CHAPTER 30. Answer to Checkpoint Questions. 1. (a), (c), (b) 2. b, c, a 3. d, tie of a and c, then b 4. (d), (a), tie of (b) and (c) (zero)

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1 800 CHAPTER 30 AMPERE S LAW CHAPTER 30 Answer to Checkpoint Questions. (a), (c), (b). b, c, a 3. d, tie of a and c, then b. (d), (a), tie of (b) and (c) (zero) Answer to Questions. (c), (d), then (a) and (b) tie. (a) into; (b) greater 3. and. (c), (a), (b) 5. (a), (b), (c) (see Eq. 30-) 6. (c), (a), (b) 7. (b), (d), (a) (zero) 8. (a), 3, ; (b) less than 9. (a), +x;, y; (b), +y;, +x 0. a, b, c. outward. a, tie of b and d, then c 3. c and d tie, then b, a. b, a, d, c (zero) 5. (d), then tie of (a) and (e), then (b), (c) 6. (a) opposite ; (b) and opposite 6; (c) and 5 opposite 3 and 6; (d) and 5 opposite, 3, and 7. 0 (scalar product is zero)

2 CHAPTER 30 AMPERE S LAW (a) a; (b) 3d Solutions to Exercises & Problems E Solve i from Eq. 30-6: i = rb 0 = (88:0 0 m)(7: T) 0 7 Tm/A = 3: A : E Use Eq. 30-6: B = 0i r = ( 0 7 Tm/A)(50 A) (:6 0 3 m=) = 7:7 0 3 T : 3E (a) The magnitude of the magnetic eld due to the current in the wire, at a point a distance r from the wire, is given by B = 0i r : Put r = 0 ft = 6:0 m. Then B = ( 0 7 Tm/A)(00 A) (6:0 m) = 3:3 0 6 T = 3:3 T : (b) This is about one-sixth the magnitude of the Earth's eld. It will aect the compass reading. E The current i due to the electron ow is i = ne = (5:6 0 =s)(:6 0 9 C) = 9:0 0 5 A. Thus B = 0i r = ( 0 7 Tm/A)(9:0 0 5 A) (:5 0 3 m) = : 0 8 T:

3 80 CHAPTER 30 AMPERE S LAW 5E Use B(x; y; z) = ( 0 =)i s r=r 3, where s = s j and r = x i + y j + zk. Thus (a) B(x; y; z) = 0 i s j (xi + y j + zk) (x + y + z ) 3= = 0i s (z i x k) (x + y s + z ) 3= : B(0; 0; 5:0 m) = ( 0 7 Tm/A)(:0 A)(3:0 0 m)(5:0 m) i [ (5:0 m) ] 3= (b) B(0; 6:0 m; 0), since x = z = 0. (c) = (: 0 ) T : B(7:0 m; 7:0 m; 0) = ( 0 7 Tm/A)(:0 A)(3:0 0 m)( [(7:0 m) + (7:0 m) + 0 ] 3= = (:3 0 k) T : 7:0 m) k (d) B( 3:0 m; :0 m; 0) = ( 0 7 Tm/A)(:0 A)(3:0 0 m)(3:0 m) k [( 3:0 m) + ( :0 m) + 0 ] 3= = (: 0 0 k) T : 6E The points must be along a line parallel to the wire and a distance r from it, where r satises B wire = 0i r = B ext ; or r = 0i = (:6 0 6 T m/a)(00 A) = :0 0 3 m : B ext (5:0 0 3 T) 7E (a) The eld due to the wire, at a point 8:0 cm from the wire, must be 39 T and must be directed due south. Since B = =r, i = rb 0 = (0:080 m)( T) 0 7 T m/a = 6 A :

4 CHAPTER 30 AMPERE S LAW 803 (b) The current must be from west to east to produce a eld to the south at points below it. 8E Set up a coordinate system as shown to the right. The B-eld at the location of the charge q is z Thus B = 0i d ( k) : F q = qv B = 0i d ( k v) : (a) Now v = v( i) so F q = 0iv d k ( i) = 0iqv d ( j) ; charge x d current i.e., F q has a magnitude of qv=d and is directed against the current direction. (b) Now the direction v is reversed so F q = qv j=d. y 9E The straight segment of the wire produces no magnetic elds at C. The eld from the two semi-circular loop cancel at C. So B C = 0. 0E Use the same coordinated system as in 8E. Then F e = ( there. (a) Now v = v( i) so e v=d)( k v), as obtained F e = 0iev j d = (3: 0 6 N) j : (c) Now v = v k so F e / k v = 0: = (:6 0 6 T m/a)(50 A)(:6 0 9 C)(:0 0 7 m/s)j (5:0 0 m) P (a) The straight segment of the wire produces no magnetic eld at C. (b) For the semicircular loop B C = 0 Z loop = ( 0i)(r) R jds rj r 3 = 0i R : = 0i Z loop r ds r 3

