in this example, we’re given that a uniform
rod ay b of mass m and length l is placed
over 2 smooth conducting rails p and q, we
can see in the figure. if the switch shown
is closed at t equal to zero, we’re required
to find the velocity of rod ay b as a function
of time. now in this situation if we just
have a look on the situation on closing the
switch, a current flows through the circuit
through the wire ay b also. and if a current
i flows through it, we can see by right hand
palm rule the wire experiences a magnetic
force f which is b i l, that is toward right,
due to which the rod will start accelerating.
but as it moves and attains velocity v, again
we can say a motional e m f will be induced,
with ay as high potential end, b as low potential
end, by right hand palm rule we can see here.
and due to which the value of current will
decrease as motional e m f will oppose the
external e m f in this situation. so here
we can see when, rod speed is v, motional
e m f, in rod is, this e m f we can write
as b v l. and at this instant, current in
circuit can be, directly given as i will be,
e minus b v l, by r, because total e m f in
this loop will be e minus the motional e m
f of rod, this is the current. so we can write
in this situation, acceleration of rod is,
ay we can write as, magnetic force by mass
which can be written as, b i l by, m. and
if we put the value of i over here, we can
see on substituting value of i we get it as,
b l upon, m r, e minus b v l. here we can
see acceleration as a function of velocity,
and acceleration we can write as d v by d
t, to obtain velocity as a function of time
and on simplifying and seperating the variables,
we get d v upon, e minus b l v is equal to,
b l by m r d t. let’s continue on the next
sheet.
now in this expression we can integrate it,
as at t equal to zero the rod started from
rest, velocity was zero and at time t, its
speed say becomes v. and on integrating we’re
getting it as, minus 1 by b l, ellen of e
minus b l v. and we apply limits from zero
to v is equal to b l by, m r. integrating
d t we’re getting it as t. so in this situation
on further simpifying we get ellen of, e minus
b l v, upon e, is equal to we can take these
terms on the other side it’ll be, minus
b square l square t by m r. on further simplifyintg
if we take anti log on both sides, so it’ll
be e minus b l v over e, is equal to e to
power, minus b square l square t over m r.
and on simplifying the expression for the
value of velocity, we’re getting as e by
b l, 1 minus e to power minus, b square l
square t by m r. so this’ll be the answer
to this problem which is giving us velocity
as a function of time. if we just have a look
on this velocity and we analyze for our knowledge,
we can see at, t turning to infinity or, we
can say after a long time, in this situation
this factor will become zero. we can see the
velocity will approach to a constant value
it is, e by b l. and we can say this’ll
remain constant and further the rod will continue
to move with this velocity. and for this velocity
we can directly state in the given situation,
when the motional e m f b v l becomes equal
to the battery e m f, we can say in this situation
current in battery becomes zero which implies
that magnetic force is zero, which implies
that acceleration is zero which implies the
velocity will become, constant. and from this
expression also we’re getting the velocity
as, e by b l. this velocity we can call as,
terminal speed because, velocity increases
upto this level and then, the rod continues
to move with this velocity only.
