E.m.f. induced in a straight conductor
When a straight conductor is
moved through a magnetic field an e.m.f. is induced between its ends. This movement must
be in such a direction that the conductor cuts through the lines of magnetic flux, and will be a
maximum when it moves at right angles to the field (Figure 1(a)).
Let the length of the conductor be L and the flux density of the
field be B.
If the conductor moves with velocity v at right angles to the field then the flux
cut per second will be BvL (since the conductor will sweep out an area vL every
second).
But the rate of cutting flux is equal to the e.m.f. induced in the conductor.
Therefore
Induced e.m.f (E) = BLv
If the conductor cuts
through the flux at an angle
q ((b) in Figure 1), where
q is the angle between the magnetic field and the direction of motion, the equation
becomes
Induced e.m.f (E) = BLv sinq
You can see that the maximum e.m.f is generated when the conductor
moves at right angles to the field.
(
q = 90
o and so sin
q = sin90 = 1).
Example problems
1. Calculate the e.m.f. generated between the wing tips of an aircraft that is flying horizontally at 200 ms-1 in a region where the vertical component of the Earth's magnetic field is 4.0x10-5 T , if the aircraft has a wingspan of 25 m.
E = 4x10-5 x 25 x 200 = 0.2 V
2. Calculate the e.m.f. generated between the fixed and the free ends of a helicopter blade 9.45 m long that is rotating at 3.5 revs per second.
The vertical component of the Earth's field has a flux density of 4.0x10-5 T.
Area swept out per second = pR2 x 3.5 = px 9.452 x 3.5
E.m.f. = BAN = 4.0x10-5 x p x 89.3 x 3.5 = 3.93x10-2 V = 39.3 mV.
