Question

The self induction ($L$) produced by solenoid of length '$l$' having '$N$' number of turns and cross sectional area '$A$' is given by the formula ($\phi = \text{magnetic flux through one turn}, \mu_0 = \text{permeability of vacuum}$)}

• A. $L = N\phi$
• B. $L = \mu_0 NAl$
• C. $L = \frac{\mu_0 N^2 A}{l}$
• D. $L = \frac{\mu_0 NA}{l}$

Hint:

Self-inductance of a solenoid is proportional to the square of the number of turns ($L \propto N^2$) and the cross-sectional area, and inversely proportional to the length.

Answer 1

The Correct Option is C

Step 1: The magnetic field $B$ inside a long solenoid with $N$ turns and length $l$ carrying current $I$ is: \[ B = \mu_0 \left(\frac{N}{l}\right) I \]
Step 2: The magnetic flux linked with each turn of the solenoid is $\phi = BA$. Substituting the expression for $B$: \[ \phi = \left(\frac{\mu_0 N I}{l}\right) A \]
Step 3: The total flux linkage $\Phi$ for the entire solenoid with $N$ turns is $\Phi = N \phi$. Therefore: \[ \Phi = N \left(\frac{\mu_0 N I A}{l}\right) = \frac{\mu_0 N^2 I A}{l} \]
Step 4: Self-inductance $L$ is defined as the ratio of total flux linkage to the current $I$: \[ L = \frac{\Phi}{I} = \frac{\mu_0 N^2 A}{l} \]