Download presentation

Presentation is loading. Please wait.

1
**Generation of Electricity: BLv sin**

A wire of length (L)pulled through a magnetic field (B) at an angle 0 at a constant speed (v) will generate a constant voltage (V). a Constant Applied Force R Fa Fb Induced magnetic Force (Fb=BILsin ) b If just the wire segment a-b is pulled, a voltage is attained but no work need be done maintaining the velocity of the wire. If a complete loop is used with a total resistance of R ohms then work must be done to pull the loop at constant velocity and heat is generated in the resistance: Fa D= VIT= -BILD sin V=BLD/T=BLv sin If 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 produced is 6 volts and the power is 18 watts. The work done moving the loop 2 m into the field while the other end is out, is 36 joules. The applied force must then be18 N.

2
**Generating electricity page 2 a b c Fa x d e f**

Another equation that can be used to calculate the voltage generated in pulling the loop through the field is: V= -^0/^t This can be derived from the diagram below: a b c V= x/t L Fa x d e f F d =work done = VIT=-BILd= Energy generated. V= - BL d/T= - B^A/^T The 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 area moving 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

3
**Rotating the loop in a field.**

Now rotate this loop 90 degrees until it is parallel to the field in 2s A 3m B 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 volts 2m axis However, this voltage is an AVERAGE by the nature of deltaB/deltaT. If the work is calculated in rotating the loop 1/4 rotation in 2 s generating 6v (average) we find its not 36 joules as in the previous problem but 44joules! What is going on here? The flux through the the loop (B A cos O ) makes the following graph of flux: Since the Rate of Change of Flux equals the induced V, we can see that it is maximum at …a,b and c and zero at points d, e and f. OB e a b c d f OB = (6)(2) Cos wt

4
**Graph of 0=12 cos x becomes O=12 cos 0.7854 T since x=wT **

Flux Slope = 9.42 2 Time Graph of 0=12 cos x becomes O=12 cos T since x=wT and w=pi/4 = Since V= - ^O/^T the generated voltage = volts

5
**Relationship between the flux moving through the loop and **

the Induced Voltage (VL) Flux Volts 12 Flux O O VL -9.42 Induced voltage

6
Power= V2/R Volts

Similar presentations

OK

Chapter 20 Induced Voltages and Inductance. Faraday’s Experiment A primary coil is connected to a battery and a secondary coil is connected to an ammeter.

Chapter 20 Induced Voltages and Inductance. Faraday’s Experiment A primary coil is connected to a battery and a secondary coil is connected to an ammeter.

© 2018 SlidePlayer.com Inc.

All rights reserved.

By using this website, you agree with our use of **cookies** to functioning of the site. More info in our Privacy Policy and Google Privacy & Terms.

Ads by Google

Ppt on self awareness books Ppt on self awareness questions Ppt on power grid failure Ppt on 9-11 conspiracy theories video Ppt on osteosarcoma Ppt on central limit theorem applet Crt display ppt on tv Free download ppt on data handling in maths for class 8 Ppt on taking notes By appt only