5 80 CHAPTER 30 AMPERE S LAW B points into the page. (c) The same as (b), since the straight segments do not contribute to B. P Since sections AH and JD do not contribute to B C, we only need to consider the two arcs. For the smaller arc B C = 0 Z i ds r r 3 = 0i Z R ds R 3 e = 0ie R ; where e is a unit vector pointing into the page. Similarly, for the larger arc B C = 0ie R : Thus B C = B C + B C = (R R )e R R : 3P First nd the magnetic eld of a circular arc at its center. Let ds be an innitesimal segment of the arc and r be the vector from the segment to the arc center. ds and r are perpendicular to each other, so the contribution of the segment to the eld at the center has magnitude ds r db = 0ids r : The eld is into the page if the current is from left to right in the diagram and out of the page if the current is from right to left. All segments contribute magnetic elds in the same direction. Furthermore r is the same for all of them. Thus the magnitude of the total eld at the center is given by B = 0is r = 0i r : Here s is the arc length and is the angle (in radians) subtended by the arc at its center. The second expression was obtained by replacing s with r. must be in radians for this expression to be valid. P

6 CHAPTER 30 AMPERE S LAW 805 Now consider the circuit of Fig. 30{a. The magnetic eld produced by the inner arc has magnitude =b and is out of the page. The eld produced by the outer arc has magnitude =a and is into the page. The two straight segments of the circuit do not produce elds at the center of the arcs because the vector r from any point on them to the center is parallel or antiparallel to the current at that point. If the positive direction is out of the page then the total magnetic eld at the center is B = 0i Since b < a the total eld is out of the page. b a : P Label the various sections of the wires as shown. Firstly, the sections a and c do not coutribute to B at point O, as point O lies on the straight line coinciding with a and c. Secondly, the B-eld due to the curved sections b and b cancel each other at point O. This leaves us just a and c. Finally, if we relocate c to c 0, its contribution to the B-eld at O will not change. Note that a and c 0 together do form an innite straight wire carrying a current i to the left. c ' b c R O b c a a i i 5P (a) The contribution to B a due to the straight sections of the wire is B a = 0ie R ; where e is a unit vector which points out of the page. For the bent section (see P) Thus B a = 0ie R : B a = B a + B a = 0i R + e = ( 0 7 Tm/A)(0 A) (5:0 0 3 m) + e = (:0 0 3 T) e : (b) Now we only need to consider the two straight wires: B b = 0ie R = ( 0 7 Tm/A)(0 A)e (5:0 0 3 m) = (8:0 0 T) e :

7 806 CHAPTER 30 AMPERE S LAW Here the factor of in the numerator is due to the fact that two wires are involved. 6P Sum the elds of the two straight wires and the circular arc. Look at the derivation of the expression for the eld of a long straight wire, leading to Eq. 30{6. Since the wires we are considering are innite in only one direction the eld of either of them is half the eld of an innite wire. That is, the magnitude is =r, where r is the distance from the end of the wire to the center of the arc. It is the radius of the arc. The elds of both wires are out of the page at the center of the arc. Now nd an expression for the eld of the arc, at its center. Divide the arc into innitesimal segments. Each segment produces a eld in the same direction. If ds is the length of a segment the magnitude of the eld it produces at the arc center is ( =r ) ds. If is the angle subtended by the arc in radians, then r is the length of the arc and the total eld of the arc is =r. For the arc of the diagram the eld is into the page. The total eld at the center, due to the wires and arc together, is B = 0i r + 0i r For this to vanish must be radians. r = 0i ( ) : r 7P Put the x axis along the wire with the origin at the midpoint and the current in the positive x direction. All segments of the wire produce magnetic elds at P that are into the page so we simply divide the wire into innitesimal segments and sum the elds due to all the segments. The diagram shows one innitesimal segment, with width dx. According to the Biot-Savart law the magnitude of the eld it produces at P is given by db = 0i sin r dx : and r are functions of x. Replace r with p x + R and sin with R=r = R= p x + R, then integrate from x = L= to x = L=. The total eld is B = 0iR Z L= L= dx (x + R ) = 0iR 3= R i dx 0 x θ x L= (x + R ) = If L R then R in the denominator can be ignored and B = 0i R L= R r = 0i R P L p L + R :

