Problems on Force Exerted by a Magnetic Fields from Ch 26 T&M


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1 Poblems on oce Exeted by a Magnetic ields fom Ch 6 TM Poblem 6.7 A cuentcaying wie is bent into a semicicula loop of adius that lies in the xy plane. Thee is a unifom magnetic field B Bk pependicula to the plane of the loop. Veify that the foce acting on the loop is zeo. Pictue the Poblem With the cuent in the diection indicated the magnetic field in the z diection, pointing out of the plane of the page, the foce is in the adial diection we can integate the element of foce d acting on an element of length dl between θ π to find the foce acting on the semicicula potion of the loop use the expession fo the foce on a cuentcaying wie in a unifom magnetic field to find the foce on the staight segment of the loop. Expess the net foce acting on the semicicula loop of wie: Expess the foce acting on the staight segment of the loop: Expess the foce d acting on the element of the wie of length dl: Expess the x y components of d: semicicul () a loop staight segment Il B IB staight segment d Idl B IBd d x d cos d y d sin Because, by symmety, the x component of the foce is zeo, we can integate the y component to d y IBsin d
2 find the foce on the wie: y IB sin d IB Substitute in equation () to obtain: IB IB 6.57 Toque on a loop with cuent A igid cicula loop of adius mass m caies a cuent I lies in the xy plane on a ough, flat table. Thee is a hoizontal magnetic field of magnitude B. What is the minimum value of B so that one edge of the loop will lift off the table? Pictue the Poblem The loop will stat to lift off the table when the magnetic toque equals the gavitational toque. Expess the magnetic toque acting on the loop: Expess the gavitational toque acting on the loop: Because the loop is in equilibium unde the influence of the two toques: mag B I gav mg I B mg B Solve fo B to obtain: B mg I
3 Poblem Magnetic moment of a loop Magnetic field calculation A wie loop consists of two semicicles connected by staight segements. The inne oute adii ae.3.5 m, espectively. A cuent I of.5 A flows in this loop with the cuent in the oute semicicle in the clockwise diection. A) What is the magnetic moment of the cuent loop? B) ind the magnetic field in P, which is at the common cente of the semicicula acs. Pictue the Poblem We can use the definition of the magnetic moment to find the magnetic moment of the given cuent loop a ighth ule to find its diection. Using its definition, expess the magnetic moment of the cuent loop: Expess the aea bounded by the loop: IA ( ) ( ) A oute inne oute inne Substitute to obtain: I ( ) oute inne Substitute numeical values evaluate : (.5A) ( ) ( ).5m.3m.377 A m [ ] Apply the ighth ule fo detemining the diection of the unit nomal vecto (the diection of ) to conclude that ì points into the page. Poblem 7. Pictue the Poblem Let out of the page be the positive x diection. Because point P is on the line connecting the staight segments of the conducto, 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 semicicles, we can use the expession fo the magnetic field at the cente of a cuent loop to find B P.
4 Expess the esultant magnetic field at P: B P B B Expess the magnetic field at the cente of a cuent loop: Expess the magnetic field at the cente of half a cuent loop: Expess B B : I B whee is the adius of the loop. I I B I B iˆ I B iˆ Substitute to obtain: B P I iˆ I ' iˆ I iˆ ' % Poblem 7.5 oce between cuent wies A long staight wie caies a cuent of A, as shown in the figue. A ectangula coil with sides paallel to the staight wie has sides 5 cm cm with the nea side at a distance cm fom the wie. The coil caies a cuent of 5 A. (a) ind the foce on each segment of the ectangula coil due to the cuent in the long staight wie. (b) What is the net foce on the coil? Pictue the Poblem Let I I epesent the cuents of A 5 A,,, 3, the foces that act on the hoizontal wie at the top of the loop, the othe wies following the cuent in a counteclockwise diection, B, B, B 3, B the magnetic fields at these wies due to I. Let the positive x diection be to the ight the positive y diection be upwad. Note that only the components into o out of the pape
5 of B, B, B 3, B contibute to the foces,, 3,, espectively. (a) Expess the foces in tems of I B B : Expess B B : Substitute to obtain: I l B I l B B I I B kˆ kˆ ˆ ' I ( Il j ) % ( l II iˆ ˆ ( I Il j ) ' l II iˆ kˆ kˆ % Substitute numeical values evaluate : 7 ( N/A )(.m)( A)( 5A) (. m) iˆ (. N)i ˆ 7 ( N/A )(.m)( A)( 5A) (.7 m) iˆ (.86 N)i ˆ (b) Expess the net foce acting on the coil: net () 3 Because the lengths of segments 3 ae the same the cuents in these segments ae in opposite diections: 3 net
6 Substitute fo in equation () simplify to obtain: net ( ) ˆ j ( ) iˆ.5 N. N (.5 N) ˆj (.86 N) iˆ (.7 N)i ˆ Pobem 7.59 Magnetic field in a solenoid A solenoid with length 3 cm, adius. cm, 3 tuns caies a cuent of.6 A. ind B on the axis of the solenoid (a) at the cente, (b) inside the solenoid at a point cm fom one end, (c) at one end. Pictue the Poblem We can use b a B x ni to find B at % b a any point on the axis of the solenoid. Note that the numbe of tuns pe unit length fo this solenoid is 3 tuns/.3 m tuns/m. Expess the magnetic field at any point on the axis of the solenoid: b a B x ni % b a Substitute numeical values to obtain: B x * ) 7 ( ( T ' m/a)( )(.6 A) % (.63 mt) b b % (. m) a (. m) b b (. m) a (. m) a a (a) Evaluate B x fo a b.5 m: B x % (.63 mt).5m.5m (.5m) (. m) (.5m) (. m) 3.5 mt (b) Evaluate B x fo a. m b. m:
7 B x (. m) (.63 mt) 3.5 mt %. m.m (. m) (. m) (.m) (. m) (c) Evaluate B x ( B end ) fo a b.3 m:.3m B (.63 mt) x %.3m. m Note that B B. end cente ( ) ( ).63 mt Conceptual Poblem 7.67 Show that a unifom magnetic field with no finging field, such as that shown in the figue, is impossible because it violates Ampee s law. Do this by applying Ampee s law to the ectangula cuve shown by the dashed line. Detemine the Concept The contou integal consists of fou potions, two hoizontal potions fo which B dl, two vetical potions. The potion within the magnetic field gives a C nonvanishing contibution, wheeas the potion outside the field gives no contibution to the contou integal. Hence, the contou integal has a finite value. Howeve, it encloses no cuent; thus, it appeas that Ampèe s law is violated. What this demonstates is that thee must be a finging field so that the contou integal does vanish. Poblem 8.5 Pictue the Poblem The feebody diagam shows the foces acting on the od as it slides down the inclined plane. The etading foce is the component of m acting up the incline, i.e., in the x diection. We can expess m using the expession fo the foce acting on a conducto moving in a magnetic field. ecognizing that only the hoizontal component of the od s velocity v poduces an induced emf, we can apply the expession fo a motional emf in conjunction with Ohm s law to find the induced cuent in the od. In pat (b) we can apply Newton s nd law to obtain an expession fo dv/ set this expession equal to zeo to obtain v t.
8 (a) Expess the etading foce acting on the od: Expess the induced emf due to the motion of the od in the magnetic field: m cos () whee IlB m I is the cuent induced in the od as a consequence of its motion in the magnetic field. Blv cos Using Ohm s law, elate the cuent I in the cicuit to the induced emf: I Blv cos Substitute in equation () to obtain: ' Blv cos % lb cos B l v cos (b) Apply x max to the od: When the od eaches its teminal velocity v t, dv/ : B l v dv mg sin cos m dv B l v g sin cos m B l vt g sin cos m Solve fo v t to obtain: Poblem 8.39 v t mgsin B l cos Pictue the Poblem We ll need to detemine how long it takes fo the loop to completely ente the egion in which thee is a magnetic field, how long it is in the egion, how long it takes to leave the egion. Once we know these times, we can use its definition to expess the magnetic flux as a function of time. We can use aaday s law to find the induced emf as a function of time.
9 (a) ind the time equied fo the loop to ente the egion whee thee is a unifom magnetic field: Letting w epesent the wih of the loop, expess evaluate φm fo < t <.7 s : ind the time duing which the loop is fully in the egion whee thee is a unifom magnetic field: Expess φm fo.7 s < t < 8.33 s : The leftend of the loop will exit the field when t.5 s. Expess φm fo 8.33 s < t <.5 s : t l side of loop v cm.7 s. cm/s m NBA NBwvt (.7 T )(.5 m )(. m/s )t (. mwb/s )t t l side of loop v cm.7 s. cm/s i.e., the loop will begin to exit the egion when t 8.33 s. m NBA NBlw (.7 T )(. m )(.5 m ) 8.5 mwb m mt b whee m is the slope of the line b is the φmintecept. o t 8.33 s φm 8.5 mwb: 8.5 mwb m(8.33 s ) b () o t.5 s φm : m(.5 s ) b () Solve equations () () simultaneously to obtain: m (. mwb/s)t 5.5 mwb The loop will be completely out of the magnetic field when t >.5 s m
10 : The following gaph of ( t) was plotted using a speadsheet pogam. m Magnetic flux (mwb) (b) Using aaday s law, elate the induced emf to the magnetic flux: t (s) d m Duing the inteval Duing the inteval.7s < t < 8.33s : Duing the inteval 8.33s < t <.5s : < t <.7s : d t d [(. mwb/s) ]. mv [ 8.5 mwb] d t. mv [( . mwb/s) 5.5 mwb] o t >.5 s: The following gaph of ε(t) was plotted using a speadsheet pogam.
11 emf (V) t (s) Poblem 8.85 The AC geneato Pictue the Poblem We can apply aaday s law the definition of magnetic flux to deive an expession fo the induced emf in the coil (potential diffeence between the slip ings). In pat (b) we can solve this equation fo ω unde the given conditions. (a) Use aaday s law to expess the induced emf: Using the definition of magnetic flux, elate the magnetic flux though the loop to its angula velocity: d m ( t ) NBA t cos m d NBab Substitute to obtain: [ NBAcos t] ( sin t) NBab sin t
12 (b) Expess the condition unde which ε ε max : Solve fo evaluate ω unde this condition: sin t max NBab ( )( T)(.m)(. m) 75 ad/s V
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