Course Hero Logo

Weber 10 15 0015web 500 0025 300 n δt ε δ δt δ n ε

Course Hero uses AI to attempt to automatically extract content from documents to surface to you and others so you can study better, e.g., in search results, to enrich docs, and more. This preview shows page 56 - 58 out of 72 pages.

weber10*1.50.015web5000.025*300NΔt*εΔΔtΔNεlaw,sFaraday'Using300V0.02510*0.75ΔtΔIpMε??Δε500,N0.025sec,Δt10A,0A-10AΔIpH,0.75M:DataGiven2ss15.7: A solenoid has 250 turns and its self-inductance is 2.4 mH. What is the flux through each turnwhen the current is 2 A? What is the induced emf when the current changes at 20As-1?48mVV10*4820*10*2.5ΔtΔI10*92.12502*10*2.5NLILINequationUsing?ε?,20A/sec,ΔtΔI2A,IH,10*2.52.5mHLinductanceSelf250,N:DataGiven3-3-53-3-Lweb15.8: A solenoid of length 8.0cm and cross sectional area 0.5cm2has 520 turns. Find the self-inductance of the solenoid when the core is air. If the current in the solenoid increases through1.5A in 0.2s, find the magnitude of induced emf in it.volt10*1.60.21.5*10*2.12ΔtΔILεH10*2.1210*/810*0.5*(520)10*4π/l)A(NμlA(N/l)μlAnμLEquationUsing?ε?Lsec,0.2Δt1.5A,ΔI520,Nm10*0.50.5cmAm,10*8cm8llength:DataGiven34424,-27-2o2o2o2-4,2-215.9: When current through a coil changes from 100 mA to 200 mA in 0.005s, an induced emf of 40mV is produced in the coil. (a) What is the self-inductance of the coil?(b) Find the increase in theenergy stored in the coil.Created in Master PDF Editor - Demo VersionCreated in Master PDF Editor - Demo Version
57mJJEL03.0310*03.0)10*100(10*221)(21E2mHH10*20.0050.13-10*40tIL?EstoredenergyinIncrease?LinductanceSelf0.005secΔtV,10*4040mVεA,10*100100mA-200mAΔI200mA,I100mA,I:DataGiven23323-3-32115.10: Like any field, the earth’s magnetic field stores energy. Find the magnetic energy stored in aspace where strength of earth’s fields isT5107, if the space occupies an area of281010mand has a height of 750m.JALBUmUmmLo9875-228-510*5.1)750*10*10(10*42)10*(721)(21?,750,m10*10AT,10*7B:DataGiven15.11: A square coil of side 16cm has 200 turns and rotates in a uniform magnetic field ofmagnitude 0.05T. If the peak emf is 12V, what is the angular velocity of the coil?47rad/secc46.9rad/se0.05*10*256*20012NABεωNωωAε?ω12V,ε0.05T,B200,N,m10*25616cm*16AArea16cm,coilsquareofSide:DataGiven4-ooo2-4215.12: A generator has a rectangular coil consisting of 360 turns. The coil rotates at 420 rev per minin 0.14 T magnetic field. The peak value of emf produced by the generator is 50V. If the coil is 5.0 cmwide, find the length of the side of the coil.0.45m0.05*43.96*360*0.1450bBNωεLb)B*Nωω(NωωAεformulatheusing?Lcoiloflength0.05m5/1005cmbwidth50V,ε0.14T,Brad/sec,43.96602π*420ω360,N:Datagivenooo15.13: It is desired to make an a.c generator that can produce an emf of maximum value 5kV with 50Hz frequency. A coil of area 1m2having 200 turns is used as armature. What should be themagnitude of the magnetic field in which the coil rotates?

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 72 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Fall
Professor
N/A
Tags
Electric charge, editor

Newly uploaded documents

Show More

Newly uploaded documents

Show More

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture