In this example, here it is given that the
figure shows a conductor O A, of length l,
place along Y Axis with one end, at the origin,
here, O is placed at the origin, and in this
region a non-uniform magnetic field exists
along Zed direction; Of which magnitude depends
only on Y coordinate which is given as B is
B knot, one plus Y square by l Square Tesla.
We can see the magnetic field exists in Z
direction here, and it is saying if the conductor
O A, starts translating with velocity V not
I cap, we are required to find the induced
E M F in this conductor.
So here in solution, as the field is non uniform
we need to consider an element in this conductor,
here we can consider if there is an element
we consider at a height Y, and this element
is of width D Y, and for this small element
of width D Y, if it is moving with a velocity
V not, a motional, e m f, will be induced
in this elelment. Direction we can obtain
by Right hand palm rule; as it is going towards
right field is in upward direction, so positive
charges will experience a force downward and
negative upward so upper end will be negative
and lower end will be positive.
So here we can directly write motional, e
m f, induced, in element D Y is, here this
motional, e m f, D can be written as B, V
Not and its length DY because motional,
e m f, we know it is given as B V L, here B
we can substitute as a function of Y which
B not, V not multiplied by 1 plus Y square
by l square D Y. This is the motional,
e m f, induced, and for such small, e m f, b d
all these are in series so we can write, total,
e m f. Induced, in conductor, O A, is, here
total, e m f, across way can be written as
integration of, D E. that’s b not v not
as constant. there is an integration of one
plus Y square by l square, d y and will integrate
in limits for the length of this conductor
which is from zero to l If we integrate here
E O is given as, b not v not. and the integration
will be. Integration of 1 will be y and that
the y square is y cube by 3 so this y cube
by 3 l square and limits we put from zero
to l, so if we substitute the values, of limits
in this, integral value this is b not v not,
multiplied by, l plus this will be l. Cube
by 3 l square so that will be this cube gets
cancelled out this will be, l by 3. On simplifying
we are getting the value 4 by 3, b not v not
l, that is the result of this problem.
