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A coil of wire of radius r has 600 turns and a self inductance of 108 mH. The self inductance of a coil with same radius and 500 turns is |
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Answer» 80mH `L=(mu_(0)N^(2)pir)/2` `LpropN^(2)r` `L_(1)/L_(2)=(N_(1)/N_(2))^(2)(r_(1)/r_(2))` `implies (108)/(L_(2))=((600)/(500))^2((r)/(r))` `thereforeL_(2)=75mH.` |
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