(b) Show that db/dx and d 2 B/dx 2 are both zero at the point midway between the

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1 . In Fgure P., the cube s 40.0 cm on each edge. Four straght segments of wre ab, bc, cd, and da form a closed loop that carres a current I 5.00 A, n the drecton shown. A unform magnetc feld of magntude T s n the postve y drecton. Determne the magntude and drecton of the magnetc force on each segment.. A nonconductng sphere has mass 80.0 g and radus 0.0 cm. A flat compact col of wre wth 5 turns s wrapped tghtly around t, wth each turn concentrc wth the sphere. As shown n Fgure P, the sphere s placed on an nclned plane that slopes downward to the left, makng an angle θ wth the horzontal, so that the col s parallel to the nclned plane. A unform magnetc feld of T vertcally upward exsts n the regon of the sphere. What current n the col wll enable the sphere to rest n equlbrum on the nclned plane? Show that the result does not depend on the value of θ. 3. Two crcular cols of radus R, each wth N turns, are perpendcular to a common axs. The col centers are a dstance R apart. Each col carres a steady current I n the same drecton, as shown n Fgure P3. (a) Show that the magnetc feld on the axs at a dstance x from the center of one col s N 0IR + Rx ( ) ( ) 3 / R + x 3 / R + x (b) Show that d/dx and d /dx are both zero at the pont mdway between the

2 cols. Ths means the magnetc feld n the regon mdway between the cols s unform. Cols n ths confguraton are called Helmholtz cols. 4. (from Hallday, problem 9.8, page 785) In Fg., part of a long nsulated wre carryng current 5.78 ma s bent nto a crcular secton of radus R.89 cm. In unt-vector notaton, what s the magnetc feld at the center of curvature C f the crcular secton (a) les n the plane of the page as shown and (b) s perpendcular to the plane of the page after beng rotated 90 counterclockwse as ndcated? Fg.. Soluton: (a) The contrbuton to C from the (nfnte) straght segment of the wre s The contrbuton from the crcular loop s C π R. C. Thus, R -7 3 ( 4p 0 T m A)( A) R π ( m) π C ponts out of the page, or n the +z drecton. In unt-vector notaton, 0 7 C C+ C T. 7 (.53 0 T)kˆ C (b) Now so C C -7 3 ( 4p 0 T m A)( A) 0 7 C C+ C T. R π m π ( )

3 and C ponts at an angle (relatve to the plane of the paper) equal to C π C tan tan In unt-vector notaton, 7? T(cos7.66 sn7.66 k) (.9 0 T)? C + + ( T)k 5. (from Hallday, problem 9.3, page 785) Fgure shows, n cross secton, two long straght wres held aganst a plastc cylnder of radus 0.0 cm. Wre carres current 60.0 ma out of the page and s fxed n place at the left sde of the cylnder. Wre carres current 40.0 ma out of the page and can be moved around the cylnder. At what (postve) angle θ should wre be postoned such that, at the orgn, the net magnetc feld due to the two currents has magntude 80.0 nt? Fg.. Soluton: y the rght-hand rule (whch s bult-nto Eq. 9-3) the feld caused by wre s current, evaluated at the coordnate orgn, s along the +y axs. Its magntude s gven by Eq The feld caused by wre s current wll generally have both an x and a y component whch are related to ts magntude (gven by Eq. 9-4) and snes and cosnes of some angle. A lttle trg (and the use of the rght-hand rule) leads us to conclude that when wre s at angle θ (shown n Fg. 9-60) then ts components are x θ y θ sn, cos. The magntude-squared of ther net feld s then (by Pythagoras theorem) the sum of the square of ther net x-component and the square of ther net y-component: ( sn θ ) + ( cos θ ) + cos θ. (snce sn θ + cos θ ), whch we could also have gotten drectly by usng the law of 3

4 cosnes. We have nt, 40 nt. πr πr Wth the requrement that the net feld have magntude 80 nt, we fnd + θ cos cos ( / 4) 04, where the postve value has been chosen. 6. (from Hallday, problem 9.58, page 787) Fgure 3 shows an arrangement known as a Helmholtz col. It conssts of two crcular coaxal cols, each of 00 turns and radus R 5.0 cm, separated by a dstance s R. The two cols carry equal currents. ma n the same drecton. Fnd the magntude of the net magnetc feld at P, mdway between the cols. Fg (from Hallday, problem 9.6, page 788) In Fg. 4a, two crcular loops, wth dfferent currents but the same radus of 4.0 cm, are centered on a y axs. They are ntally separated by dstance L 3.0 cm, wth loop postoned at the orgn of the axs. The currents n the two loops produce a net magnetc feld at the orgn, wth y component y. That component s to be measured as loop s gradually moved n the postve drecton of the y axs. Fgure 4b gves y as a functon of the poston y of loop. The curve approaches an asymptote of y 7.0 T as y. The horzontal scale s set by ys 0.0 cm. What are (a) current n loop and (b) current n loop? 4

