let us study about e m f induced in a rotating
conductor in uniform magnetic field. here
we can see, a uniform magnetic field exist
in inward direction, and a straight rod p
q is rotating, about its point p as axis of
rotation, with an angular speed omega. here
also we can see that this conductor is cutting
magnetic flux in magnetic field so there must
be an e m f induced in it. let’s calculate
the e m f by the 2 methods, 1 by motional
e m f and other by faraday’s law. in this
situation as, at every point of the rod, the
velocity is different, we consider at a distance
x from the center an element of width d x.
the velocity of this element is v which we
can write as x omega. this element d x is
revolving in a circle of radius x with the
same angular speed omega. now for this element
which is of width d x , and velocity v it
is moving, we can write, motional e m f, induced,
in element d x is, this we write as d e and
the value will be b v d x, as we know motional
e m f is b v l. so here in the element of
width d x motional e m f is b v d x. and by
using right hand palm rule we can say that
for this element d x, here its left hand will
be high potential, and right end will be low
potential. or we can say this element d x
will behave like, a battery of e m f d e and,
several such batteries are connected in series
between points p and q. so here in this situation
we can write, total e m f induced, across
p and q is, this can be written as e p q which
is integration of d e and this can be given
as integration of b, v we can write as x omega
multiplied by d x with limits from zero to
l, as l is the length of this rod. so in this
situation this e m f induced across ends p
q can be written as b omega as a constant.
and x d x will be integrated as x square by
2 with limits from zero to l. so here induced
e m f across p q is given as, half b omega
l square, which itself is a very important
result, and in many cases this can also be
used directly. and here by right hand palm
rule we can directly say that point p will
be, high potential end, and point q in this
situation will be the low potential end. on
the next sheet we’ll continue, and the same
expression, we’ll calculate by using faraday’s
law without using the direct result of motional
e m f.
let’s continue to understand the same analysis
by using faraday’s law. or this is an alternative
analysis to get the same result. here we can
see as the conductor is rotating at an angular
speed omega, in time d t we can say the conductor
will reach a position where it is rotated
by an angle d-theta in time d t. so we can
write, the angle of rotation, by conductor,
in time d t is, this can be written as d-theta
of which the value is omega d t as the angular
speed is omega. now in this situation at this
time the conductor swept out this sector area,
and we can directly write area of, sector
swept, by conductor, in time d t is, d s we
can write as, half of r square d-theta. and
here we can substitute the value of d-theta
it is, half r square omega d t. and if we
calculate, magnetic flux cut, in time d t
by conductor is, this flux d phi we can write
as b d s. and the area we already calculate
so this’ll be half b omega, r square d t.
and by faraday’s law we can directly write,
e m f induced, in conductor is e p q which
can be written as mod of d phi by d t, and
here on substituting this we’re getting
it as half b omega r square. here we can see
the result is same which we’ve obtained,
by using faraday’s law and on the previous
sheet, i’ve calculated the same expression,
the same result by using motional e m f result.
so we must be careful while analyzing, as
the most important thing here is the e m f
induced in the rotating conductor given by
half b omega r square.
