In this example, we are given a square coil
of edge L, which is carrying a current, I
2, is placed near to a long straight wire,
which is carrying a current I 1, as shown
in the figure.
Here we are required to find
the work required to rotate the coil A B C
D, about the axis, this is the given axis
about the edge BC, by an angle 180 degree.
So we are going to flip this coil from this
position to, the dotted position. And, in
this situation, this new position is A dash
D dash, for the edge A D, after rotation by
one eighty degree. So here we are required
to find the work done, and here in solution
we can write, the work done in the process,
we have already studied in the basic theory
videos, the work done in rotation of a coil
can be given by current in coil, multiplied
by the change in flux, phi 1 minus phi 2,
if phi 1 is the initial flux, and phi 2 is
the final flux through the coil. So, we are
going to calculate the flux through the coil
in initial and final stage, and then we substitute
over here, so here we can write, initial flux,
through coil is, here we can calculate phi
1, here the flux through coil can be calculated
by considering an element because the magnetic
field due to I1, will be varying in this coil.
So, at a distance X, from the wire, we consider
an element of width DX, if this element is
of width DX, then the magnetic field at the
location of this element can be given as,
B is Mu not I1 by 2 pie X. So here phi 1 which
is the total flux through the coil, can be
written as integration of, B dot D S, where
we can see the magnetic field due to this
current in inward direction, the area vector
of this square is also in inward direction
so flux will be positive, and here, dot product
will be giving us the same value so, here
it is integration of, we just omit the dot
product because the angle between B vector
and area vector is 0 degree. So here it is
integration of B d S, which is integration
of, d we write as mu not I 1 by 2 pie X, multiplied
by the area of this element we write as L
d S, and we integrate it with its limits from
R 2 R plus L. so on integrating this is giving
us mu not I1L upon 2 pie, and 1 by x d x will
be integrated to Lon X, and limits we substitute
from r to r plus l. and here, the result we
are getting is mu not I 1 L by 2 pie, substituting
the limits from r 2 r plus l this will be
giving us Lon of, r plus l by r. This is the
value of magnetic flux through the coil, in
its initial stage. Similarly if we calculate
the final flux, through coil then this can
be given as phi 2 which is again integration
of B dot d S, but when you consider and element
over here, in this final state, we consider
again an element is a strip, again at a distance
x from the wire, here you can see, when the
coil will flip, the current would be flowing
in this stage in anticlockwise sense, as seen
from above. So magnetic field is still in
downward direction or inward direction due
to I 1, but its area vector is in upward direction.
So here b dot ds will be negative integration
of B d S, which is minus integration of mu
not I by 2 pie x, l d x, integration limits
will be from this point to this point, so
x value from to here to here will be r plus
l to r plus 2 l. So this will be from r plus
l to r plus 2 l. if we integrate this mu not
I 1 l upon 2 pie, 1 by x d x will be Lon x,
and limits we substitute from r plus l to
r plus 2 l, and if we substitute the limits,
here there is a negative sign also which we
miss, so here it will be minus of mu not I
1 l by 2 pie, lon of r plus 2l by r plus l,
that’s the final flux which is negative.
Soif we calculate the work done, in rotating,
the coil, by one eighty degree, is, this work
doen can be given as I 2 multiplied by phi
1 minus phi 2. So here if we substitute the
values, phi 1 is this which is, I can take
this mu not I 1 l upon 2 pie, I can take this
term common out, and inside am having, Lon
of r plus l by r, and minus of phi 2, so this
minus and minus will become plus, this will
be lon of r plus 2 l, by r plus l. This is
what we are getting, and further on simplifying
this value, we are getting the result, mu
not I 1 I 2 l upon 2 pie, and, this lon A
plus lon b will be, lon AB, so here the value
we are getting is lon of r plus 2 l by r.
That’s the result we are getting for this
problem. That is the total work done in flipping
the coil, from initial state to final state.
