QJEEM23Ph9afa: JEE Main 2023 Physics - Class 12 JEEM23Ph9afa Solution: The magnetic intensity at the center of a long current carrying …

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 $2 \, \text{A}$ .