8 CHAPTER 30 AMPERE S LAW 807 is obtained. This is the eld of a long straight wire. For points close to a nite wire the eld is quite similar to that of an innitely long wire. 8P Follow the same steps as in the solution of 7P above but replace R with D, change the lower limit of integration to L, and change the upper limit to 0. The magnitude of the total eld is B = 0iD Z 0 L dx (x + D ) = 0iD 3= D x (x + D ) = 0 L = 0i L p D L + D : 9P You can easily check that each of the four sides produces the same magnetic eld B at the center of the square. Apply the result of 7P for B and let R = a= and L = a to obtain B = B = 0i a (a=) (a + a ) = p : = a 0P The B-eld produced by the four sides of the rectangle have the same direction. For each of the two longer sides (see 7P) B L = Similarly for each of the shorter sides Thus For L W B W = B = B L + B W = = 0i(L + W ) = LW L (W=) (L + W ) : = 0i W (L=) (L + W ) : = L W (L + W ) = + : B! 0iL LW = 0i W ; which is consistent with Eq for W = d and x = 0. 0 W L(L + W ) =

9 808 CHAPTER 30 AMPERE S LAW P When the elds of the four sides are summed vectorially the horizontal components add to zero. The vertical components are all the same, so the total eld is given by Thus B total = B cos = Ba R = B total = Ba p x + a : a (x + a ) p x + a : For x = 0 the expression reduces to B total = 0ia a p a = p a in agreement with the result of 9P. ; P The square has sides of length L=. The magnetic eld at the center of the square is given by the result of 9P, with a = L= and x = 0. It is B sq = 8p L = :3 0i L : The radius of the circle is R = L=. Use Eq. 30{8 of the text, with R = L= and z = 0. The eld is B circ = L = 9:87 0i L : The square produces the larger magnetic eld. 3P Since ds is parallel to r, B Q / Z ids r r 3 = 0 : O a P Let e be a unit vector unit vector pointing into the page. At point P a i r B P = 0 Z ids r r Z 3 a = 0ie a dx 0 r 3 p e = : 8a = 0ie = 0ie Z sin ds Z a 0 r a dx (a + x ) 3= θ ds x

10 CHAPTER 30 AMPERE S LAW 809 P Obviously, B P 3 = B P 6 = 0, and (see 3P). Thus B P = B P = B P = B P 5 = p 8a ; p 8(a) i 3 P B P = 6X n= B P n = (B P + B P )e (B P + B P 5 )e p p p e = e = ; 8a 6a 8a where e is a unit vector pointing into the page P Let e be a unit vector pointing into the page. Use the results of 8P and 3P to calculate B P through B P 8 : p p B P = B P 8 = 8(a=) = a ; p p B P = B P 5 = 8(3a=) = 6a ; B P = B P 7 = 0i (a=) = 3a= [(3a=) + (a=) ] = = 3 0i p ; 0a and Finally, B P 3 = B P 6 = B P = 8X n= = 0i a B P n e (3a=) = a= [(a=) + (3a=) ] = = p p p = ( 0 7 Tm/A)(0 A) (8:0 0 m) = (:0 0 T) e ; e 3 p 0 p P 0i 3 p 0a : p p p 0 e 5 i 3

11 80 CHAPTER 30 AMPERE S LAW where e is a unit vector pointing into the page. 6P Consider a section of the ribbon of thickness dx located a distance x away from point P. The current it carries is di = i dx=w, and its contribution to B P is Thus B P = and B P points upward. Z db P = 0di x = 0idx xw : Z d B P = d+w 0i w d dx x = 0i w ln + w ; d 7E (a) If the currents are parallel, the two elds are in opposite directions in the region between the wires. Since the currents are the same the total eld is zero along the line that runs halfway between the wires. There is no possible current for which the eld does not vanish. (b) If the currents are antiparallel, the elds are in the same direction in the region between the wires. At a point halfway between they have the same magnitude, =r. Thus the total eld at the midpoint has magnitude B = =r and i = rb 0 = (0:00 m)( T) 0 7 Tm/A = 30 A : 8E The point P at which B P = 0 form a line parallel to both currents passing through the line joining the two wires, as shown. Note that B P / i =r, and B P / i =(d r ). Let B P = B P to obtain i =r = i =(d r ). Solve for r : i =3i i =i r d-r d r = i d i + i = 3id 3i + i = 3d : 9E The current i carried by wire must be out of the page. Since B P / i =r where i = 6:5 A and r = 0:75 cm + :5 cm = :5 cm, and B P / i =r where r = :5 cm, from B P = B P we get :5 cm i = i r r = (6:5 A) :5 cm = :3 A :

