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# On making a coil of copper wire of length I and coil radius $r$, the value of self-inductance is obtained as L. If the coil of same wire, but of coil radius $\dfrac{r}{2}$ is made, the value' of self-inductance will be-IA.$2L$B. 4 LC.$\dfrac{1}{2}$D. $\dfrac{L}{3}$

Last updated date: 18th Sep 2024
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Hint: In electromagnetism and electronics, inductance is an electrical conductor's tendency to oppose a change in the electrical current that flows through it. A magnetic field around the conductor is created by the flow of electric current. We may also increase inductance by increasing the diameter of the coils, as well as increasing the number of coil turns, or making the core longer.

Formula used:
$\text{L = }\dfrac{\text{N }\left( \text{ }\!\!\varphi\!\!\text{ B} \right)}{\text{I}}$

Complete solution:
We know the formula for self-inductance of the coil “L” is given as,
$\text{L = }\dfrac{\text{N }\left( \text{ }\!\!\varphi\!\!\text{ B} \right)}{\text{I}}$
$\text{L = }\dfrac{\text{N }\left( \text{AB} \right)}{\text{I}}$
$\text{L = }\dfrac{\text{N }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ N }\!\!\times\!\!\text{ I }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}{\text{2r }\!\!~\!\!\text{ }\!\!\times\!\!\text{ I}}$
$\text{L = }\dfrac{{{\text{N}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ r}}{\text{2}}$-(I)
Where
N = no. of turns of coil
I = current
φB = magnetic flux
B = magnetic field

Case 1:
Initial coil radius = $r$
Here let the initial self-inductance be denoted as “${{I}_{1}}$”.
${{\text{l}}_{\text{1}}}\text{ = }\dfrac{{{\text{N}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ r}}{\text{2}}$-(II)

Case 2:
Here we have the coil of same wire but the coil radius = $\dfrac{r}{2}$
Let the self- inductance after the change in the radius of the coil be denoted as “${{I}_{2}}$”.
${{\text{l}}_{\text{2}}}\text{ = }\dfrac{{{\text{N}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ }\dfrac{\text{r}}{\text{2}}}{\text{2}}$- (III)

Now, on dividing the equation (iii) by (ii), we get
$\text{}\dfrac{{{\text{l}}_{\text{2}}}}{{{\text{l}}_{\text{1}}}}\text{=}\dfrac{\text{ }\dfrac{{{\text{N}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ }\dfrac{\text{r}}{\text{2}}}{\text{2}}}{\dfrac{{{\text{N}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\!\mu\!\!\text{ }}_{0}}\text{ }\!\!\times\!\!\text{ }\!\!\pi\!\!\text{ r}}{\text{2}}}$
$\dfrac{{{l}_{2}}}{{{l}_{1}}}=\dfrac{1}{2}$
$\therefore {{l}_{2}}=\dfrac{{{l}_{1}}}{2}$

Thus, the value of self-inductance after the coil radius becomes $\dfrac{r}{2}$. ${{l}_{2}}=\dfrac{{{l}_{1}}}{2}$

Hence, the correct option is (c).

Note:
In electromagnetism and electronics, an electrical conductor's tendency to oppose a change in the electrical current that flows through it is inductance. A magnetic field around the conductor is created by the flow of electric current. Long solenoid inductance depends solely on its physical characteristics (such as the number of wire rotations per unit length and volume) and not on the magnetic field or current.