E.m.f. generated in a rotating coil
Consider a coil of N
turns and area A being rotated at a constant angular velocity q in a
magnetic field of flux density B, its axis being perpendicular to the field (Figure 1).
When the normal to the coil is at an angle q to the field the flux through the coil is BAN
cosq = BAN cos(w)t, since q = wt.
Therefore the e.m.f E generated between the ends of the coil
is:
E = -d(
f)/dt = -d(BANcos
q)/dt
Therefore:
E = BANwsinq = BANwsin(wt)
The
maximum value of the e.m.f (E
o) is when
q (=
wt) = 90
o (that is, the coil is in the plane of the field, Figure 2) and
is given by
Maximum e.m.f (Eo) = BANw
At this
point the wires of the coil are cutting through the flux at right angles – they chop through the field
lines rather than slide along them.
The r.m.s value of the e.m.f is (Er.m.s) = BANw/21/2
Example problem
Example problem
Calculate the maximum value of the e.m.f generated in a coil with 200 turns and of area 10 cm2 rotating at 60 radians per second in a field of flux density 0.1 T.
E= BANw = 0.1x10-3 x 200 x 60 = 1.2 V
Notice the use of radians per second.
