos6 ˆi + sin6 ˆj] = ( 5 N/m)i ˆ+ (47 N/m)j ˆ 38 (a) The fact that the curve in Fig 9-64(b) passes through zero implies that the currents in wires and 3 exert forces in opposite directions on wire Thus, current i points out of the page When wire 3 is a great distance from wire, the only field that affects wire is that caused by the current in wire ; in this case the force is negative according to Fig 9-64(b) This means wire is attracted to wire, which implies (by the discussion in Section 9-) that wire s current is in the same direction as wire s

12 4 current: out of the page With wire 3 infinitely far away, the force per unit length is given (in magnitude) as 67 N/m We set this equal to F = μ ii /πd When wire 3 is at x = 4 m the curve passes through the zero point previously mentioned, so the force between and 3 must equal F there This allows us to solve for the distance between wire and wire : d = (4 m)(75 A)/(5 A) = m Then we solve 67 N/m= μ o i i /πd and obtain i = 5 A (b) The direction of i is out of the page 39 Using a magnifying glass, we see that all but i are directed into the page Wire 3 is therefore attracted to all but wire Letting d = 5 m, we find the net force (per meter length) using Eq 9-3, with positive indicated a rightward force: which yields F / = 8 N/m F μ3 i i i i4 i5 = π d d d d 4 Using Eq 9-3, the force on, say, wire (the wire at the upper left of the figure) is along the diagonal (pointing toward wire 3, which is at the lower right) Only the forces (or their components) along the diagonal direction contribute With θ = 45, we find the force per unit meter on wire to be μi μi 3 μi F= F+ F3+ F4 = F cosθ + F3= cos 45 + = πa πa π a ( 4π T m A)( 5A) 3 3 = = N/m π ( 85 m) The direction of F is along r ˆ = (i ˆ ˆj)/ In unit-vector notation, we have -3 ( N/m) ˆ ˆ -4 (i j) (794 N/m)i ˆ -4 F = = + ( 794 N/m)j ˆ 4 The magnitudes of the forces on the sides of the rectangle that are parallel to the long straight wire (with i = 3 A) are computed using Eq 9-3, but the force on each of the sides lying perpendicular to it (along our y axis, with the origin at the top wire and +y downward) would be figured by integrating as follows:

13 4 CHAPTER 9 F =za+ b sides a iμ i πy dy Fortunately, these forces on the two perpendicular sides of length b cancel out For the remaining two (parallel) sides of length L, we obtain μ ii L μ iib F = = π a a+ d πa a+ b ( ) ( π )( )( )( )( ) 4 T m/a 3A A 8cm 3 m 3 3 N, = = π cm 8cm ( + ) and F points toward the wire, or + ĵ That is, 4 The area enclosed by the loop L is A= ( 4d)( 3d) = 6d Thus c 3 F = (3 N) ˆj in unit-vector notation ds μ i μ ja ( )( )( )( ) = = = π = T m A 5A m 6 m 45 T m 43 We use Eq 9- = μ ir /πa for the -field inside the wire ( r< a) and Eq 9-7 = μ i/πr for that outside the wire (r > a) (a) At r =, = (b) At r = m, μ ir π πa π(m) (4 T m/a)(7a)(m) 4 = = = 85 T (c) At r= a= m, μ ir π πa π(m) (4 T m/a)(7a)(m) 3 = = = 7 T μ i (4π T m/a)(7a) 4 (d) At r = 4m, = = = 85 T πr π(4m) z 44 We use Ampere s law: ds = μ i, where the integral is around a closed loop and i is the net current through the loop (a) For path, the result is ds = μ ( + ) = π ( ) = (b) For path, we find 7 6 5A 3A (4 T m/a) A 5 T m

14 43 ds = μ ( ) = π ( ) = 7 5 5A 5A 3A (4 T m/a) 3A 6 T m 45 (a) Two of the currents are out of the page and one is into the page, so the net current enclosed by the path is A, out of the page Since the path is traversed in the clockwise sense, a current into the page is positive and a current out of the page is negative, as indicated by the right-hand rule associated with Ampere s law Thus, ds = μ = ( ) = 6 i (4π T m/a) A 5 T m (b) The net current enclosed z by the path is zero (two currents are out of the page and two are into the page), so ds = μ ienc = 46 A close look at the path reveals that only currents, 3, 6 and 7 are enclosed Thus, noting the different current directions described in the problem, we obtain ds = μ ( i i+ i+ i) = μ i= ( π )( ) = T m/a 45 A 83 T m 47 For r a, μenc i μ r μ r r μjr ( r) = = J ( r) rdr J rdr r r p = = p p p p a 3a (a) At r =, = (b) At r= a/, we have ( ) r 3 μjr (4π T m/a)(3a/m )(3 m / ) = = = 3 3a 3(3 m) (c) At r= a, 3 μja (4π T m/a)(3a/m )(3 m) ( r= a) = = = 3 3 T 4 T 48 (a) The field at the center of the pipe (point C) is due to the wire alone, with a magnitude of μwire i μwire i C = = π 3R 6πR ( ) For the wire we have P, wire > C, wire Thus, for P = C = C, wire, i wire must be into the page:

