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