Question

The permeability of the material of the core used in a solenoid of length 1.4 m, radius 7 cm having $10^3$ turns and a self-inductance of 2.2 H is

• A. $1 \times 10^{-4}\,Hm^{-1}$
• B. $2 \times 10^{-4}\,Hm^{-1}$
• C. $3 \times 10^{-4}\,Hm^{-1}$
• D. $4 \times 10^{-4}\,Hm^{-1}$

Hint:

For a solenoid, \[ L=\frac{\mu N^2A}{l} \] Always remember that inductance increases with permeability and square of the number of turns.

Answer 1

The Correct Option is B

Concept: The self inductance of a long solenoid is \[ L=\frac{\mu N^2A}{l} \] where \[ \mu=\text{permeability of the core material} \] \[ N=\text{number of turns} \] \[ A=\pi r^2 \] \[ l=\text{length of solenoid} \]

Step 1: Calculate the cross-sectional area.
Given, \[ r=7\,cm=0.07\,m \] \[ A=\pi r^2 \] \[ A=\frac{22}{7}(0.07)^2 \] \[ A=0.0154\,m^2 \]

Step 2: Use the self inductance formula.
\[ L=\frac{\mu N^2A}{l} \] \[ \mu=\frac{Ll}{N^2A} \] Substituting values, \[ \mu= \frac{2.2\times1.4} {(10^3)^2\times0.0154} \] \[ \mu= \frac{3.08}{15400} \] \[ \mu=2\times10^{-4}\,Hm^{-1} \]

Step 3: State the answer.
\[ \boxed{2\times10^{-4}\,Hm^{-1}} \]