15 44 CHAPTER 9 μ i μ i = = πr π wire P P,wire P,pipe ( R) Setting C = P we obtain i wire = 3i/8 = 3 3 3(8 A) / 8 3 A = (b) The direction is into the page 49 (a) We use Eq 9-4 The inner radius is r = 5 cm, so the field there is ( 4π T m/a)( 8 A)( 5) π( ) μ in πr 5 m 4 = = = 533 T (b) The outer radius is r = cm The field there is ( 4π T m/a)( 8 A)( 5) π( ) μ in πr m 4 = = = 4 T 5 It is possible (though tedious) to use Eq 9-6 and evaluate the contributions (with the intent to sum them) of all loops to the field at, say, the center of the solenoid This would make use of all the information given in the problem statement, but this is not the method that the student is expected to use here Instead, Eq 9-3 for the ideal solenoid (which does not make use of the coil radius) is the preferred method: F N = in= i H G I μ μ K J where i = 36 A, = 95 m, and N = This yields = 57 T 5 It is possible (though tedious) to use Eq 9-6 and evaluate the contributions (with the intent to sum them) of all loops to the field at, say, the center of the solenoid This would make use of all the information given in the problem statement, but this is not the method that the student is expected to use here Instead, Eq 9-3 for the ideal solenoid (which does not make use of the coil diameter) is the preferred method: F N = in= i H G I μ μ K J where i = 3 A, = 5 m, and N = This yields 4 = 3 T 5 We find N, the number of turns of the solenoid, from the magnetic field = μ in = μoin / : N = / μi Thus, the total length of wire used in making the solenoid is

16 45 πrn 3 πr π 6 m 3 T 3 m = = μ i 4π T m/a 8 A c hc hb g c hb g = 8 m 53 The orbital radius for the electron is which we solve for i: mv i = = eμ nr = 7A mv r = mv e = e μ ni 3 8 ( 9 kg)( 46)( 3 m s) 9 ( 6 C)( 4π T m A)( m)( 3 m) 54 As the problem states near the end, some idealizations are being made here to keep the calculation straightforward (but are slightly unrealistic) For circular motion (with speed, v, which represents the magnitude of the component of the velocity perpendicular to the magnetic field [the field is shown in Fig 9-9]), the period is (see Eq 8-7) T = πr/v = πm/e Now, the time to travel the length of the solenoid is t = L/ v where v is the component of the velocity in the direction of the field (along the coil axis) and is equal to v cos θ where θ = 3º Using Eq 9-3 ( = μ in) with n = N/L, we find the number of revolutions made is t /T = (a) We denote the fields at point P on the axis due to the solenoid and the wire as s and w, respectively Since s is along the axis of the solenoid and w is perpendicular to it, s w For the net field to be at 45 with the axis we then must have s = w Thus, μ iw s = μ isn= w = π d, which gives the separation d to point P on the axis: (b) The magnetic field strength is iw 6A d = = = 477 cm π in π A turns cm s 3 ( )( ) ( )( )( ) = = π = 3 5 s 4 T m A A turns m 355 T

17 46 CHAPTER 9 56 We use Eq 9-6 and note that the contributions to P from the two coils are the same Thus, P ( π ) ( ) μir N 8μNi 8 4 T m/a () A = = = = 3 R + ( R ) 5 5R 5 5( 5m) P is in the positive x direction T 57 (a) The magnitude of the magnetic dipole moment is given by μ = NiA, where N is the number of turns, i is the current, and A is the area We use A = πr, where R is the radius Thus, μ = NiπR = 3 4 A π 5m = 4A m b gb g b g (b) The magnetic field on the axis of a magnetic dipole, a distance z away, is given by Eq 9: = μ μ 3 π z We solve for z: 3 3 μ ( 4 T m A)( 36A m ) μ π z = = = 46cm 6 π π ( 5 T) 58 (a) We set z = in Eq 9-6 (which is equivalent using to Eq 9- multiplied by the number of loops) Thus, () i/r Since case b has two loops, b i Rb Ra = = = 4 i R R a a b (b) The ratio of their magnetic dipole moments is b iab Rb a iaa Ra μ = = = = = 5 μ 59 The magnitude of the magnetic dipole moment is given by μ = NiA, where N is the number of turns, i is the current, and A is the area We use A = πr, where R is the radius Thus, μ = 3A π 5 m = 47A m b gb g b g 6 Using Eq 9-6, we find that the net y-component field is