12 CHAPTER 30 AMPERE S LAW 8 30E Lable these wires through 5, left to right. Then F = 0i d + d + 3d + d j = 5 d = (3)( 0 7 Tm/A)(3:00 A) (:00 m)j (8:00 0 m) = (: T) j ; F = 0i d + 5 j = 3d d F 3 = 0 (because of symmetry); F = F ; and F 5 = F. j = (: T) j ; j 3E Consider, for example, x > d. Then in Fig the direction of B b is reversed, and B(x) = Ba (x) B b (x) = (d + x) (x d) = d (d x ) : 3E (a) Refer to the gure to the right. We have B P = jb L + B R j = B L cos = d p cos = ( 0 7 Tm/A)(00 A)(cos 5 ) p (0 0 m) = :0 0 T : B L B θ P θ B R d d/ / B P points to the left. (b) Reverse the direction of B R in the gure above but keep its magnitude unchanged. Obviously B P still has a magnitude of :0 0 T but now points up the page. 33P For 0 < x < d B(x) = B a (x) B b (x) = (d + x) (d x) = x (x d ) ;

13 8 CHAPTER 30 AMPERE S LAW and for x > d B(x) = B a (x) + B b (x) = (x + d) + (x d) = x (x d ) : 0 0 B(x) (mt) x (cm) The same expression for B(x) is valid for x < 0 due to symmetry. 3P Each wire produces a eld with magnitude given by B = =r, where r is the distance from the corner of the square to the center. According to the Pythagorean theorem the diagonal of the square has length p a, so r = a= p and B = = p a. The elds due to the wires at the upper left and lower right corners both point toward the upper right corner of the square. The elds due to the wires at the upper right and lower left corners both point toward the upper left corner. The horizontal components cancel and the vertical components sum to B total = 0i p a cos 5 = 0i a = ( 0 7 T m/a)(0 A) (0:0 m) = 8:0 0 5 T : In the calculation cos 5 was replaced with = p. The total eld points up the page.

14 CHAPTER 30 AMPERE S LAW 83 35P Refer to the gure as shown. For example, the force on wire is F = jf + F 3 + F j θ θ=5o F F 3 = F cos + F 3 = cos 5 + 0i a p a = 0:338 : a F F points from wire to the center of the square. 3 36P Use F = F + F + F 3 : The components of F are then given by F x = F 3 F cos = 0i a = 3 0i a cos 5 p a y F and Thus F y = F = 0i a = 0i a : F sin sin 5 p a F = (F x + F y) = = F 3 θ F θ=5 o 3 " 3 # = p 0 + = a a a ; x and F makes an angle with the positive x axis, where Fy = tan = tan F x 3 = 6 :

15 8 CHAPTER 30 AMPERE S LAW 37P (a) From the gure to the right B P = jb + B j = B cos d= = r r d = [R + (d=) ] d = (R + d ) : (b) B P points to the right, as shown. θ d θ r r P B θ θ B B P 38P The forces on the two sides of length b cancel out. For the remaining two sides F = 0i i L a a + d = i i b a(a + b) = (:6 0 6 T m/a)(30 A)(0 A)(8:0 cm)(30 0 m) (:0 cm + 8:0 cm) = 3: 0 3 N ; and F points toward the wire. 39P Consider a segment of the projectile between x and x + dx. Use Eq to nd the magnetic force on the segment: df = df + df = i(dx i) B (x) + i(dx i) B (x) = i[b (x) + B (x)] j dx = i x + (R + w x) j dx ; where j is a unit vector pointing to the right. Here we used the expression for the magnetic eld of a semi-innite wire. Thus R W R O x projectile i i i wire j wire y

