in this example, we’re given that a conductor
p q of length l, is moving with uniform velocity
v as shown, parallel to a long straight wire
carrying a current i. and we’re required
to find the induced e m f in conductor p q.
here we can see that due to this long straight
conductor magnetic field in this region is
in inward direction. all these magnetic lines
are cut by the conductor p q. so a motional
e m f will be induced in it. and, which can
be easily calculated by considering an element
of width d x, from the wire at a distance
x. and for this small element d x which is
also moving with a velocity v, we can see
the free electrons in the element d x, will
experience a magnetic force towards right
by using right hand palm rule. so this element
d x the right side will behave as a negative
side of small battery, and left side will
behave as small positive terminal. or we can
say this element d x behaves like small elemental
induced e m f d e, and all such elemental
induced e m f are connected in series between
p and q. so total e m f induced across p and
q can be calculated by simple integration
of, the e m f induced in this element. so
here we can write, the motional e m f, in
the element, is, this motional e m f we write
as b v l, so here we write magnetic induction
b the speed v and its length is d x. and a
magnetic induction at the location of this
element due to the wire we can directly write,
which is given as mu not i by 2 pi x, this
is the magnetic induction due to wire at a
distance x, multiplied by v into d x. and
total e m f induced in the conductor p q,
can be given as integration of, d e, within
limits of x from r to r plus l for conductor.
so here we can write, this e p q, mu not i
v upon 2 pi, we can take them common out as
they are constants. it is integration of 1
by x d x, which will be integrated from r
to r plus l. so this can be given as mu not
i v by, 2 pi, integration of 1 by x is lon
of x within limits from r to r plus l. if
we substitute the limits e p q, we’re getting
as mu not i v by, 2 pi. this is ellen of r
plus l minus ellen r which can be written
as ellen of r plus l by, r. this’ll be the
answer to this problem.
