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A conducting rod MN moves with a speed v parallel to a long straight wire which carries a constant current i, as shown in Fig. The length of the rod is normal to the wire. Find the emf induced in the total length of the rod. State which end will be at a lower potential. |
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Answer» Solution :The magnetic field induction due to current i is different at different sections of the rod, because they are at different DISTANCES from the wire.Let us, first of all, subdivide the entire length of the conductor MN into elementary sections. Consider a section (shown shaded in the figure (b)) of thickness dx at a DISTANCE X from the wire. As all the three, V, B and (dX) are mutually normally to each other, so the emf induced in it is de = Bvdx. (from N to M by Fleming.s right hand rule)  For the rest of sections, the induced emf is in the same sense, (i.e, from N to M) `THEREFORE ` Total emf induced in the conductor is ` e = int de = int_b^(b+a) Bvdx` Substituting for `B = (mu_0 i)/(2pi x)`,the above equation gets changed to ` e =int_b^(b+a) (mu_0 iv dx)/(2pi x)` ` e = (mu_0iv)/(2pi)[ln x]_(b)^(b-a) " or" e = (mu_0 iv)/(2pi) ln (1+alb)` |
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