16 CHAPTER 30 AMPERE S LAW 85 F = Z df = i Z R+w R x + R + w (b) Use K = mv f = W ext = R Fds for the projectile: = = " x j dx = 0i # = ln + w R Wext Z L v f = m m 0 ln + w dy R ( 0 7 Tm/A)( A) (:0 m) ln( + : cm=6:7 cm) = (0 0 3 kg) = :3 0 3 m/s : j : = 0E (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 :0 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 I Bds = = (:0 A)( 0 7 Tm/A) = :5 0 6 T m : (b) The net current enclosed by the path is zero (two currents are out of the page and two are into the page), so H Bds = = 0. E A close look at the path reveals that only currents no. and no.6 are enclosed. Thus I B ds = 0 (6i 0 ) = 5 : E The area enclosed by the loop L is A = (d)(3d) = 6d. Thus I c B ds = = 0 ja = ( 0 7 Tm/A)(5 A/m )(6)(0:0 m) = :5 0 6 Tm : 3E Use Eq. 30- for the B-eld inside the wire and 30-9 for that ouside the wire. The plot is shown in the next page.

17 86 CHAPTER 30 AMPERE S LAW B (mt) 0.8 inside outside r (cm) P Use Ampere's law: H Bds =, where the integral is around a closed loop and i is the net current through the loop. For path and for path I I B ds = 0 ( 5:0 A + 3:0 A) = ( :0 A)( 0 7 Tm/A) = :5 0 6 Tm ; B ds = 0 ( 5:0 A 5:0 A 3:0 A) = ( 3:0 A)( 0 7 Tm/A) = :6 0 5 Tm : 5P Use Ampere's law. For the dotted loop shown on the diagram i = 0. The integral R Bds is zero along the bottom, right, and top sides of the loop. Along the right side the eld is zero, along the top and bottom sides the eld is perpendicular to ds. If ` is the length of the left edge then direct integration yields H Bds = B`, where B is the magnitude of the eld at the left side of the loop. Since neither B nor ` is zero, Ampere's law is contradicted. We conclude that the geometry shown for the magnetic eld lines is in error. The lines actually bulge outward and their density decreases gradually, not precipitously as shown. 6P (a) For the circular path L of radius r concentric with the conductor I L B ds = rb(r) = enclosed = (r b ) (a b ) :

18 CHAPTER 30 AMPERE S LAW 87 Thus (b) At r = a; B(a) = B(r) = At r = b; B(b) / r b = 0. For b = 0 (a b ) r b a b (a b ) a r B(r) = 0i a r r = 0ir a : : = 0i a : 7P H Use s db = rb = enclosed, or B = enclosed =r. (a) r < c: r (b) c < r < b: (c) b < r < a: (d) r > a: B (mt) B(r) = 0i encl r B(r) = 0i encl r = 0i r B(r) = 0i encl r = 0i r c = 0i r : = 0ir c : (r b ) = 0i(a r ) (a b ) r(a b ) : B(r) = 0i encl = 0 : r (e) For example, check what happens if b = c. In this case the expressions in (a), (b) and (c) above should yield the same result at r = b = c. This is indeed the case. (f) r (cm)

19 88 CHAPTER 30 AMPERE S LAW 8P For r < a, B(r) = 0i enclosed r Z = r Z 0 J(r)r dr = r 0 r J 0 r dr = 0J 0 r : r 0 0 a 3a 9P The eld at the center of the pipe (point C) is due to the wire alone, with a magnitude of B C = 0i wire (3R) = 0i wire 6R : For the wire we have B P; wire > B C; wire, so for B P = B C = B C; wire ; i wire must be into the page: B P = B P; wire B P; pipe = 0i wire R (R) : Let B C = B P to obtain i wire = 3i=8. 50P (a) Take the magnetic eld at a point within the hole to be the sum of the elds due to two current distributions. The rst is the solid cylinder obtained by lling the hole and has a current density that is the same as that in the original cylinder with the hole. The second is the solid cylinder that lls the hole. It has a current density with the same magnitude as that of the original cylinder but it is in the opposite direction. Notice that if these two situations are superposed the total current in the region of the hole is zero. Recall that a solid cylinder carrying current i, uniformly distributed over a cross section, produces a magnetic eld with magnitude B = r=r a distance r from its axis, inside the cylinder. Here R is the radius of the cylinder. For the cylinder of this problem the current density is J = i A = i (a b ) ; where A [= (a b )] is the cross-sectional area of the cylinder with the hole. The current in the cylinder without the hole is I = JA = Ja = ia a b and the magnetic eld it produces at a point inside, a distance r from its axis, has magnitude B = 0I r r a = a a (a b ) = r (a b ) :

