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

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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?

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