Physics 114 Exam 3 Spring Name:


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1 Physcs 114 Exam 3 Sprng 015 Name: For gradng purposes (do not wrte here): Queston Problem 4. Answer each of the followng questons. Ponts for each queston are ndcated n red. Unless otherwse ndcated, the amount beng spread among parts (a,b,c etc) are equal. Be sure to show all your work. Use the back of the pages f necessary.
2 Queston 1. (10 ponts) Fnd the drecton of the mnmum magnetc feld actng on the postvely charged partcle movng n the varous stuatons shown n the fgure below f the drecton of the magnetc force actng on t s as ndcated. By mnmum magnetc feld we mean the smallest one capable of producng the force shown. Soluton.. Snce the partcle s postvely charged, use the rght hand rule. In ths case, start wth the fngers of the rght hand n the drecton of and the thumb pontng n the drecton of. As you start closng the hand, the fngers pont n the drecton of after they have moved (the mnmum feld to provde a gven force would be perpendcular to the velocty). The results are: (a) nto the page; (b) toward the rght; (c) toward the bottom of the page.
3 Queston. (10 ponts) (a) In the fgure below fnd the drecton of the current n the wre that would produce a magnetc feld drected as shown. State ts drecton (to the rght, left, top of the page, bottom of the page, out of the page, nto the page) Soluton: Imagne graspng the conductor wth the rght hand so the fngers curl around the conductor n the drecton of the magnetc feld. The thumb then ponts along the conductor n the drecton of the current. The results are: (a) toward the left; (b) In the fgure below fnd the drecton of the current n the wre that would produce a magnetc feld drected as shown. State ts drecton (to the rght, left, top of the page, bottom of the page, out of the page, nto the page) (b) out of the page; (c) In the fgure below fnd the drecton of the current n the wre that would produce a magnetc feld drected as shown. State ts drecton (towards the upper rght, lower left, top of the page, bottom of the page, out of the page, nto the page, towards the upper left, lower rght). (c) lower left to upper rght.
4 Queston 3. (10 ponts) (a) A rectangular conductng loop s placed near a long wre carryng a current I as shown n the fgure below. If I decreases n tme, what can be sad of the current nduced n the loop (s there one, f so s clockwse or counter clockwse as vewed from above)? Soluton. The feld through the loop s nto the page and t decreases n tme due to the decreasng current. Thus, the flux decreases nto the page and the counter flux needs to be nto the page and so the current s clockwse. (b) A crcut conssts of a conductng movable bar and a lghtbulb connected to two conductng rals as shown n the fgure below. An external magnetc feld s drected perpendcular to the plane of the crcut. Answer the followng (True or False) ) When the bar s moved to the left, the bulb lghts up ) When the bar s moved to the rght, the bulb lghts up ) When the magntude of the magnetc feld ncreases the lght bulb lghts up v) When the magntude of the magnetc feld decreases the lght bulb lghts up v) An appled force s requred to mantan the moton of the bar to the rght one t s gven some ntal velocty. Soluton. These are all true. (c) Two rectangular loops of wre le n the same plane as shown n the fgure below. If the current I n the outer loop s counterclockwse and ncreases wth tme, s there an nduced current n the nner loop and (f so) what s ts drecton (clockwse or counter clockwse). Does the magntude of the nduced feld (f there s one) depend on the dmensons of the loops? Soluton: There s an nduced current. The feld due to the outer loop wthn the nner loop s ncreasng out of the page so the flux s ncreasng out of the page through the nner loop. The nduced current wll be nto the page, whch means the current must be clockwse. The dmensons do matter. The flux depends on the area of the nner loop and on how close the outer loop s to the nner one.
