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# In a circular coil, when the number of turns is double and resistance becomes half of the initial then inductance becomes.A) $4$ $time$sB) $2$ $times$C) $8$ $times$D) $\text{No Change}$

Last updated date: 17th Apr 2024
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Hint: The circular coil resistance here is half and the number of turns is doubled, so we can calculate the inductance using the solenoid formula with magnetic field of flux. Then We can also assume that self-inductance is a phenomenon in which a voltage is caused in a wire carrying current.

Formula used:
The magnetic flux per turn,
$B = {\mu _0}\dfrac{N}{l}I$
Where, A magnetic field $B$
Current $I$
Solenoid having $N$ turns
Length be $l$
Magnetic material of permeability ${\mu _0}$

Complete step by step solution:
Given by,
Number of turns is double, $N \to 2N$
At any given point, there will be a magnetic field $B$ in the solenoid.
The magnetic flux per turn will, therefore, be equal to $B$
We know that,
$A = \pi {r^2}$
Now, through every turn and the total number of turns, the total magnetic flux $\left( \phi \right)$associated with the solenoid will be supplied by the flux product current.
$\phi = \dfrac{{{\mu _0}NI}}{l} \times \pi {r^2}$
If $L$ is the self-inductance of the solenoid,
Then, $N\phi = LI$
Substituting the given value in above equation,
$\Rightarrow$ $N\dfrac{{{\mu _0}NI}}{l} \times \pi {r^2} = LI$
Common factor is cancelled,
We get,
$\Rightarrow$ $L = \dfrac{{{N^2}\mu A}}{l}$
According to the solenoid formula,
$\Rightarrow$ $L\propto {N^2}$
Then the number of turns is doubled,
We get,
$\Rightarrow$ $L\propto 4{N^2}$
then inductance becomes four times,

Hence, option A is the correct answer.

Note: For the shifting current, which is the alternating current and not for the direct or steady current, the above property of the coil exists only. Self-inductance is often opposed to the changing current in the Henry SI unit and is measured. The induced current is always opposed to the current shift in the circuit, whether the current change is a rise or a decrease.