in this example, the figure shows a semi circular
wire loop of resistance r, and radius a, hinged
at point o, and rotating at an angular speed
omega which we can see in the figure. it is
saying that point o is located on the boundary
of a uniform magnetic induction b, here we
can see on the right side of the boundary,
a uniform magnetic induction b exists, on
left side there’s no magnetic induction.
it is asking to plot the current as a function
of time in the loop, taking clock wise direction
of current as positive. we need to take here,
theta-is equal to zero at t equal to zero.
now here we can see, when the loop is rotating,
its semi circular segment of wire is not cutting
any magnetic line, only the radial length
o b is cutting the magnetic lines. so an e
m f is induced in o b, which can be given
by that of, motional e m f of a rotating rod
about 1 point. so we can write here, e m f
induced, in o b is, e m f we can write as
half b omega ay square. and here if we just
have a look on the free electrons of the wire
due to the motion towards right, they experience
a magnetic force in inward direction, b will
be high potential, o will be, low potential
point. so due to this e m f we can see an
anti clock wise current is produced, and here
induced current in loop can be given as, i
will be e by r that can be written as b omega,
ay square by 2 r. and when-theta becomes pi,
that is after duration t equal to pi by omega
that is half rotation, this o b will come
out of the field and o ay will enter into
the field. and as soon as o ay enters, ay
will be high potential point and due to that
a clock wise current starts flowing. so initially
it was anti clock wise, and we can write after
t is equal to pi by omega, current will be
same that is b omega ay square by, 2 r, but
its direction will be clock wise. and again
after half rotation, this current will start.
so if we plot the graph, of current as a function
of time, we can write from, initial time zero
to pi by omega, the current will remain negative
constant, and the value will be at minus b
omega, ay square by 2 r. and at t equal to
pi by omega it changes to positive value,
magnitude remains same b omega, ay square
by 2 r. and again after half cycle at t equal
to 2 pi by omega, it changes to negative,
then again at 3 pi by omega, it changes to
positive, and so on, so this’ll be, the
answer to our problem, that is the plot of
current as a function of time in the loop.
