Question

Three solenoid coils of the same dimension, the same number of turns and the same number of layers of winding are taken. Coil $1$ with inductance$L_1$ was wound using a $\mathrm{Mn}$ wire of resistance $11\Omega m^{-1}$, coil $2$with inductance$L_2$ was wound using a similar wire but the direction of winding was reversed in each layer, coil $3$ with inductance $L_3$ was wound using a superconducting wire. The self-inductance of the coils $L_1,L_{2}and{L}_3$ are

• A. $L_1=L_2=L_3$
• B. $L_1=L_2=L_3=0$
• C. $L_1=L_2,L_3=0$
• D. $L_1>L_2>L_3$

Answer 1

The correct option is C

$L_1=L_2,L_3=0$

Step 1: Given information

$\mathrm{Mn}$ wire of resistance $11\Omega m^{-1}$

The self-inductance coil 1, coil 2 and coil 3= $L_1,L_{2}and{L}_3$

Step 2: Formula

$\Phi =B.Awhere\Phi ismagneticfluxdensity,Bismagneticfeild$

$\Phi ={\mu }_{0}NAi/l$

Step 3: Calculation

The self-inductance of a coil,

$$\begin{gathered} L=\frac{N\Phi }{i} \\ \Rightarrow L=\frac{\left[{\displaystyle \frac{{\mu }_{0}NAi}{l}}\right]}{i} \end{gathered}$$

$L=\frac{{\mu }_{0}NA}{l}$

$\Rightarrow L=\frac{{\mu }_{0}NAl}{l}xl$

$\Rightarrow L={\mu }_0\left(\frac{N}{l}\right)2Al$

$\Rightarrow L={\mu }_{0}n2Al\left(if\frac{N}{l}=n\right)$

Where,

1. ${\mu }_0$ = permeability of air
2. $n$ = number of turns per unit length
3. $l$= length of the solenoid
4. $A$ = area of cross-section
5. $i$ = current

Explanation

1. The coil's self-inductance is affected by its cross-sectional area, length, and number of turns. It will not be affected by the material utilized.
2. The current flow will be continuous if the superconductor is below the critical temperature.
3. The inductance will no longer have the property of inductance. As a result,$L_1=L_2,L_3=0$.

Hence, the correct option is (B).