20 CHAPTER 30 AMPERE S LAW 89 The current in the cylinder that lls the hole is I = Jb = ib a b and the eld it produces at a point inside, a distance r from the its axis, has magnitude B = 0I r b = r b b (a b ) = r (a b ) : At the center of the hole this eld is zero and the eld there is exactly the same as it would be if the hole were lled. Place r = d in the expression for B and obtain B = d (a b ) : for the eld at the center of the hole. The eld points upward in the diagram if the current is out of the page. (b) If b = 0 the formula for the eld becomes B = 0id a : This correctly gives the eld of a solid cylinder carrying a uniform current i, at a point inside the cylinder a distance d from the axis. If d = 0 the formula gives B = 0. This is correct for the eld on the axis of a cylindrical shell carrying a uniform current. (c) The diagram shows the situation in a cross-sectional plane of the cylinder. P is a point within the hole, A is on the axis of the cylinder, and C is on the axis of the hole. The magnetic eld due to the cylinder without the hole, carrying a uniform current out of the page, is labeled B and the magnetic eld of the cylinder that lls the hole, carrying a uniform current into the page, is labeled B. The line from A to P makes the angle with the line that joins the centers of the cylinders and the line from C to P makes the angle with that line, as shown. B is perpendicular to the line from A to P and so makes the angle with the vertical. Similarly, B is perpendicular to the line from C to P and so makes the angle with the vertical. The x component of the total eld is B y B θ θ P r r x θ θ A d C B x = B sin B sin = r (a b ) sin = (a b ) [r sin r sin ] : r (a b ) sin

21 80 CHAPTER 30 AMPERE S LAW As the diagram shows r sin = r sin, so B x = 0. The y component is given by B y = B cos + B cos = = r (a b ) cos + (a b ) [r cos + r cos ] : The diagram shows that r cos + r cos = d, so B y = d (a b ) : r (a b ) cos This is identical to the result found in part (a) for the eld on the axis of the hole. It is independent of r, r,, and, showing that the eld is uniform in the hole. 5P (a) Suppose the eld is not parallel to the sheet, as shown in the upper diagram. Reverse the direction of the current. According to the Biot-Savart law the eld reverses, so it will be as in the second diagram. Now rotate the sheet by 80 about a line that perpendicular to the sheet. The eld, of course, will rotate with it and end up in the direction shown in the bottom diagram. The current distribution is now exactly as it was originally, so the eld must also be as it was originally. But it is not. Only if the eld is parallel to the sheet will be nal direction of the eld be the same as the original direction. If the current is out of the page any innitesimal portion of the sheet, in the form of a long straight wire, produces a eld that is to the left above the sheet and to the right below the sheet. The eld must be as drawn in Fig. 30{65. Integrate the tangential component of the magnetic eld B around the rectangular loop shown with dotted lines. The L upper and lower edges are the same distance from the current sheet and each has length L. This means the eld has the same magnitude along these edges. It points to the left along the upper edge and to the right along the lower. If the integration is carried out in the counterclockwise sense the contribution of the upper edge is BL, the contribution of the lower edge is also BL, and the contribution H of each of the sides is zero because the eld is perpendicular to the sides. Thus Bds = BL. The total current through the loop is L. Ampere's law yields BL = 0 L, so B = 0 =. B

22 CHAPTER 30 AMPERE S LAW 8 5P* (a) I L B ds = = Z Z d Z Z Z B ds 3 [3: + 8:0(x =d ) j] i dxj y=0 Z d 0 Z 0 d Z 0 d [3: + 8:0(x =d ) j] j dyj x=d [3: + 8:0(x =d ) j] i dxj y=d [3: + 8:0(x =d ) j] j dyj x=0 = (3:0d + 8:0d 3:0d) mt = (8:0d) mt ; y 3 (d, d, 0) path L x (d, 0, 0) where d is in meters. (b) From H L B ds = (8:0d) mt = 0i enclosed, we get i enclosed = (8:0d) mt 0 = (8:0)(0:50)(0 3 T) 0 7 Tm/A = 3: 03 A: (c) Since H L B ds > 0; i enclosed > 0, so the current is in the positive k direction. 53E The magnetic eld inside an ideal solenoid is given by Eq. 30{5. The number of turns per unit length is n = (00 turns)=(0:5 m) = 800 turns/m. Thus B = 0 ni 0 = ( 0 7 T m/a)(800 m )(0:30 A) = 3:0 0 T : 5E B = 0 n = ( 0 7 Tm/A)(3:60 A)(00=0:950 m) = 5:7 0 3 T : 55E Solve N, the number of turns of the solenoid, from B = n = N=L: N = BL= : Thus the length of wire is l = rn = rbl = (:60 0 m)(3:0 0 3 T)(:30 m) (:6 0 6 T m/a)(8:0 A) = 08 m :

