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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-2 Demo Videos

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Hello, students, now let’s analyse the case of Motional EMF when B is not uniform along the length of the wire. See find the induced EMF in the rod. Current in the wire is constant and the rod is moving with constant velocity v in the vertical direction.

Now see due to the current carrying wire there will be a magnetic field but at a distance X we know magnetic field is Mu0i by 2 pi x . So basically magnetic field is varying along the length of the wire. It is constant with respect to time but along the length of the wire it is changing. Now we know induced EMF or motional EMF is integral v cross B.dl. Now B at different parts of rod is different. So what we have to do we have to find the small induced EMF at small element of the rod. So what we can do, we know v, B and dl are mutually perpendicular, so we can use Bv into dl. Net induced EMF will be the integral of v B into dl. Now see velocity of all the parts of the rod is v, what about the magnitude of B, B is equal to Mu0i by 2pi x. So in the element at a distance x whose length is dx, B at a distance x from the wire will be Mu0i by 2 pi x. So integral v instead of B for that element B will be Mu0i by 2 pi x. And what is the length of that element, the length is dx. So e is equal to integral v B and dx. Now what will be the limits of this integration? See we have to find the net induced EMF in the entire rod, so x will begin from a, it will come at the first end of the rod, so initial limit will be a. Now x has to be a plus L to cover the entire length. So initial was a to the limit a plus L. So this will be our limit, from a to a plus L. Now integration of 1 by dx is lnx constant and we can take it out. So vMu0i by 2 pi lnx, a to a plus L. Now lna minus lnb is lna by v so lna plus L minus lna will be lna plus L by a. So induced EMF will be Mu0iv by 2 pi lna plus L by a.

So when v is not constant over the length what we have to do, we have to take an element, find the small induced EMF in that part, integrate from the initial to final to get the total motional EMF in the rod.

Now suppose we have to find the direction, so direction we can use the same method, v cross b. So if we do v cross b we see left side will be the positive, right side will be the negative. Now suppose it is semi circular, now what we will do, integration will be a bit difficult, so we can use the concept of L effective c. v is constant for all the parts. Now B at different points is different but it is constant with respect to time, see suppose we have taken two points, B is constant now as the wire moves up, the distance of the point remains same. That means B at a point remains same. B at different points will be different, but as the wire moves B at one point will remain same. So B remains same so basically B is constant over the length. And v and B is perpendicular. So we can use the concept of L effective. So v cross B into L effective, L effective will be the same as the previous one. So the answer to this question will again be Mu0iv by 2 pi ln a plus L by a. So we have taken this as L effective. But we can only do this if v is constant, B at different points can be different, but B at one point should remain constant and v and B are perpendicular.

Suppose the situation was like this, this wire is moving towards right. Now if you take a point basically B will change, see if you take L effective and you write the answer as E is equal to 0, because L effective and v is along the length, but this will be incorrect. Why? See, because at same point when the wire moves B changes, suppose we take a point, initial the distance was x then when the wire moves like this the distance changes, so B at the same point changes so we cannot use directly the formula of L effective. So in these situations we have to solve it using integration which will be very complex. But we have to remember but we cannot use L effective in these cases. We have to check that v should be constant, B should be constant and they should be perpendicular.

Thank you.

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