( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.

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David Eaton
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1 Homewok #8. hapte 7 Magnetic ields. 6 Eplain how ou would modif Gauss s law if scientists discoveed that single, isolated magnetic poles actuall eisted. Detemine the oncept Gauss law fo magnetism now eads The flu of the magnetic field though an closed suface is equal to zeo. Just like each electic pole has an electic pole stength (an amount of electic chage, each magnetic pole would have a magnetic pole stength (an amount of magnetic chage. Gauss law fo magnetism would ead The flu of the magnetic field though an closed suface is popotional to the total amount of magnetic chage inside. A small cuent element at the oigin has a length of. mm and caies a cuent of. A in the +z diection. ind the magnitude of the magnetic field due to this element and indicate its diection on a diagam at (a. m, 4. m, z and (b. m,, z 4. m. Pictue the Poblem We can substitute fo and d l l in the iot-savat law dl ( d, evaluate and fo the given points, and substitute to find d. Appl the iot-savat law to the dl d given cuent element to obtain ( ( ( 7.A.mmk N/A k (.4 nt m (a ind and fo the point whose coodinates ae (. m, 4. m, Evaluate d at (. m, 4. m, (. m + ( 4. mj. 5 m, and j , j d (.m,4.m, (.4 nt m + 5 ( 5 m 5 j ( 8pT + ( 8.9pTj

2 The diagam is shown to the ight d (. m, 4. m, (b ind and fo the point whose coodinates ae (. m,, 4. m z (.m + ( 4.m. 5 m, and , Evaluate d d at (. m,, 4. m (.m,,4.m (.4 nt m (. 5 m ( 8.9pTj The diagam is shown to the ight z d (. m,, 4. m 3 The cuent in the wie shown in igue 7-5 is 8. A. ind the magnetic field at point P. Pictue the Poblem Note that the cuent segments a-b and e-f do not contibute to the magnetic field at point P. The cuent in the segments b-c, c-d, and d-e esult in a magnetic field at P that points into the plane of the pape. Note that the angles bpc and epd ae 45 and use the epession fo due to a staight wie segment to find the contibutions to the field at P of segments bc, cd, and de.

3 c d a b Epess the esultant magnetic field at P Epess the magnetic field due to a staight line segment P e f + + bc cd de ( sinθ + sinθ ( Use equation ( to epess de Use equation ( to epess bc and cd bc cd ( sin 45 + sin sin 45 ( sin 45 + sin 45 sin 45 Substitute to obtain Substitute numeical values and evaluate sin 45 + sin 45 + sin 45 4 sin A 4( T m/a sin 45. m.3mt into the page 36 An infinitel long wie lies along the ais, and caies cuent in the + diection. A second infinitel long wie lies along the ais, and caies cuent in the + diection. At what points in the z plane is the esultant magnetic field zeo? Pictue the Poblem Let the numeal denote the cuent flowing in the positive diection and the magnetic field esulting fom it and the numeal denote the cuent flowing in the positive diection and the magnetic field esulting fom it. We can epess the magnetic field anwhee in the plane using and the ight-hand ule and then impose the condition that to detemine the set

4 of points that satisf this condition. Epess the esultant magnetic field due to the two cuent-caing wies Epess the magnetic field due to the cuent flowing in the positive diection + ( Epess the magnetic field due to the cuent flowing in the positive diection Substitute fo and in equation ( and simplif to obtain because. o. evewhee on the plane that contains both the z ais and the line in the z plane. 46 n igue 7-55, one cuent is 8. A into the page, the othe cuent is 8. A out of the page, and each cuve is a cicula path. (a ind d l fo each path, assuming that each integal is to be evaluated in the counteclockwise diection. (b Which path, if an, can be used to find the combined magnetic field of these cuents? Pictue the Poblem We can use Ampèe s law, d l, to find the line integal dl fo each of the thee paths. (a Noting that the angle between and d l ( 8.A dl is 8, evaluate dl The positive tangential diection on fo is counteclockwise. Theefoe, in accod with convention (a ight-hand ule, the positive nomal diection fo the flat suface bounded b is out of the page. dl is negative because the

