in this example, figure shows a fixed coil
of n turns and radius ay which carries a current
i. and at a distance x from its center another
small coaxial coil of radius b, and resistance
r, is moving toward the 1st coil at a uniform
speed v. we’re required to find the induced
current in the smaller coil. then to find
the current in smaller coil, we need to 1st
calculate the e m f induced in it, for which,
we find out magnetic induction, at the location
of, smaller coil, which will be certainly
due to the bigger coil, and for this we use
the magnetic induction at an axial point of
a circular coil. the result we can recall
that we studied in previous topic that is,
mu not n i, ay square by twice of, x square
plus ay square to power, 3 by 2. and if we
know the magnetic induction at the location
of smaller coil we can calculate the magnetic
flux, through smaller coil, which can be written
as phi is equal to b multiplied by the area
of coil, as being a small coil we can azume
that it is uniform in this coil. so it is
b pi b square, and if we know the magnetic
flux as a function of x, then we can directly
calculate the e m f induced, in smaller coil,
which can be written as e as, mod of d phi
by d t, for this magnitude we can take all
the constant terms as it is, and differentiate
the variable terms, so this’ll be, mu not,
pi n, i ay square, b square by 2. we differentiate
this 1 by x square plus ay square to power
3 by 2. on taking magnitude of this derivative
it’ll be 3 by 2, multiplied by 1 by x square
plus ay square to power 5 by 2, multiplied
by 2 x multiplied by d x by d t as we’re
differentiating with respect to time. now
on simplifying the value of e m f we’re
getting as, here one 2 gets cancelled out,
this’ll be 3 by 2, mu not pi n, i ay square,
b square x, d x by d t we can write as v,
divided by, x square plus ay square to power,
5 by 2. and directly now we can write, current
in smaller coil, as this current is induced
e m f divided by r as it is the total loop
e m f in the coil due to variation in magnetic
field. so on substituting the values we get
it as, 3 by 2 mu not, pi n, i ay square, b
square, v x, divided by, r multiplied by,
x square plus ay square to power 5 by 2. that’ll
be the answer to this problem.
