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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Anshum Sharma Best Answer
First you will have to see what we will have to find .. so in this Question you have to find the induced emf or current produced by the wire on the square loop..... we know that tha total flux on the loop is integration of (B.ds) ds means the surface area. now how we take the limit ... for that we assume that (I)current is in the x axis and z axis towards upward. Now for dx surface of the loop and the wire is parrallel so we take the side of the loop which is from 0 to a ..... and in the dz direction it is perpendicular to the wire so we take the distance from wire to the end of the loop which is s + a from a .. in this way we take the limits in integration.