Presentation on theme: "Generation of Electricity: BLv sin"— Presentation transcript:
1Generation of Electricity: BLv sin A wire of length (L)pulled through a magnetic field (B) at an angle 0at a constant speed (v) will generate a constant voltage (V).aConstantAppliedForceRFaFbInduced magneticForce (Fb=BILsin )bIf just the wire segment a-b is pulled, a voltage is attained but no workneed be done maintaining the velocity of the wire. If a complete loop isused with a total resistance of R ohms then work must be done to pullthe loop at constant velocity and heat is generated in the resistance:Fa D= VIT= -BILD sinV=BLD/T=BLv sinIf a loop 2m long and 3 m wide a-b moves to the left at 1 m/s,at right angles to a magnetic field of 2 T, the Voltage producedis 6 volts and the power is 18 watts. The work done movingthe loop 2 m into the field while the other end is out, is36 joules. The applied force must then be18 N.
2Generating electricity page 2 a b c Fa x d e f Another equation that can be used to calculate the voltage generated in pullingthe loop through the field is: V= -^0/^tThis can be derived from the diagram below:abcV= x/tLFaxdefF d =work done = VIT=-BILd= Energy generated.V= - BL d/T= - B^A/^TThe flux 0 through the loop is BA. 01 is the B times the area bounded by a-b-e-f.As the loop moves further into the field distance x (e to d) the additional areamoving into the field is equal to the area bounded by b-c-d-e. Therefore^0=B^A and V= - B^A/^T and V= -( 2 tesla)(3m)1m= -6 volts
3Rotating the loop in a field. Now rotate this loop 90 degrees until it is parallelto the field in 2sA3mB is still 2 tesla; 01= BA= 2(6)02=0 since no field passes through the loop.V= -^0/^T= /I= - (0-12)/2= - 6 volts2maxisHowever, this voltage is an AVERAGE by the nature ofdeltaB/deltaT. If the work is calculated in rotating theloop 1/4 rotation in 2 s generating 6v (average) we findits not 36 joules as in the previous problem but 44joules!What isgoing onhere?The flux through the theloop (B A cos O ) makesthe following graph of flux:Since the Rate of Change ofFlux equals the induced V,we can see that it is maximumat …a,b and c and zero atpoints d, e and f.OBeabcdfOB = (6)(2) Cos wt
4Graph of 0=12 cos x becomes O=12 cos 0.7854 T since x=wT FluxSlope = 9.422TimeGraph of 0=12 cos x becomes O=12 cos T since x=wTand w=pi/4 =Since V= - ^O/^T the generated voltage = volts
5Relationship between the flux moving through the loop and the Induced Voltage (VL)FluxVolts12FluxOOVL-9.42Inducedvoltage