JEE Main 2023 — Magnetic Properties Of Matter Question with Solution

In a solenoid, the magnetic field intensity ($H$) is given by the product of the number of turns per unit length ($n$) and the current ($I$) flowing through the solenoid. This can be represented mathematically as:

$H=nI$

This is actually derived from Ampere's law applied to the special case of a solenoid, where the magnetic field is uniform and directed along the axis of the solenoid.

If we rearrange this equation to solve for the current ($I$), we get:

$I=\frac{H}{n}$

In this problem, we're given that the magnetic field intensity ($H$) at the center of the solenoid is $1.6\times 10^3{\text{Am}}^{-1}$ and the number of turns per unit length ($n$) is 8 per cm (which is equal to $8\times 10^{-2}$ per meter, as there are 100 cm in a meter).

Substituting these values into the equation gives:

$I=\frac{1.6\times 10^3{\text{Am}}^{-1}}{8\times 10^{-2}\text{turns/m}}=2\text{A}$

So, the current flowing through the solenoid is $2\text{A}$.