5 Fg.. Soluton: Usng Hallday Eq. 9-6, we fnd that the net y-component feld s y R R π( R z ) ( R z ) 0 0 3/ 3/ + π + where z L (see Fg. (a)) and z y (because the central axs here s denoted y nstead of z). The fact that there s a mnus sgn between the two terms, above, s due to the observaton that the datum n Fg. (b) correspondng to y 0 would be mpossble wthout t (physcally, ths means that one of the currents s clockwse and the other s counterclockwse). (a) As y, only the frst term contrbutes and (wth y T gven n ths case) we can solve for. We obtan (45/6π) Α 0.90 A. (b) Wth loop at y 0.06 m (see Fg. (b)) we are able to determne from We obtan (7 3 /50π) Α.7 A. R R ( R + L ) ( R + y ) 8. (from Hallday, problem 9.65, page 788) / 3/ Fgure 5 shows a cross secton of a long cylndrcal conductor of radus a 4.00 cm contanng a long cylndrcal hole of radus b.50 cm. The central axes of the cylnder and hole are parallel and are dstance d.00 cm apart; current 5.5 A s unformly dstrbuted over the tnted area. (a) What s the magntude of the magnetc feld at the center of the hole? (b) Dscuss the two specal cases b 0 and d 0., 5

6 Fg. 5. Soluton: (a) The magnetc feld at a pont wthn the hole s the sum of the felds due to two current dstrbutons. The frst s that of the sold cylnder obtaned by fllng the hole and has a current densty that s the same as that n the orgnal cylnder (wth the hole). The second s the sold cylnder that flls the hole. It has a current densty wth the same magntude as that of the orgnal cylnder but s n the opposte drecton. If these two stuatons are superposed the total current n the regon of the hole s zero. Now, a sold cylnder carryng current whch s unformly dstrbuted over a cross secton, produces a magnetc feld wth magntude r 0 pr at a dstance r from ts axs, nsde the cylnder. Here R s the radus of the cylnder. For the cylnder of ths problem the current densty s J A ca bh, π where A π(a b ) s the cross-sectonal area of the cylnder wth the hole. The current n the cylnder wthout the hole s a I JA p Ja a b and the magnetc feld t produces at a pont nsde, a dstance r from ts axs, has magntude 0Ir ra r pa pa a b p a b The current n the cylnder that flls the hole s c h c h. 6

7 I b p Jb a b and the feld t produces at a pont nsde, a dstance r from the ts axs, has magntude 0Ir rb r pb pb a b p a b c h c h. At the center of the hole, ths feld s zero and the feld there s exactly the same as t would be f the hole were flled. Place r d n the expresson for and obtan ( a b ) ( ) 7 d (4 0 T m/a) 5.5A (0.000m) T π π π[(0.0400m) (0.050m) ] for the feld at the center of the hole. The feld ponts upward n the dagram f the current s out of the page. (b) If b 0 the formula for the feld becomes d 0. pa Ths correctly gves the feld of a sold cylnder carryng a unform current, at a pont nsde the cylnder a dstance d from the axs. If d 0 the formula gves 0. Ths s correct for the feld on the axs of a cylndrcal shell carryng a unform current. Note: One mght apply Ampere s law to show that the magnetc feld n the hole s unform. Consder a rectangular path wth two long sdes (sde and, each wth length L) and two short sdes (each of length less than b). If sde s drectly along the axs of the hole, then sde would be also parallel to t and also n the hole. To ensure that the short sdes do not contrbute sgnfcantly to the ntegral n Ampere s law, we mght wsh to make L very long (perhaps longer than the length of the cylnder), or we mght appeal to an argument regardng the angle between and the short sdes (whch s 90 at the axs of the hole). In any case, the ntegral n Ampere s law reduces to z ds enclosed rectangle ds + ds z z sde sde d L 0 sde sde 0 n hole where sde s the feld along the axs found n part (a). Ths shows that the feld at off-axs ponts (where sde s evaluated) s the same as the feld at the center of the hole; therefore, the feld n the hole s unform. 9. (from Hallday, problem 9.8, page 789) Fgure 6 shows a cross secton of a hollow cylndrcal conductor of rad a and b, carryng a 7

8 unformly dstrbuted current. (a) Show that the magnetc feld magntude (r) for the radal dstance r n the range b < r < a s gven by (b) Show that when r a, ths equaton gves the magnetc feld magntude at the surface of a long straght wre carryng current ; when r b, t gves zero magnetc feld; and when b 0, t gves the magnetc feld nsde a sold conductor of radus a carryng current. Fg.. Soluton: (a) For the crcular path L of radus r concentrc wth the conductor Thus, πca 0 b F hhg z π r b ds πr enc L πca bh. r b r (b) At r a, the magnetc feld strength s I KJ. F hhg I KJ a b. π a b a πa At r b, r b 0. Fnally, for b 0 r r πa r πa whch agrees wth Eq c c h 8