Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 1 Magnetic Fields & Frce 1. A pint charge, q= 510 C and m=110-3 m kg, travels with a velcit f: v = 30 ˆ s i then enters a magnetic field: = 110 T ˆj. a. What is the kinetic energ f the pint charge? 1 K = mv = 0.45 J b. What is the magnitude f the magnetic frce that acts n the charge nce it has entered the field? -10 F = qv = 1.510 N c. What is the magnetic frce vectr eerted n the charge just as it enters the field? -10 F = qv = 1.510 N kˆ d. Wh des the magnetic frce eerted n the pint charge nt change its kinetic energ? The magnetic frce is alwas t the directin f travel.. Initiall at rest, a charged particle, q=+1.610-19 C and m=1.6710-7 kg, is accelerated thrugh a regin f cnstant electric field ( E = E ˆ j), acrss a ptential difference f V = -3 100V. The charged particle then enters a magnetic field: = 10 T ˆi. a. What is the kinetic energ f the particle just as it enters the magnetic field? Appl the Cnservatin f Energ t the particle. -17 K = K = U = q V = 1.610 J b. Determine the magnetic frce vectr eerted n the charge, in cmpnent frm, as it enters the field? K 5 m -17 v = = 1.3810 ˆ s F = qv = -.110 N k m c. In which directin is the particle deflected nce it enters the field? Calculate the radius f the particles path. Clckwise
Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 3. A pint charge, q=110 C and m=, travels with a velcit f: m ˆ m ˆ m v = 30 i + 50 j + 40 kˆ, in a magnetic field: = 0.01T ˆj - 0.05T kˆ. s s s a. Determine the magnetic frce vectr eerted n the charge, in cmpnent frm, as it enters the field? F =qv =(v - v )i+(v ˆ - v )j+(v ˆ - v )k=(.9n)i+(1.5n)j+(0.3n)k ˆ ˆ ˆ ˆ b. What is the magnitude f the magnetic frce vectr? F = (.9N) +(1.5N) +(0.3N) = 3.8N 4. A current carring cpper wire, i=a and length L=1.5 m, riented alng the directin is psitined in a cnstant magnetic field, as shwn belw: = 0.1T ˆi i=a a. Determine the magnetic frce vectr eerted n the wire. F ˆ =il =-0.3N k b. Verif that the relatin F =il is equivalent t F =Nqv, where N is the number f charge carriers in the wire affected b the magnetic field, i.e. il = Nqv. d dl F =il = NeL = Ne = qv dt dt
Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 3 Mass Spectrmeter 5. A mass spectrmeter is used t measure the mass f charged particles. Initiall at rest, a beam f charged particles, each with q=+1.610-19 C and m= 1.6710-7 kg, is accelerated thrugh a small aperture acrss a charged capacitr (V = 1000V), fllwing which the particle then enters a magnetic field. a. Determine the magnetic frce vectr eerted n the charge, in cmpnent frm, as it enters the field? q V 5 m v = = 4.3810 s m -15 F =qv =-7.0010 N kˆ b. The particle cmpletes a semicircular path within the field and its displacement is measured with a sensr. Hw far frm the entr pint des the particle reach the sensr? mv -15 F ˆ = =7.0010 N k r mv 4K d=r= = F F d=0.0914 m c. A secnd particle, same charge, is then measured with the device. In rder fr this particle t reach the sensr (same psitin as abve), the field is increased t 0.6T. What is the mass f the particle? m m 1 1 = m =m 1 =6.410 kg 1 d. An imprtant eperiment in develpment f electrmagnetism was the measurement f the mass t charge rati fr an electrn, m e /e. Derive an equatin fr the rati f electrn mass t charge in terms f r,, V using this mass spectrmeter. mv m r 1 ev = ev = since ev = mv v = r e v m m m m r = r = e ev e V + - e. What magnetic field strength wuld be needed t direct the electrn s path t the sensr? q V=1000V Sensr = 0.10T ˆi (ut f page)
Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 4 m m = = = 0.003 T m m e e 1 p 1 p f. What is the electric field vectr, in cmpnent frm, that must be applied t the particle beam, while in the magnetic field, that keep it frm deflecting s that it travels directl t the ppsite side? ee = ev E = v E = v k ˆ 6. A circular lp f wire f radius 0.5 m, area vectr A = A ˆj + A kˆ, and current i=3a (cunterclckwise as viewed frm the right), is psitined within a cnstant magnetic field, as shwn in the figure belw. The angle between the area vectr and the field is 30 in the - plane. a. What is the magnetic mment fr the current lp? ˆ µ = NiA(cs30 j + sin30 k) ˆ ˆ µ =.04 A m j + 1.18 A m kˆ ( ) ( ) b. Determine the trque vectr eerted n the current lp. 30 = 0.110 T ˆj ˆi ˆj kˆ = µ =0.04 A m 1.18 A m -7 = -1.1810 N m ˆi 0 0.110 T 0 c. What is the magnitude and directin f the trque vectr? -7 = 1.1810 N m in the - directin, this will prduce clckwise rtatin d. Fr a net trque vectr f magnitude 5.0 N. m, determine the radius f the lp. µ = Ni( r ) = sin30 3 r = = 3.610 N m Ni sin30
Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 5 7. In the hr Mdel f the hdrgen atm, the electrn mves arund the prtn at a speed f. 10 6 m/s in a circle f radius 5.3 10-11 m. a. Treat the rbiting electrn as a small current lp. Determine the magnetic mment assciated with this mtin. {Hint: The electrn travels arund the circle in a time equal t the perid f mtin.} e er µ = ia = A = = 9.3310 A m T r v -4 prtn electrn b. Determine the magnitude f the magnetic field at the center f the electrn s rbit. µ e µ e µ ev 3 using equatin 9.9 frm Ch9: = = = = 1.610 T Tr r 4r r v c. This atm is placed in a cnstant magnetic field directed initiall 30 with respect t the area vectr f the circular rbit f the electrn. What is the magnitude and directin f the mmentar trque vectr eerted n the electrn b this magnetic field? Assume the area vectr is in the - plane. = µ = -µ sin30 ˆi -3 = -1.1710 N m ˆi r 30 v =.5 T ˆj
Ph 13: General Phsics III 6/14/007 Chapter 8 Wrksheet 6 DC Mtr 8. A.5A current flws thrugh the rtating wire lp, called an armature, f a DC mtr, as shwn in the figure. The dimensins f the armature are 0.m b 0. m. The cnstant magnetic field between the magnets is has a magnitude f 1.010-4 T. a. Determine the magnetic frce n either f the armature segments that is parallel t the magnetic field. -5 F = il = il = 5.010 N b. At what rtatinal psitin f the armature is the frce eerted n the wire maimum? What abut the trque? Trque is maimum when A is t c. Calculate the trque eerted n the armature fr the figure shwn. L 0. m ( ) -5-5 = r1 F 1 + r F = F = 5.010 N = 1.010 N m d. Derive the general equatin fr the trque eerted n the armature. Assume that when the armature rtates 180, the directin f the electric current instantaneusl reverses directin t that it never resists the directin f the rtatin. (t) = il sin ωt+ φ = il sinθ ( ) e. Estimate the average trque generated b the DC mtr when it is in peratin. avg il sin θ dθ 0 il il = =- cs θ = = 6.3710 N m 0 0 dθ f. When the number f lps is 100, what is the average trque generated b the DC mtr while it is in peratin? avg -4 =N avg = 100(6.3710 N m)=6.3710 N m N=1