1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?


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1 CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 33, with x, gives the magnetic field at the center of a circular loop, B µ a (with direction along the axis of the loop, consistent with the sense of circulation of the current and the righthand rule). Thus, the radius is 7 a µ B ( π 1 N/ A )( 15 A ) ( 8 µ T ) cm. 3. A.mlong wire carrying 3.5 A is wound into a tight, loopshaped coil 5. cm in diameter. What is the magnetic field at its center? The magnetic field at the center of each of the tightlywound turns is the same (Equation 33 with x ), so the net field is B N( µ a), where N is the number of turns. f the wire has length L and the coil has diameter D, then N Lπ D and 7 B ( L π D)( µ D) µ L π D ( 4 1 N/A )( 3. 5 A )(. m ) ( 5 cm ) 1. 3 mt. 5. Suppose Earth s magnetic field arose from a single loop of current at the outer edge of the planet s liquid core (core radius 3 km), concentric with Earth s center. What current would be necessary to give the observed field strength of 6 µ T at the north pole? (The currents responsible for Earth s field are more complicated than this problem suggests.) The north pole is on the axis of the hypothetical circular current loop, with radius a 3. Mm, at a distance 3 x Mm from the earth s center. Equation 33 gives a current of B( x + a ) µ a R E 3 7 ( 6 µ T )[( Mm ) + ( 3. Mm ) ] ( 4π 1 T m/ A) ( 3. Mm ) GA. 6. Earth s magnetic dipole moment is 8. 1 A m. What is the magnetic field strength on Earth s surface at either pole? The magnetic field strength at points on the axis of a magnetic dipole is given by Equation 34. The magnetic poles are two such points on the Earth s surface; see ig Thus, µ µ B 1 3 π (The main component of the Earth s magnetic field is a dipole.) R E H 7 N A K ( 8 1 A m ) T. 6 G. 6 3 ( m) 1. A single piece of wire is bent so that it includes a circular loop of radius a, as shown in ig A current flows in the direction shown. ind an expression for the magnetic field at the center of the loop.
2 76 CHAPTER 3 a GURE The field at the center is the superposition of fields due to current in the circular loop and straight sections of wire. The former is µ a out of the page (Equation 33 at x for CCW circulation), and the latter is µ π a out of the page (Equation 35 at y a for the very long, straight sections). Their sum is B ( 1 + π ) µ π a out of the page. 1. our long, parallel wires are located at the corners of a square 15 cm on a side. Each carries a current of.5 A, with the top two currents into the page in ig and the bottom two out of the page. ind the magnetic field at the center of the square. GURE The magnetic field from each wire has magnitude µ π( a ) (from Equation 35, with y a, the distance from a corner of a square of side a to the center). The righthand rule gives the field direction along one of the diagonals, as shown superposed on ig. 349, such that the fields from currents at opposite corners are parallel. The net field is 7 B1 + B + B3 + B4 ( B1 + B ) [ µ π ( a )]( cos 45 î) µ î π a ( 8 1 T )(. 5) î (. 15) µ Tî. 15. Part of a long wire is bent into a semicircle of radius a, as shown in ig A current flows in the direction shown. Use the BiotSavart law to find the magnetic field at the center of the semicircle (point P). The BiotSavart law (Equation 3) written in a coordinate system with origin at P, gives B( P) ( µ 4π) wire dl rɵ r, where ɵr is a unit vector from an element dl on the wire to the field point P. On the straight segments to the left and right of the semicircle, dl is parallel to ɵr or ɵ, r respectively, so dl ɵr. On the semicircle, dl is perpendicular to ɵr and the radius is constant, r a. Thus, µ dl µ π a µ B( P). 4π semicircle a 4π a 4a The direction of B(P), from the cross product, is into the page.
3 CHAPTER 3 77 GURE igure 351 shows a conducting loop formed from concentric semicircles of radii a and b. f the loop carries a current as shown, find the magnetic field at point P, the common center. GURE The BiotSavart law gives B( P) ( µ 4π ) loop dl rɵ r. dl ɵr r on the inner semicircle has magnitude dl a and direction out of the page, while on the outer semicircle, the magnitude is dl b and the direction is into the page. On the straight segments, dl rɵ, so the total field at P is ( µ 4π )[( π a a ) ( π b b )] µ ( b a) 4ab out of the page. (Note: the length of each semicircle is dl π r. ) 1. The structure shown in ig. 35 is made from conducting rods. The upper horiontal rod is free to slide vertically on the uprights, while maintaining electrical contact with them. The upper rod has mass g and length 95 cm. A battery connected across the insulating gap at the bottom of the lefthand upright drives a 66A current through the structure. At what height h will the upper wire be in equilibrium? GURE f h is small compared to the length of the rods, we can use Equation 36 for the repulsive magnetic force between the horiontal rods (upward on the top rod) µ l π h. The rod is in equilibrium when this equals its weight, mg, 7 hence h µ l π mg ( 1 N/ A )( 66 A ) (. 95 m ) ( N ) mm. (This is indeed small compared to 95 cm, as assumed.)