5 cuent though the suface is in the negative diection (into the page. Noting that the net cuent bounded b is zeo, evaluate dl d l ( 8.A8.A Noting that the angle between and dl is, Evaluate dl fo 3 d l 3 + ( 8.A (b None of the paths can be used to find because the cuent configuation does not have clindical smmet, which means that cannot be factoed out of the integal. 5 [SSM] A tightl wound -tun tooid has an inne adius. cm and an oute adius. cm, and caies a cuent of.5 A. The tooid is centeed at the oigin with the centes of the individual tuns in the z plane. n the z plane (a What is the magnetic field stength at a distance of. cm fom the oigin? (b What is the magnetic field stength at a distance of.5 cm fom the oigin? Pictue the Poblem The magnetic field inside a tightl wound tooid is given b N π, whee a < < b and a and b ae the inne and oute adii of the ( tooid. Epess the magnetic field of a N tooid π (a Substitute numeical values and evaluate (. cm 7 ( ( N/A ( (.5A. cm π(. cm 7.3mT (b Substitute numeical values and evaluate (.5 cm 7 ( ( N/A ( (.5A.5 cm π(.5 cm. mt 8 The closed loop shown in igue 7-64 caies a cuent of 8. A in the counteclockwise diection. The adius of the oute ac is.6 m and that of the inne ac is.4 m. ind the magnetic field at point P. Pictue the Poblem Let out of the page be the positive diection and the numeals 4 and 6 efe to the cicula acs whose adii ae 4 cm and 6 cm. ecause point P is on the line connecting the staight segments of the conducto,

6 these segments do not contibute to the magnetic field at P. Hence the esultant magnetic field at P will be the sum of the magnetic fields due to the cuent in the two cicula acs and we can use the epession fo the magnetic field at the cente of a cuent loop to find P. Epess the esultant magnetic field at P P ( Epess the magnetic field at the cente of a cuent loop Epess the magnetic field at the cente of one-sith of a cuent loop whee is the adius of the loop. 6 Epess 4 and and 6 6 Substitute fo 4 and 6 in equation ( and simplif to obtain P Substitute numeical values and evaluate P o P 7 ( N/A ( 8.A P.6m.4m.7 T into the page (.7 T 94 igue 7-68 shows a squae loop that has -cm long sides and is in the z plane with its cente at the oigin. The loop caies a cuent of 5. A. An infinitel long wie that is paallel to the ais and caies a cuent of A intesects the z ais at z cm. The diections of the cuents ae shown in the figue. (a ind the net toque on the loop. (b ind the net foce on the loop. Pictue the Poblem The field due to the -A cuent is in the z plane. The net foce on the wies of the squae in the diection cancel and do not contibute to a net toque o foce. We can use τ l, l, and the epession fo the magnetic field due to a long staight wie to epess the toque acting on each of the wies and hence, the net toque acting on the loop.

7 z A 5. A cm l cm l 5. A cm (a The net toque about the ais is the sum of the toques due to the foces and Substituting fo τ net τ + τ and τ - ields τ net l + l The foces acting on the wies ae given b Substitute fo and to obtain τ net The leve ams fo the foces acting on the wies at cm and cm ae τ - whee the subscipts efe to the positions of the cuent-caing wies. ( l and ( l [( l ] l [( l ] l + l ( (. mj and l (. mj The magnetic field at the wie at cm is given b 4 π ( j whee (.m + (.m.4m. Substitute numeical values and evaluate 7 N/A ( ( j (. T( j A.4m

8 Poceed similal to obtain (. T( j Substitute in equation ( and simplif to obtain τ net + (. m j [( 5. A(. m (. T( j ] (. m j [( 5. A(. m( (. T( j + ] (. N m (b The net foce acting on the loop is the sum of the foces acting on its fou sides net + ( Evaluate to obtain ( l ( ( ( ( j 5.A.m. T ( N [ ( j ] ( N( + j Evaluating ields ( l ( ( ( ( j 5. A.m. T + ( N [ ( j + ] ( N( + j Substitute fo and in equation ( and simplif to obtain ( N( + j + ( N( + j ( Nj ne t

( ) ( )( ) ˆ. Homework #8. Chapter 28

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