InterviewSolution
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InterviewSolution
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A length L of a wire carries a current I. Show that if the wire is formed into a circular coil, the maximum torque in a given magnetic fiel B, is developed when the coil has one turn only and the maximum torque has the value, `tau_(max)=L^2IB//4pi`. |
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Answer» Torque on the current loop is `tau=NIAB sin alpha` …(i) If there are N turns of circular coil, each of radius r, then `L=2pirN` or `r=L//2piN` Area of the coil, `A=pir^2=piL^2//(4pi^2N^2)=L^2//(4piN^2)` Putting this value in (i),we get `tau=NI[(L^2)/(4piN^2)]B sin alpha=(L^2IBsinalpha)/(4piN)`...(ii) From (ii), it is clear that `tau` is maximum if N is minimum, i.e., the number of turns `N=1`. Torque is maximum if `sin alpha=1` and `N=1` `:. tau_(max)=L^2IB//4pi`. |
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