4 78 CHAPTER 3 4. A long, straight wire carries A. A 5.cm by 1cm rectangular wire loop carrying 5 ma is located. cm from the wire, as shown in ig ind the net magnetic force on the loop. At any given distance from the long, straight wire, the force on a current element in the top segment cancels that on a corresponding element in the bottom. The force on the near side (parallel currents) is attractive, and that on the far side (antiparallel currents) is repulsive. The net force is the sum, which can be found from Equation 36 (+ is attractive): µ 1l 1 1 H K H 7 N ( A) A ( 1 cm) π 7 K H K H14 K cm cm A cm 6 N. GURE A solenoid 1 cm in diameter is made with.1mmdiameter copper wire wound so tightly that adjacent turns touch, separated only by enamel insulation of negligible thickness. The solenoid carries a 8A current. n the long, straight wire approximation, what is the net force between two adjacent turns of the solenoid? Equation 36 provides an approximate value of the force per unit length between adjacent turns of l µ π d 7 ( 1 N/ A )( 8 A ) (. 1 mm ) mn/m. The length of one turn is π r 1π cm. 314 m, so the force on one turn is approximately (74.7 mn/m)(.314 m) 3.5 mn. 6. A long, flat conducting ribbon of width w is parallel to a long straight wire, with its near edge a distance a from the wire (ig. 354). Wire and ribbon carry the same current ; this current is distributed uniformly over the ribbon. Use integration to show that the force per unit length between the two has magnitude µ a + w ln. π w a The magnitude of the force per unit length on a thin strip of ribbon, of width dx, carrying current dx w, is given by Equation 36: d l µ ( dx w) π( a + x), where x is the distance from the near edge. ntegrating from x to w, we find: H K
5 CHAPTER 3 79 The force is attractive since the currents are parallel. w dx l µ µ π w a + x π w H K a + w ln. a GURE n ig. 355, 1 A flowing out of the page; 1 A, also out of the page, and 3 A, into the page. What is the line integral of the magnetic field taken counterclockwise around each loop shown? (a) (b) (e) (c) (d) GURE Ampère s law says that the line integral of B is equal to µ encircled, where, for CCW circulation, currents are positive 7 7 out of the page. (a) B dl µ + π π a ( 1 ) ( 4 1 N/ A )( 3 A ) 1 1 T m. (b) B dl µ b ( ). (c) B dl µ π c T m. (d) d B dl µ ( 3) 4π 1 T m. (e) e B dl 7 µ 4π 1 T m. 9. The magnetic field shown in ig has uniform magnitude 75 µ T, but its direction reverses abruptly. How much current is encircled by the rectangular loop shown?
