in this example we’re given that figure
shows 2 smooth rails ay b and c d, which are
inclined to horizontal at an inclination alpha.
and here the ends ay and c are joined with
a resistance r and a rod p q of mass m is
horizontally placed on the rails, and released
at t equal to zero. we’re required to find
the velocity of rod, as a function of time.
now in this situation when we release the
rod it’ll experience a downward force, m
g sine alpha because of which it’ll start
moving. in this situation at any instant if
it’ll attain a speed v, we can see that
the magnetic flux, in downward direction is
cut by the horizontal component of the rod’s
velocity which is v coz-alpha as this angle
is alpha. so we can write, that motional e
m f, induced, in rod p q is, this e p q we
can write as, b v coz alpha, multiplied by
l because motional e m f is b v l. and here
v sine alpha is the velocity component which
is parallel to magnetic induction due to which
no motional e m f will be induced in rod.
it is only due to v coz alpha that’s why
we take, e p q as b v l coz alpha. and here
by right hand palm rule we can see, magnetic
induction is in downward direction and, velocity
is toward right, so in this situation we can
see that free electrons of the rod will move
towards p. so the direction of induced e m
f is such that point q will be high potential,
and p will be low potential. due to which
a current i starts flowing in the rod as well
as in the circuit of resistance, so here we
can write, induced current, in rod is, this
i we can write as, this e m f divided by,
its resistance. so this’ll be b v l, coz
of alpha by r. and due to this current, here
we can see, the rod p q will experience a
magnetic force which is in horizontally leftward
direction which is b i l. by right hand palm
rule we can see that field is in downward
direction, and current is in inward direction
so field will be horizontally toward left
at an angle alpha with the inclined plane.
so in this situation the upward component
of this magnetic force on rod will be b i
l coz alpha. so here we can write that if
ay be the acceleration, of rod, here acceleration
we can write as m g sine alpha minus, b i
l coz alpha, divided by m. and here if we
put the value of current as b v l coz alpha
by r, the relation will be given as, m, g
r sine alpha minus, b square, l square v coz
square alpha, divided by m r. this is the
acceleration that we get and, just to find
the velocity of rod as a function of time
here acceleration we can write as, d v by
d t, and further we can separate the variables,
let’s continue on the next sheet.
now in this situation we can separate the
variables as, d v by, m g r sine alpha minus,
b square, l square, v coz square alpha, is
equal to d t by m r. and now if we integrate
this, within limits from t equal to zero to
t and speed it attains, will be from zero
to v. then on integration we get, 1 by, b
square, l square, coz square alpha, with a
negative sign, ellen of, m g r sine alpha,
minus b square, l square, v coz square alpha,
within limits from zero to v is equal to,
t by m r. on further simplifying on substituting
limits here we’ll get, ellen of, m g r sine
alpha minus, b square l square, v coz square
alpha, divided by m g r sine alpha, is equal
to negative of b square, l square t, coz square
alpha, by m r. if we take anti log on both
sides and simplify, we get the speed of rod,
finally as a function of time as, m g r sine
alpha, by b square l square coz square alpha
multiplied by, 1 minus e to power minus, b
square l square t, coz square alpha by, m
r. now this is the answer to this problem
and in this situation we can write, at t tending
to infinity or after a long time, when terminal
speed is attained we can see the terminal
speed of rod we can write as m g r, sine alpha,
by b square, l square, coz square alpha. that’ll
be another answer of terminal speed for this
rod. and this terminal speed as we’ve discussed
earlier, can be directly obtained when, the
total force acting on the sliding rod will
be zero that is when, m g sine alpha will
be balanced by, b i l coz alpha. putting the
value of current we can get this relation
directly, without simplifying this expression.
so if we’ve to find the terminal speed there’s
a direct way to get it.
