Generation of Electricity: BLv sinA 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).

ConstantAppliedForce

a

bInduced magneticForce (Fb=BILsin )

R

If 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 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 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 moving

the loop 2 m into the field while the other end is out, is 36 joules. The applied force must then be18 N.

Fa Fb

Generating electricity page 2

Another equation that can be used to calculate the voltage generated in pulling

the loop through the field is: V= -^0/^tThis can be derived from the diagram below:

a b c

defF 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

V= x/t

FaL

x

Rotating the loop in a field.

2m

3m

axis

A

Now rotate this loop 90 degrees until it is parallel to the field in 2s

B is still 2 tesla; 01= BA= 2(6)02=0 since no field passes through the loop.V= -^0/^T= - 02-01/I= - (0-12)/2= - 6 volts

However, this voltage is an AVERAGE by the nature of deltaB/deltaT. If the work is calculated in rotating theloop 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?

OB

0

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 maximumat …a,b and c and zero at

points d, e and f.

a b c

d

e

fOB = (6)(2) Cos wt

Slope = 9.42

12

2

Flux

Time

Graph of 0=12 cos x becomes O=12 cos 0.7854 T since x=wTand w=pi/4 = 0.7854

Since V= - ^O/^T the generated voltage = - 9.42 volts

Flux

Inducedvoltage

Relationship between the flux moving through the loop andthe Induced Voltage (VL)

VL

FluxVolts

12

O

-9.42

O

Power= V2/R

Volts