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Shown a circular coil of N turns and radius a, connectedto a battery of emf epsilon through arheostat. The rheostat has a totallength L and resistance R. The resistance of the coil is r. Asmall circular loop of radius a' and resistance r' is placed coaxially with the coil. The centre of the loop is at a distance x from the centre of hte coil. In the beginning, the sliding contact of the rehostat is at the left end and then on wards it is moved towards right at a constant speed v. Find the emf induced int he small circular loop at the instant (a) the contact begins to slide and (b) it has slid thrugh half the length of the rheostat. |
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Answer» Solution :Magneticfield due to the coil (1) at the CENTER of (2) is ` B=( mu_0 Nia^2)/(2(a^2+x^2)^3//2)` Flux linked with the second, `phi=B.A'=( mu_0 Nia^2)/(2(a^2+x^2)^3//2) pi a'^2 ` `E.m.f.induced ` `=(d phi)/(dt)=( mu_0 Na^2 a'^2 pi)/(2(a^2+x^2)^3//2) (di)/(dt)` ` ( mu_0 N pi a^2 a'^2)/(2(a^2+x^2)^3//2) (d)/(dt) (e)/((R/L)x+r)` ` ( mu_0 N pia^2 a'^2)/(2(a^2+x^2)^3//2)e.(-1.(R)/(L).v)/((R)/(L)x+r)^2)` `(a) for x= L` `e=( mu_0 N pi a^2a'^2)/RvE)/(2L(a^2+x^2)^3//2 (R+r)^2)` ` (b) e=( mu_0 N pi a^2a'^2)/(2L(a^2+x^2)^3//2) (ERV)/(L((R/L)+r)^2)` `[for x = l/2,R/L, x=R/2]` |
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