in this example, here the figure shows a rectangular
wire loop ay b c d with length l and breadth
b. and we’re given that a wire is having
a resistance lambda per unit length. if the
loop is pulled out from the magnetic field
at a uniform speed v as we can see in the
figure. we’re required to find the potential
differences v b minus v c and v ay minus v
d. here the length of loop is given as l and
say its breadth is b. now in this situation
we can see, only wire ay d is cutting these
magnetic lines, wire ay b and d c, are not
having any associated magnetic flux, so e
m f will be induced only in wire ay d, and
by using right hand palm rule we can see that
free electrons of wire will move toward end
d. so ay will be the high potential and d
will be the low potential end. and in this
situation we can write, e m f induced, across
ay d is, e ay d we can write as, b v b, as
the motional e m f we can directly write as
b v l, here the breadth is b, in which the
e m f is induced. now, as we’re given that
the wire is having a resistance lambda per
unit length, the resistance of length l will
be lambda l and that of b is lambda b. so
we can draw the equivalent circuit, of the
loop as, between ends ay and b there’s a
resistance of, value lambda l. and between
points ay and b, there’s an e m f induced
which is having an internal resistance lambda
b. here we can consider the resistance of
wire ay d as internal resistance of the e
m f, b v b. and similarly, across points d
and c, the resistance is lambda l, and that
across b and c is lambda b. no e m f is induced
across b c because it is out of the magnetic
field. now in this situation we can directly
see, it is a single loop circuit in which
a current i flows, here we can write induced
current, in circuit, i we can simply write
as, b v b divided by, total resistance, that’ll
be 2 lambda b plus l. now if we know the current
in the circuit and the e m f induced across
ay and d, we can easily find out the potential
difference, like potential difference across
terminals b and c we can directly write v
b minus v c is, i r that is, i lambda b which
can be written as, b v lambda, b square divided
by twice of lambda b plus l. lambda gets cancelled
out and the result is, b v b square, by twice
of b plus l that’ll be 1 answer to this
problem. and another is a potential difference
across ay d, for which we can write it is
a potential difference across, the terminals
of the induced e m f which can be either written
as b v b minus i lambda b, or the potential
differnce across these 3 resistances, which
can be written as i multiplied by, lambda
multiplied by, b plus 2 l. so if we substitute
the value of i, the result we’re getting
is, b v b, multiplied by b plus 2 l, divided
by twice of, b plus l. that’ll be another
answer to this problem.