23 8 CHAPTER 30 AMPERE S LAW 56E (a) Use Eq The inner radius is r = 5:0 cm so the eld there is B = 0iN r = ( 0 7 Tm/A)(0:800 A)(500) (0:50 m) = 5:33 0 T : (b) The outer radius is r = 0:0 cm. The eld there is B = 0iN r = ( 0 7 Tm/A)(0:800 A)(500) (0:00 m) = :00 0 T : 57E In this case L = r is roughly the length of the toroid so B = 0 N r = 0 ni 0 : This result is expected, since from the perspective of a point inside the toroid the portion of the toroid in the vicinity of the point resembles part of a long solenoid. 58P Use B = 0 n and note that ni 0 =. Thus B = 0 : Also from 5P we have B = 0 ( 0) = 0 as we move through an innite plane current sheet of current density. 59P Consider a circle of radius r, inside the toroid and concentric with it. The current that passes through the region between the circle and the outer rim of the toroid is Ni, where N is the number of turns and i is the current. The current per unit length of circle is = Ni=r and 0 is 0 Ni=r, the magnitude of the magnetic eld at the circle. Since the eld is zero outside a toroid, this is also the change in the magnitude of the eld encountered as you move from the circle to the outside. The equality is not really surprising in light of Ampere's law. You are moving perpendicularly to the magnetic eld lines. Consider an extremely narrow loop, with the narrow sides along eld lines and the two long sides perpendicular H to the eld lines. If B is the eld at one end and B is the eld at the other end then Bds = (B B )w, where w is the width of the loop. The current through the loop is w, so Ampere's law yields (B B )w = 0 w and B B = 0. 60P (a) Denote the B-elds at point P on the axis due to the solenoid and the wire as B s

24 CHAPTER 30 AMPERE S LAW 83 and B w, respectively. Since B s is along the axis of the solenoid and B w is perpendicular to it, B s? B w, respectively. For the net eld B to be at 5 with the axis we then must have B s = B w. Thus B B s B s = s n = B w = 0i w d ; which gives the separation d to point P on the axis: d = (b) i w i s n = 6:00 A (0:0 0 3 A)(0 turns/cm) = :77 cm : B w 5 o axis P B = p B s = p ( 0 7 Tm/A)(0:0 0 3 A)(0 turns=0:000 m) = 3: Tm : 6P Let and solve for i: r = mv eb = mv e 0 ni i = mv e 0 nr (9: 0 3 kg)(0:060)(3: m/s) = (: C)( 0 7 Tm/A)(00=0:000 m)(:30 0 m) = 0:7 A : 6E The magnitude of the dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. Use A = R, where R is the radius. Thus = (00)(0:30 A)(0:050 m) = 0:7 Am : 63E (a) Set z = n Eq. 30-8: B(0) / i=r. Since case b has two loops, B b B a = i=r b i=r a = R a R b = :

25 8 CHAPTER 30 AMPERE S LAW (b) b a = ia b ia a = R b R a = = : 6E Use Eq and note that the contribute to B P from the two coils are the same. Thus B P = B P is in the positive x direction. R N [R + (R=) ] 3= = 8 0Ni 5 p 5R : 65E (a) Similar to 6E, the magnitude of the dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. Use A = R, where R is the radius. Thus = NiR = (300)(:0 A)(0:05 m) = : Am : (b) The magnetic eld on the axis of a magnetic dipole, a distance z away, is given by Eq. 30{9: B = 0 z : 3 Solve for z: z = 0 =3 ( 0 7 Tm/A)(:36 Am =3 ) = = 6 cm : B (5:0 0 6 T) 66E (a) For x a the result of P reduces to B(x) a (x )(x ) = 0(ia ) ; = x 3 indeed the B-eld of a magnetic diploe (see Eq. 30-9). (b) = ia, by comparison between Eq and the result above. 67P (a) The two straight segments of the wire do not contribute to B P. For the larger semicircular loop B P = =b (see P), and for the smaller one B P = =a. Thus B p = B P + B P = 0i a + b :

