Matter & Interactins Chapter 7 Slutins are int the page ( z directin, and and 6 are ut f the page (+z directin. Due t symmetry, 3 6 4. µ q v ˆr sin θ 4π r 2 T ( m 0 7 A (.6 0 9 C(5 0 6 m/s( sin 25 (0.08 m 2 5.28 0 8 T Thus, 6 0, 0, 5.28 0 8 T and 3 4 0, 0, 5.28 0 8 T. P22: Slutin: (a q E 4πɛ 0 r 2ˆr r 4 0 3, 4 0 3, 3 0 3 m r (4 0 3 m 2 + (4 0 3 m 2 + (3 0 3 m 2 6.40 0 3 m ˆr r r < 0.625, 0.625, 0.469 > E (9 09 N m2 C 2 (.6 0 9 C(< 0.625, 0.625, 0.469 > (6.4 0 3 m 2 2.20 0 5, 2.20 0 5,.65 0 5 N C (b µ q v ˆr 4π r 2 ( 0 7 T m A (.6 0 9 C( 3 0 4, 2 0 4, 7 0 4 m/s 0.625, 0.625, 0.469 (6.4 0 3 2.34 0 7, 2.26 0 7,.22 0 7 T P23: Slutin: (a Directin f E is tward the electrn. Thus, 794
Matter & Interactins Chapter 7 Slutins Ê <, 0, 0 > Directin f is given by the right-hand rule. Since the particle is an electrn, pint yur thumb ppsite v, and is tangent t a circle arund an axis thrugh yur thumb. Thus, is ut f the page, in the +z directin. ˆ < 0, 0, >. (b E 4πɛ 0 q r 2 (9 0 9 N m2 C 2 (.6 0 9 C (5 0 0 m 2 5.76 0 9 N C µ 4π q v sin θ r 2 T ( m 0 7 A (.6 0 9 C(3 0 6 m/s sin 60 (5 0 0 2 0.66 T P24: Slutin: i 3.7 0 8 electrns per secnd, t the right. The cnventinal current is in the ppsite directin f the electrn current. The cnventinal current is P25: Slutin: I 2 A t the right. (a The electrn current is I ( ( 8 electrns.6 0 9 C 3.7 0 s electrn 0.59 A, t the left ( electrn i (2 C/s.6 0 9 C 7.5 0 9 electrns/s (b Electrns are traveling t the left, ppsite the cnventinal current. 795
Matter & Interactins Chapter 7 Slutins µ side 4π LI r(r 2 + ( L 2 2 2 ( 0 7 T m A 2.05 0 6 T (0.083 m(0.62 A (0.024 m((0.024 m 2 + ( 0.083 m 2 2 2 lp 3side 6.5 0 6 T cil N lp 3(6.5 0 6 T.84 0 5 T The cmpass deflectin frm nrth is θ tan ( cil θ 43 P37: Slutin: (a It is 0 because l ˆr l ˆr sin θ, θ 80, and sin(80 0 (b A piece f is at lcatin x and has a length dx in the directin <, 0, 0 >. The vectr r t lcatin P is r r P r < w, h, 0 > < x, 0, 0 > < w x, h, 0 > r ((w + x 2 + h 2 2 ˆr r r 802
Matter & Interactins Chapter 7 Slutins d µ Id l ˆr 4π r 2 µ Id l r 4π r 3 d l r < dx, 0, 0 > < (w + x, h, 0 > < 0, 0, hdx > d is thus in the +z directin, cnsistent with the right-hand rule. d z µ Ihdx 4π ((w + x 2 + h 2 3 2 z,p xd µ x0 z,p µ 4π Ih Ihdx 4π ((w + x 2 + h 2 3 2 d 0 dx ((w + x 2 + h 2 3 2 (c As shwn by cmputing d l ˆr, P is in the +z directin. This can als be fund using the right-hand rule. P38: Slutin: Use the cmpass deflectin t find the current in the and its directin. y the right-hand rule, the current flws N (in the +x directin. tan θ (2 0 5 T tan( 3.89 0 6 T µ 2I 4π r I r ( µ 4π 2 (3.89 0 6 T(0.003 m 2( 0 7 T m A 0.0583 A 803