18 47 y μ ir μ ir = π( R z ) ( R z ) 3/ 3/ + π +, where z = L (see Fig 93(a)) and z = y (because the central axis here is denoted y instead of z) The fact that there is a minus sign between the two terms, above, is due to the observation that the datum in Fig 93(b) corresponding to y = would be impossible without it (physically, this means that one of the currents is clockwise and the other is counterclockwise) (a) As y, only the first term contributes and (with y = 7 6 T given in this case) we can solve for i We obtain i = (45/6π) Α 9 A (b) With loop at y = 6 m (see Fig 93(b)) we are able to determine i from μ ir μ ir ( R + L ) ( R + y ) = 3/ 3/ We obtain i = (7 3 /5π) Α 7 A 6 (a) We denote the large loop and small coil with subscripts and, respectively μ i 4π TmA 5A = = R m c hb g b g = 79 (b) The torque has magnitude equal to τ = μ = μ sin9 = N i A = πn i r ( )( )( 3A 8 m) ( 79 5 T) 5 = π 5 = 6 N m 6 (a) To find the magnitude of the field, we use Eq 9-9 for each semicircle (φ = π rad), and use superposition to obtain the result: ( ) μ (4 T m/a) 56A iπ μiπ μi π = + = + = + 4πa 4πb 4 a b 4 57m 936m = 497 T (b) y the right-hand rule, points into the paper at P (see Fig 9-6(c)) T (c) The enclosed area is has magnitude A a b = ( π + π )/, which means the magnetic dipole moment

19 48 CHAPTER 9 πi π (56A) = + = + = 3 μ ( a b ) [(57m) (936m) ] 6 A m (d) The direction of μ is the same as the found in part (a): into the paper 63 y imagining that each of the segments bg and cf (which are shown in the figure as having no current) actually has a pair of currents, where both currents are of the same magnitude (i) but opposite direction (so that the pair effectively cancels in the final sum), one can justify the superposition (a) The dipole moment of path abcdefgha is ˆ ˆ ˆ μ = μ j i i ˆ bc f gb + μabgha + μcde f c = ia + = ia j ( )( ) ( )( ) ˆ = 6A m j = (6 A m ) ˆj (b) Since both points are far from the cube we can use the dipole approximation For (x, y, z) = (, 5 m, ), μ π y π(5 m) 6 μ (6 T m/a)(6 m A) j (, 5 m, ) = = (96 T ) ˆj (a) The radial segments do not contribute to P, and the arc segments contribute according to Eq 9-9 (with angle in radians) If k^ designates the direction "out of the page" then μi(7 π /4 rad) ˆ μi(7 π /4 rad) k kˆ P = 4 π(4 m) 4 π( m) ˆ where i = A This yields = 5 8 k^ T, or = 75 8 T (b) The direction is ˆk, or into the page 65 Using Eq 9-, μ i π R = r, we find that r = 8 m gives the desired field value 66 (a) We designate the wire along y = r A = m wire A and the wire along y = r = 5 m wire Using Eq 9-4, we have

20 5 CHAPTER 9 68 We take the current (i = 5 A) to flow in the +x direction, and the electron to be at a point P, which is r = 5 m above the wire (where up is the +y direction) Thus, the field produced by the current points in the +z direction at P Then, combining Eq 9-4 with Eq 8-, we obtain Fe = eμ i pr v k (a) The electron is moving down: v b ge j = v j (where v = 7 m/s is the speed) so or 6 F e = 3 N F e 6 ( ) eμiv = ˆi = (3 N)i ˆ, pr (b) In this case, the electron is in the same direction as the current: v = v i so or 6 F e = 3 N F e 6 ( ) eμiv = ˆj = (3 N)j ˆ, π r (c) Now, v =± vk so F e k k = 69 (a) y the right-hand rule, the magnetic field (evaluated at a) produced by wire (the wire at bottom left) is at φ = 5 (measured counterclockwise from the +x axis, in the xy plane), and the field produced by wire (the wire at bottom right) is at φ = y symmetry d = i we observe that only the x-components survive, yielding μ i = ˆ + = = π 5 cos 5 i ( 346 T)i where i = A, = m, and Eq 9-4 has been used To cancel this, wire b must carry current into the page (that is, the k direction) of value r 5 (87 m) ib π ( 346 T) π = = = 5 A μ 4π T m/a where r = 3 = 87 m and Eq 9-4 has again been used (b) As stated above, to cancel this, wire b must carry current into the page (that is, the direction) ˆ z

21 57 to the sheet and only has a horizontal component That is, the field at P must be purely horizontal, as drawn in Fig 9-83 z (b) The path used in evaluating ds is rectangular, of horizontal length Δx (the horizontal sides passing through points P and P' respectively) and vertical size δy > Δy The vertical sides have no contribution to the integral since is purely horizontal (so the scalar dot product produces zero for those sides), and the horizontal sides contribute two equal terms, as shown next Ampere s law yields Δ x= μλδx = μλ 8 Equation 9-7 applies for each wire, with r = R + d / g (by the Pythagorean theorem) The vertical components of the fields cancel, and the two (identical) horizontal components add to yield the final result μ i d / μ id = π = = r r π R + d/ 6 5 T ( ( ) ) where (d/)/r is a trigonometric factor to select the horizontal component It is clear that this is equivalent to the expression in the problem statement Using the right-hand rule, we find both horizontal components point in the +x direction Thus, in unit-vector 6 notation, we have = (5 T)i ˆ 83 The two small wire segments, each of length a/4, shown in Fig 9-85 nearest to point P, are labeled and 8 in the figure (below left) Let ˆk be a unit vector pointing into the page b, We use the result of Problem 9-7: namely, the magnetic field at P (shown in Fig 9-43 and upper right) is