Question

A long solenoid of $N$ turns has a self inductance $L$ and area of cross section $A$. When a current $i$ flows through the solenoid, the magnetic field inside it has magnitude $B$. The current $i$ is equal to:

• A. $\frac{BNA}{L}$
• B. $BANL$
• C. $\frac{BN}{AL}$
• D. $\frac{B}{ANL}$

Answer 1

The correct option is A $\frac{BNA}{L}$

Number of turns of solenoid $=N$

$\therefore$ Number of turns per unit length is,

$n=\frac{N}{l}$

Where, $l$ is the length of the solenoid.

Magnetic field inside the solenoid is,

$B={\mu }_{0}ni$

$B={\mu }_0\frac{Ni}{l}...\left(1\right)$

Also, self inductance of solenoid is,

$L={\mu }_0{n}^{2}Al$

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

$L=\frac{{\mu }_0{N}^{2}A}{l}...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$\frac{B}{L}=\frac{i}{NA}$

$\Rightarrow i=\frac{BNA}{L}$

​​​​​​​Hence, $\left(A\right)$ is the correct answer.