Matter & Interactins Chapter 7 Slutins net,z,z + 2,z µ ( 4π Iπ + R R 2 Its magnitude is ( µ net 4π Iπ + R R 2 Its directin is ˆ < 0, 0, >. P46: Slutin: is the superpsitin f + lp. is in the z directin and has a magnitude µ 2I 4π r ( 0 7 T m A 2(3.5 A 0.058 m.2 0 5 T s 0, 0,.2 0 5 T lp is in the z directin and has a magnitude lp µ 2IA 4π R 3 ( 0 7 T m Aπ(0.058 m2 2(3.5 A (0.058 m 3 3.79 0 5 T s lp 0, 0, 3.79 0 5 T net lp + 0, 0, 3.79 0 5 T + 0, 0,.2 0 5 T 0, 0, 5.00 0 5 T 809
Matter & Interactins Chapter 7 Slutins P47: Slutin: The magnetic field alng the axis f the cil is µ 2IAN 4π (r 2 + R 2 3 2 Where r is the distance alng the axis frm the center f the cil. Align the cil s that its plane is N-S and the magnetic field alng the axis f its cil is E-W. Measure the deflectin f the cmpass needle frm Nrth, θ. Then, tan θ cil cil tan θ Calculate I. I cil (r2 + R 2 3 2 µ 4π 2AN I ( tan θ(r2 + R 2 3 2 µ 4π 2(πR2 N Substitute measured values f r, R, N, and θ t calculate I. P48: Slutin: Current flws clckwise arund the cil, thus at the center f the cil is in the -z directin. The needle will be deflected inward, and the tp view f the cmpass will appear like the figure belw. W S N E 80
Matter & Interactins Chapter 7 Slutins Assume it is a thin cil. At the cil s center, µ 4π 2IAN R 3 ( 0 7 T2(0.25 Aπ(0.5 m 2 (3 (0.5 m 3 3.4 0 6 T tan θ cil ( 3.4 0 θ tan 6 T 2 0 5 T 8.9 P49: Slutin: (a Neglect the upper straight line segments because they cntribute zer field. Neglect the side segments because they are relatively shrt and far away and cntribute negligible field. Thus, the net magnetic field is a superpsitin f the magnetic field due t the hemisphere and the lwer straight line segment. Applicatin f the right-hand rule shws that the field due t each f these parts f the is int the page (-z directin. (b net hemisphere + straight / net,z z,hemisphere + z,straight/ µ Iπ 4π R + µ 2I 4π h where we assume that h << L. Thus, µ ( π 4π I R + 2 h P50: Slutin: net + 2. Since and 2 are t the right, net is t the right. 8
Matter & Interactins Chapter 7 Slutins (a µ cil 2IAN 4π (r 2 + R 2 3 2 ( 0 7 T m 2(2Aπ(0.03 m 2 (0 A ((0.0 m 2 + (0.03 m 2 3 2 9.94 0 6 T net 9.94 0 6, 0, 0 T + 9.94 0 6, 0, 0 T.99 0 5, 0, 0 T net.99 0 5 T Directin is ˆ <, 0, 0 >. (b Apprximate frmula gives cil µ 2I(πR 2 4π r 3 ( 0 7 T m A 2(2Aπ(0.03 m2 (0 (0. m 3.3 0 5 T % errr (.3 0.994 4% 0.994 00 In this case, z 3R, thus z is nt much greater than R. If z 0R, then the apprximatin is better and the errr is less. (c In this case, 2 is in the -x directin. Since 2 and is in the +x directin, then net + 2 0. P5: Slutin: The straight segments cntribute zer magnetic field at pint C. Thus net is the superpsitin f ( due t the large hemisphere and (2 due t the small hemisphere and (3 due t the lp. In all cases, is in the -z directin. due t a hemisphere is 82