6 71 CHAPTER 3 B 15 cm cm GURE Ampère s law applied to the loop shown in ig (going clockwise) gives B dl Bl ( 75 µ T )(. m) µ encircled (.4 π µ T m/ A) encircled (since the sides of the loop perpendicular to B give no contribution to the line integral). Thus, encircled ( 75π ) A 3. 9 A. As explained in the text, the current flows along the boundary surface between the regions of oppositely directed B, positive into the page in ig. 356, for clockwise circulation around the loop. 35. A long conducting rod of radius R carries a nonuniform current density given by J J r R, where J is constant and r is the radial distance from the rod s axis. ind expressions for the magnetic field strength (a) inside and (b) outside the rod. The magnetic field from a long conducting rod is approximately cylindrically symmetric, as discussed in Section 34, so (b) the field outside ( r R) is given by Equation 38, and has direction circling the rod according to the righthand rule. The total current can be related to the current density by integrating over the crosssectional area of the rod, R J da J ( r R) π r dr π JR 3. Here, area elements were chosen to be circular rings of radius r, thickness dr, and area da π r dr. Equation 38 can then be written as B µ π r µ JR 3r, for r R. (a) nside the rod, Ampère s law can be used to find the field, as in Example 34, by integrating the current density over a smaller crosssectional area, corresponding to encircled for an amperian loop with r R. Then J ( r R) π r dr π J r 3R, r 3 and Ampère s law gives π rb µ encircled, or B µ J r 3R, for r R. encircled 36. A long, hollow conducting pipe of radius R carries a uniform current along the pipe, as shown in ig Use Ampère s law to find the magnetic field strength (a) inside and (b) outside the pipe. R GURE (b) Since the pipe is assumed to have cylindrical symmetry, the field outside is given by Equation 38. (a) or an amperian loop with r < R,, hence B inside. (The thickness of the pipe is considered negligible.) encircled
7 CHAPTER The coaxial cable shown in ig. 36 consists of a solid inner conductor of radius a and a hollow outer conductor of inner radius b and thickness c. The two carry equal but opposite currents, uniformly distributed. ind expressions for the magnetic field strength as a function of radial position r (a) within the inner conductor, (b) between the inner and outer conductors, and (c) beyond the outer conductor. c a b GURE See the solution to 61, with Ra a, Rb b, and Rc b + c. 45. You have 1 m of.5mmdiameter copper wire and a battery capable of passing 15 A through the wire. What magnetic field strengths could you obtain (a) inside a.cmdiameter solenoid wound with the wire as closely spaced as possible and (b) at the center of a single circular loop made from the wire? (a) The length of a solenoid, with one layer of N turns of closely spaced wire of diameter d, is l Nd, so the number of turns per unit length is n Nl d (. 5 mm ) m. (Although not needed, the value of N 1 m π cm 159 turns can be used to check that l mm cm is barely long enough, compared to cm, to justify the use of 7 1 Equation 311 as an approximation to B at the solenoid s center.) Then B µ n ( 4π 1 N/ A )( m ) ( 15 A ) T. (b) The magnetic field at the center of a flat, circular current loop can be found from Equa 7 6 tion 33, B µ a. Here, π a 1 m, so B µ π 1 m ( 4π 1 N/ A )( 15 A ) 1 m T. 46. A toroidal coil of inner radius 15 cm and outer radius 17 cm is wound from 1 turns of wire. What are (a) the minimum and (b) the maximum magnetic field strengths within the toroid when it carries a 1A current? The magnetic field strength inside a toroid is given by Equation 31, and shows a 1r variation. or this particular toroid, 7 therefore (a) the minimum B is µ N π rmax ( 1 N/ A )( 1)( 1 A ) (. 17 m ) 141 G, and (b) the maximum field strength is B ( 17 15) B 16 G. max min 47. A toroidal fusion reactor requires a magnetic field that varies by no more than 1% from its central value of 1.5 T. f the minor radius of the toroidal coil producing this field is 3 cm, what is the minimum value for the major radius of the device?
8 71 CHAPTER 3 The central value of the toroidal field is given by Equation 31 with r R maj. At other values of r inside the toroid, the percent difference in field strength is 1( B Bmaj ) Bmaj 1( Rmaj r) r. The extremes of r are Rmaj ± R min, so it is required that 1 1 [ Rmaj ( Rmaj ± Rmin )] ( Rmaj ± Rmin ), or Rmaj ( 1 1) Rmin. The minimum major radius (corresponding to the limit using the smallest value of r, or the lower sign above) is Rmaj 11Rmin 11 3 cm 3. 3 m. (Since 1% is not infinitesimal, differentiation w.r.t. r gives an alternative approximate limit: db B dr r Rmin R maj 1%, or R maj 3 m. ) 57. igure 364 shows a wire of length l carrying current fed by other wires that are not shown. Point A lies on the perpendicular bisector, a distance y from the wire. Adapt the calculation of Example 3 to show that the magnetic field µ l at A due to the straight wire alone has magnitude. What is the field direction? π y l + 4y GURE To find the magnetic field at point A, one may use the same argument as in Example 3, except that for a wire of finite length, one integrates from x l to x l. or current flowing in the positive x direction, B is out of the page at point A, with strength l l y dx µ y x µ l B( A). l µπ 3 4 x + y ) 4π y x + y π y ( l + 4y l 58. Point B in ig lies a distance y perpendicular to the end of the wire. Show that the magnetic field at B has µ l magnitude. What is the field direction? 4π y l + y The direction of the magnetic field at point B is still out of the page, but its strength is y dx x B( B) µ l l µ µ l π 3 4 x + y ) π y ( 4 x + y 4π y l + y (see Example 3 and previous solution).,
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