in this example, we’re given that on a smooth
horizontal table, a disc is placed with non
conducting ring, with uniformly distributed
charge q fixed on its sircumference. both
disc and ring are of mass m and radius r.
if a uniform magnetic field of induction b,
which is symmetric with center of disc is
switched on, in vertically downward direction.
we’re required to find the angular speed
attained by the disc due to this. now in this
situation, if we draw the situation we can
directly state, on a smooth horizontal table,
there’s a disc placed, on which on the sircumference
a ring is fixed which is having a uniformly
distributed charge q. and if in this situation
in downward direction, a symmetric uniform
magnetic field b is switched on, then suddenly
the magnetic flux through, the ring will increase,
due to which an electric field is induced
for a small duration, in which the field is
switched on. and due to the impulse of this
field, the ring will start rotating. now in
this situation if we calculate the electric
field induced, at the sircumference, of disc,
then this electric field we can write as,
half r, d b by d t. in this situation say
the switched on duration of the field is delta
t. so it is increasing from zero to b in time
delta t we can write it as, half r, b by delta
t. as the field is increased from zero to
b in time delta t, and as the process is very
quick we can consider it b by delta t as d
b by d t. and if we calculate the angular
impulse, on ring, due to the torque of this
electric field which we denote as j-theta.
then this can be written as, tau delta t and
the torque on the ring can be written as,
q e r, multiplied by delta t is the angular
impulse on it. and whatever angular impulse
is imparted will, increase the angular momentum
of the ring and that can be written as i omega.
in this situation as magnetic field is increasing
in downward direction, the ring will rotate
in anti clock wise manner so as to oppose,
the increase in magnetic induction by lens
law, so it’ll attain angular speed omega
in anti clock wise fashion. so here we can
directly write, angular speed, attained, this
omega can be written as, q e r, delta t by,
moment of inertia, and the total moment of
inertia of this disc plus ring system we can
directly write as, for the disc it is half
m r square, and for ring it is m r square.
so this can be given as, q e r, delta t divided
by, 3 by 2, m r square. if we substitute the
value of this induced electric field, we get
omega as, q r, delta t divided by, 3 by 2,
m r square, multiplied by half r, b by delta
t. in this situation this delta t gets cancelled
out and this r square will also get cancelled
out, so the final result we’re getting is,
this 2 also, is getting cancelled, so the
result of omega we get as, q b, by 3 m,that’ll
be the answer to this problem.
