in this example, the figure shows a wire loop
made up of a wire having resistance r ohm’s
per meter. and a normal magnetic field is
applied on the loop of which magnitude varies
with time as, 4 k t square tesla, where k
is a constant of value 2. we’re required
to find the current in the loop at t equal
to 4 seconds. in this situation, we can directly
state, that the total resistance of this loop
if we calculate, resistance of loop can be
directly given as r is total length multiplied
by small r, so here total length can be written
as, 4 times, ay plus b mutiplied by r, this
much ohm’s will be the total resistance
of the loop. now due to variation in this
magnetic field in inward direction, which
is increasing here with time, so by lens law
we can say if inward direction magnetic field
is increasing, in both of these loops e m
f induced will be anti clock wise in nature
so as to oppose, the external increment in
magnetic induction. so if we calculate e m
f induced, in bigger loop, then this bigger
loop which is of side edge b, the e m f induced
can be written as e 1, which is area multiplied
by d v by d t. so here the area of this loop
is b square and d b by d t we can write it
as, 8 k t. and we’re required to find the
current at t equal to 4 seconds if we put
t equals to 4, and k is equal to 2, then this’ll
be 64 b square, that is at time t equal to,
4 seconds. similarly if we calculate e m f
induced, in smaller loop, then for loop e
m f induced in smaller loop we again use the
formula ay d b by d t. area of this smaller
loop is ay square d b by d t is again 8 k
t. on substituting the values this is 64 ay
square. here we’ve seen that as magnetic
induction is increasing in inward direction,
so the e m f induced will be anti clock wise
here and, here also, this’ll be e 1 this’ll
be e 2 . and we can see, the current induced
due to this, and current induced due to this
e m f, both will oppose each other. so we
can directly write, current in, loop is, it
is a combined loop in which both the loops
are connected to each other. so total current
we can write as, e 1 minus e 2 divided by,
total resistance because, the 2 e m f currents
are opposing each other. so in this situation
if we substitute the values, it’ll be 64,
b square minus ay square, divided by the total
resistance which is 4 times, ay plus b r.
and further simplifying the result we get
is, 16, b minus ay, divided by r, that’ll
be the answer to this problem.
