The magnetic intensity at the center of a long current carrying solenoid is found to be $1.6 \times 10^{3} \mathrm{Am}^{-1}$. If the number of turns is 8 per cm, then the current flowing through the solenoid is __________ A.

<p>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:</p> <p>$H = nI$</p> <p>This is actually derived from Ampere&#39;s law applied to the special case of a solenoid, where the magnetic field is uniform and directed along the axis of the solenoid.</p> <p>If we rearrange this equation to solve for the current ($I$), we get:</p> <p>$I = \frac{H}{n}$</p> <p>In this problem, we&#39;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).</p> <p>Substituting these values into the equation gives:</p> <p>$I = \frac{1.6 \times 10^{3} \, \text{Am}^{-1}}{8 \times 10^{-2} \, \text{turns/m}} = 2 \, \text{A}$</p> <p>So, the current flowing through the solenoid is $2 \, \text{A}$.</p>