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Physics-

General

Easy

Question

A circular coil of radius $5 blank c m$ has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be

$25 m i l l i h e n r y$
$25 \times 10^{- 3} m i l l i h e n r y$
$50 \times 10^{- 3} m i l l i h e n r y$
$50 \times 10^{- 3} m i l l i h e n r y$

The correct answer is: $25 blank m i l l i h e n r y$

$ϕ equals L i rightwards double arrow N B A equals L i$
Since magnetic field at the centre of circular coil carrying current is given by $B equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 pi N i over denominator r end fraction$
$therefore N. fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 pi N i over denominator r end fraction. pi r to the power of 2 end exponent equals L i rightwards double arrow L equals fraction numerator mu subscript 0 end subscript N to the power of 2 end exponent pi r over denominator 2 end fraction$
Hence self inductance of a coil
$equals fraction numerator 4 pi cross times 10 to the power of negative 7 end exponent cross times 500 cross times 500 cross times pi cross times 0.05 over denominator 2 end fraction equals 25 blank m H$

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parallel

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