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N = 2.5.10^(3) wire turnsare uniformly wound on a woodern toroidal core of very smalll cross-section. A current I flowsthroughthe wire. Find the magenticinductioninsidethe coreto thatat the centre ofthetoroid. |
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Answer» Solution :Suppose `a` is the radius of cross section of the core. The windinghas a pitch `2piR//N`, so the SURFACE current DENSITY is `vec(J_(s)) = (I)/(2pi R//N) vec(e_(1))+ (1)/(2pi a) vec(e_(2))` where `vec(e_(1))` is a unit vector along the CROSSSECTION of the core and `vec(e_(2))`is a unit vector along its length. The magentic field insidethe crosssection of thecoreis due to first term above, and is given by `B_(varphi) 2pi R = mu_(0) NI` (`NI` is TOTAL current due to the abovesurface current(first term)) Thus , `B_(varphi) = mu_(0) NI//2pi R` The magentic field at the centre of the corecan be obtainedfrom the basic formula. `d vec(B) = (mu_(0))/(4pi) (vec(J_(s)) xx vec(r_(0)))/(r_(0)^(3)) dS` and is due to the second term. So, `vec(B) = B_(s) vec(e_(z)) = vec(e_(z)) (mu_(0))/(4pi) (I)/(2pi a) int (1)/(R^(3)) Rd varphi xx 2pi a` or, `B_(x) = (mu_(0) I)/(2R)` The ratio of the two magentic field, is `= (N)/(pi)` |
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