Question

A wire of fixed length is wound on a solenoid of length l and radius r. Its self- inductance is found to be L. Now, if the same wire is wound on a solenoid of length $\dfrac {l}{2}$ and a radius $\dfrac{r}{2}$, then the self-inductance will be
A.2L
B.L
C.4L
D.8L

Answer 1

Use the formula for self-inductance. Substitute area of the solenoid and length of the wire in expression for self-inductance. Thus, find the value of L for the solenoid with length l and radius r. Then, substitute the value of length and radius and find the self-inductance of the wire with length $\dfrac {l}{2}$ and a radius $\dfrac{r}{2}$.

Complete answer:
The expression for self-inductance is given by,
$L= \dfrac {{\mu}_{0}{N}^{2}A}{l}$ …(1)
Where, ${\mu}_{0}$ is the magnetic permeability of free space
${N}^{2}$ is the number of turns per unit length
A is the area of the solenoid
l is the length of the solenoid
Area of a solenoid is given by,
$A=\pi {r}^{2}$
Where, r is the radius of cross-section of the solenoid
Substituting this value in the equation. (1) we get,
$L= \dfrac {{\mu}_{0}{N}^{2} \pi {r}^{2}}{l}$ …(2)
Now, length of the wire is given by,
$length= circumference \times number \quad of \quad turns$
$\Rightarrow l= 2 \pi r \times N$
Since, the same wire is taken in both the cases, then the circumference and number of turns will be the same for both the wires.
$\Rightarrow L= 2 \pi r \times N$
Rearranging the above equation we get,
$r= \dfrac {L}{2\pi N}$ …(3)
Now, substituting equation.(3) in equation.(2) we get,
$L= \dfrac {{\mu}_{0}{N}^{2} \pi {\dfrac {L}{2\pi N}}^{2}}{l}$
$\Rightarrow L= \dfrac {{\mu}_{0} {L}^{2}}{4\pi l}$
This will be the value of L when the length of the wire is l and radius is r.
For, length $\dfrac{l}{2}$ and radius $\dfrac{r}{2}$ value of L will be,
${L}_{0}= \dfrac {{\mu}_{0} {L}^{2}}{4\pi \dfrac {l}{2}}$
$\Rightarrow {L}_{0}=2 \dfrac {{\mu}_{0} {L}^{2}}{4\pi l}$
Thus, the self- inductance of the coil will be doubled.

Hence, the correct answer is option A i.e. 2L.

Note:
The expression for self-inductance of a solenoid is dependent on the physical properties like number of turns of wire per unit length and volume of the solenoid. And it does not depend on the magnetic field or current. Self-inductance is always opposing the changing current.