in this example, this figure shows a conducting
loop of which, the semi circular part lies
in the magnetic induction b. we can see it
is in outward direction, which varies with
the time given as, ay t cube plus, c t square
plus, f tesla. and here the wire is having
a resistance capital r ohm per meter. and
we’re required to find the current in the
loop at time 2 seconds. now in this situation
here we can see, when in outward direction
magnetic induction is increasing with time,
then in this semi circular loop, an e m f
is induced which will be clock wise, so as
to produce an magnetic induction in inward
direction, to decrease this magnetic induction.
so if this e m f induced is e, we can see
the direction of current by this e m f is
opposing the direction of current produced
by the external battery. so here we can directly
write, induced, e m f e, in the semi circular
part, can be given as, area multiplied by
d b by d t, as it is time varying magnetic
field, and the area in this situation as pi
r squareby 2 that of a semi circle. and d
b by d t from this expression we can write,
3 ay t square plus, 2 c t. now this e m f
is opposing the external battery, so we can
directly write, current in loop, i, this we
can write as battery e m f, minus the induced
e m f divided by the total resistance, that
we denote by r t. now in this situation this
r t we can calculate as, here we can seperately
write, total resistance of loop is, in this
situation total resistance r t we can write,
capital r multiplied by the total length of
this loop. and for this total length we can
see, in this situation, this length here it
is r by 2, so it is r by 2 plus, r by 2 plus,
2 r is the diametrical length, and this semi
circular part, we can use as, plus pi r. so
in this situation, we can see the total resistance
we get is small r capital r, and if we take
remaining terms common this’ll be 3 plus
pi. now on substituting the value of r t over
here, we can see the value of current we get
is e minus, pi r square by 2 multiplied by,
3 ay t square plus 2 c t, divided by, small
r capital r into 3 plus pi. so this’ll be
the total current in the loop that’ll be
the answer to this problem.