5 Problem 1. A partcle wth a charge q = 3 C moves through a regon where there s a magnetc feld B = (3 ˆ+ k)t ˆ. (a) (6 ponts) Determne the force on the charge when t has a velocty of Soluton: F qv x B =3[v(3j) ˆ (3 ˆ+ k) ˆ ]= 3(9kˆ 3) ˆ (97 ˆ k) ˆ v = (3j) ˆ m/s (b) (3 ponts) What s the angle between ths force (calculated n part a) and the magnetc feld? Soluton: The angle s always 90 degrees. (c) (6 ponts) Now say that the velocty of the partcle s (somehow) changed so that t s movng wth a speed (magntude of the velocty) of 5 m/s. It now experences a force gven by F = (7 ˆj) N. Determne the ycomponent of the velocty of the charge when t experences ths force (that s, fnd vy,  f ths component s undetermned state so). Soluton: 7 ˆj = 3 ( v ˆ + v ˆj v k) ˆ (3 ˆ+ k) ˆ x y z 3(0 v ˆj3 v k+ ˆ v ˆ 3v ˆj 0) x y y z For both sdes to be equal, vy must be zero. (d) (3 ponts extra credt) Fnd the other two components (that s, fnd vx, and vz  f one component s undetermned state whch). 7 ˆj = 3( v ˆj+3 ˆ x vz j) 9v 3v x z. vx = 3 vz 9 Snce the speed s 5 m/s, we have 5 = vx + vz. 5 = 9 vz vz+ vz 5 = 10 vz vz 0 = 10 vz vz =(vz  4)(10vz 14). vz = 4 or vz = 1.4; vx = 3 or vx = 4.8 m/s
6 Problem. (15 ponts) A conductor conssts of a crcular loop of radus R= 0.m and two straght, long sectons as shown below. The wre les n the plane of the paper and carres a current I = 5A. R Soluton a. What s the magntude and drecton of the magnetc feld due to the long wre at the center of the crcular loop b. What s the magntude and drecton of magnetc feld due to current n the loop at the center of the loop? c. What s the magntude and drecton of the total magnetc feld at the center of the loop? (a) The magnetc feld due to the long straght wre has magntude and s drected nto the page. Wth I = 5 A and R = 0. m we have B = 5T. 0I d ˆ (b) For the crcular loop s obtaned from the Bot Savart law. B 4 s r. Here, ds s perpendcular to ˆr r so the cross product s just ds and snce r s constant we get that ntegral s just r/r 0I = /r. Thus B s = R 15.7 T and s nto the page. (c) = 0.7 T (drected nto the page)
7 Problem 3. (15 ponts) Consder the long cylndrcal shell for whch a cross secton s shown below. There s a unform current n the shell (blue) that s gong nto the paper as ndcated by the yellow. The magntude of the current s A. The nner radus of the conductng shell s 0.03 m and the outer radus of the outer conductor s m. Use Ampere s law to fnd the magnetc feld a dstance r = (a) 0.05 m, and (b) 0.0 m. Soluton (a) Generally, f we apply Ampere s law ( B d s 0I ) at a pont at radus ra, we have the ntegral wll gve B*ra and, where s the net current through the area of the crcle of radus. In ths case, Ia = A nto the page (the current n the shell), r = 0.05 m so 7 (4 10 T m/a)(a) B = 8 T (0.05 m) 0.03 m (b) From Ampere's law, the magnetc feld s B d s 0I, where I refers to the current enclosed. If we draw an Amperan loop wth radus 0.0 m, then the total current enclosed s zero and so s the magnetc feld.
8 Problem 4. (15 ponts) Consder A conductng bar of length moves to the rght on two frctonless rals as shown n the fgure below. A unform magnetc feld drected nto the page has a magntude of 0.30 T. Assume R = 9.00 Ω and = m. (a) When the bar s 0.1 m from the resstor, what s the flux through the square loop enclosed by the crcut comprsed of the resstor, rals, and movng bar? (b) At what constant speed should the bar move to produce an 8.50mA current n the resstor? (c) What s the drecton of the nduced current? (d) What s around square loop enclosed by the crcut comprsed of the resstor, rals, and movng bar? Soluton (a) B BdA., and here we just have = B*A = (0.3)(0.35)(0.1) = Wb. (b) The motonal emf nduced n the bar must be where I s the current n ths seres crcut. The nduced emf s gven by the change n flux whch wll be BdA/dt = Bldw/dt, so, the speed of the movng bar must be (c) The flux through the closed loop formed by the rals, the bar, and the resstor s drected nto the page and s ncreasng n magntude. To oppose ths change n flux, the current must flow n a manner so as to produce flux out of the page through the area enclosed by the loop. Ths means the current wll flow counterclockwse. (d), so t s just Blv = (0.3)(0.35)(0.79) = V.
9 Possbly Useful Informaton 1 q1 q F ( C / Nm ) r e = C E F q 0 q E, E = / 00EdA. q enc 4 0r x = x  x1, t = t  t1 v = x / t s = (total dstance) / t v = dx/dt a = v / t a = dv/dt = d x/dt v = vo + at g = 9.8 m/s xxo = vot + (½)at r = x + yj + zk v = vo + a(xxo) r=r r 1 xxo = ½( vo+ v)t r = (x  x1) + (y  y1) j + (z  z1) k xxo = vt 1/at v = r / t, v=dr/dt a = dv / dt a = v / t U = Uf  U = W U=W V = Vf  V = W/q0 = U/q0 V = W /q0 V V f E f. ds V E. ds f V 1 40 q r Uf + Kf = U +K V V n V K = ½ mv V Es V E s x E V y E V x ; y ; z z V E 1 q1q U W s 40 r1 Q = CV A C 0 d l ab C 0 C 4 0 ln( b / a) b a C 4 0R Ceq Cj (parallel) 1 1 C eq C (seres) Q U 1 j CV C u 1 0E C = C0 I= dq/dt dq r n q r
10 1 L R A V = IR P = IV P = I R=V /R I ( R r) Pemf = I R R (seres) 1 1 R R (parallel) eq j eq q(t)= Q(1e t/rc ) I = (R)e t/rc q(t) = Qe t/rc I = (QRC)e t/rc, I0 = (QRC) = Q/L, = Q/A, = Q/V E = o F qv x B r = mv/qb, = qb/m F IL x B df Ids x B x B U B NI A db 0 Ids x r 4 3 r, = 4 x 107 T.m/A B = I/( r) B = ni (solenod) F/l = ( I1I)/ a B.ds 0Ienc j d Eds. N B = dt = Blv B BdA. BdA. 0 B
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