JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 19)
Explanation
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 $2 \, \text{A}$.
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