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Physics-

General

Easy

Question

There are two solenoids of same length and inductance L but their diameters differ to the extent that one can just fit into the other. They are connected in three different ways in series. (1) They are connected in series but separated by large distance, (2) they are connected in series with one inside the other and senses of the turns coinciding, (3) both are connected in series with one inside the other with senses of the turns opposite as depicted in figures 1, 2 and 3 respectively. The total inductance of the solenoids in each of the case 1, 2 and 3 are respectively

$0, 4 L_{0}, 2 L_{0}$
$4 L_{0}, 2 L_{0}, 0$
$2 L_{0}, 0, 4 L_{0}$
$2 L_{0}, 4 L_{0}, 0$

The correct answer is: $2 L subscript 0 end subscript comma blank 4 L subscript 0 end subscript comma blank 0$

When two solenoids of inductance $L subscript 0 end subscript$ are connected in series at large distance and current $i$ is passed through them, the total flux linkage $ϕ subscript t o t a l end subscript$ is the sum of the flux linkages $L subscript 0 end subscript i$ and $L subscript 0 end subscript i comma blank i e comma$
$ϕ subscript t o t a l end subscript equals L subscript 0 end subscript i plus L subscript 0 end subscript i$
If $L$ be the equivalent inductance of the system, then
$ϕ subscript t o t a l end subscript equals L i$
$therefore L i equals L subscript 0 end subscript i plus L subscript 0 end subscript i$
or $L equals 2 L subscript 0 end subscript$
When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be mutual inductance $L$ between them. In this case the resultant induced emf in the coils is the sum of the emf $s blank e subscript 1 end subscript$ and $e subscript 2 end subscript$ in the respective coils, $i e comma$
$e equals e subscript 1 end subscript plus e subscript 2 end subscript$
$equals open parentheses negative L subscript 0 end subscript fraction numerator d i over denominator d t end fraction plus-or-minus L subscript 0 end subscript fraction numerator d i over denominator d t end fraction close parentheses plus open parentheses negative L subscript 0 end subscript fraction numerator d i over denominator d t end fraction plus-or-minus L subscript 0 end subscript fraction numerator d i over denominator d t end fraction close parentheses$
Where (+) sign is for positive coupling and (-) sign for negative coupling.
But, $e equals negative L. fraction numerator d i over denominator d t end fraction$
$therefore blank minus L fraction numerator d i over denominator d t end fraction equals negative L subscript 0 end subscript fraction numerator d i over denominator d t end fraction minus L subscript 0 end subscript fraction numerator d i over denominator d t end fraction plus-or-minus 2 L subscript 0 end subscript fraction numerator d i over denominator d t end fraction$
$i e comma L equals L subscript 0 end subscript plus L subscript 0 end subscript plus 2 L subscript 0 end subscript$
$equals 4 L subscript 0 end subscript$ (for positive coupling)
When solenoids are connected in series with one inside the other with senses of the turns opposite, then their is negative coupling.
So, $L equals L subscript 0 end subscript plus L subscript 2 end subscript minus 2 L subscript 0 end subscript equals 0$

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