26 CHAPTER 30 AMPERE S LAW 85 B P points into the page. (b) = A loop i = (a + b )i, pointing into the page. 68P Use Eq to obtain B(x) = 0iR " 3= + R + (x R=) R + (x + R=) 3= # : The plot of B(x) is as follows. Note that B(x) is almost a constant in the vicinity of x = B (mt) x (cm) 69P Denote the large and small loops with subscripts and, respectively. (a) B = 0i = ( 0 7 Tm/A)(5 A) = 7:9 0 5 T : R (0: m) (b) = j B j = B sin 90 = N i A B = N i r B = (50)(:3 A)(0:8 0 m) (7:9 0 5 T) = : 0 6 Nm : 70P (a) Contribution to B C from the straight segment of the wire: B C = 0i R :

27 86 CHAPTER 30 AMPERE S LAW Contribution from the circular loop: Thus B C points out of the page. (b) Now B C? B C so B C = 0i R : B C = B C + B C = 0i + : R B C = qb + C B = 0i C R and B C points at an angle out of the page, where r = tan BC = tan B C + : = 8 : 7P (a) By imagining that both bg and cf have a pair of currents of the same magnitude (= i) and opposite direction, you can justify the superposition. (b) = bcfgb + abgha + cdefc = (ia )( j = (6:0 A)(0:0 m) j = (6:0 0 m A) j : i + i) = ia j (c) Since both points are far from the cube we can use the dipole approximation. (x; y; z) = (0; 5:0 m; 0) For B(0; 5:0 m; 0) 0 y 3 = (:6 0 6 Tm/A)(6:0 0 m A) j (5:0 m) 3 = (9:6 0 T) j : For (x; y; z) = (5:0 m; 0; 0), note that the line joining the end point of interest and the location of the dipole is perpendicular to the axis of the dipole. You can check easily that if an electric diploe is used, the eld wold be E (= 0 )(p=x 3 ), which is half of the magnitude of E for a point on the y axis the same distance from the dipole. By analogy, in our case B is also half the value or B(0; 5:0 m; 0), i.e., B(5:0 m; 0; 0) = B(0; 5:0 m; 0) = (9:6 0 T) = :8 0 T :

28 CHAPTER 30 AMPERE S LAW 87 Just like the electric dipole case, B(5:0 m; 0; 0) points in the negative y direction. 7P (a) The magnitude of the magnetic eld on the axis of a circular loop, a distance z from the loop center, is given by Eq. 30{8: B = N R (R + z ) 3= ; where R is the radius of the loop, N is the number of turns, and i is the current. Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense and the elds they produce are in the same direction in the region between them. Place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the eld of the left-hand loop set z = x in the equation above. The chosen point on the axis is a distance s x from the center of the right-hand loop. To calculate the eld it produces put z = s x in the equation above. The total eld at the point is therefore B = N 0iR Its derivative with respect to x is db dx = N 0iR (R + x ) + : 3= (R + x sx + s ) 3= 3x (R + x ) + 3(x s) 5= (R + x sx + s ) 5= When this is evaluated for x = s= (the midpoint between the loops) the result is db = dx N 0iR 3s= 3s= = 0 ; (R + s =) 5= (R + s = s + s ) 5= s= independently of the value of s. (b) The second derivative is At x = s=, d B dx = N 0iR d B dx s= = N 0iR = N 0R " 3 (R + x ) 5= + 5x (R + x ) 7= # 3 (R + x sx + s ) + 5(x s) : 5= (R + x sx + s ) 7= 6 (R + s =) + 30s = 5= (R + s =) 7= 6(R + s =) + 30s = (R + s =) 7= = 3N R s R (R + s =) 7= : :

29 88 CHAPTER 30 AMPERE S LAW Clearly, this is zero if s = R. 73 (a) B from sum: 7: T; n = 5: T; 0% dierence; (b) B from sum: :03 0 T; n = :005 0 T; % dierence; (c) B from sum: :506 0 T; n = :53 0 T; 0:3% dierence 7 (a) each side contributes 3:0 7 Tm; the total is :60 6 Tm; = :60 6 Tm; (b) the sides contribute : T m, :8 0 7 T m, : T m, and : 0 7 T m; the total is :6 0 6 Tm; (c) the sides contribute :5 0 7 T m, : T m, :5 0 7 T m, and 5:9 0 7 T m; the total is :6 0 6 Tm; (d) the sides contribute : Tm, :7 0 7 Tm, : Tm, and 5:9 0 7 Tm; the total is essentially zero 75 (a) B = ( 0 =)[i =(x a) + i =x] j; (b) B = ( 0 =)(i =a)( + b=) j