Question 146: Magnetism and Matter

Question

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

Accepted Answer

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 $1.6 	imes 10^{3} \, 	ext{Am}^{-1}$ and the number of turns per unit length ($n$) is 8 per cm (which is equal to $8 	imes 10^{-2}$ per meter, as there are 100 cm in a meter).
Substituting these values into the equation gives:
$I = \frac{1.6 	imes 10^{3} \, 	ext{Am}^{-1}}{8 	imes 10^{-2} \, 	ext{turns/m}} = 2 \, 	ext{A}$
So, the current flowing through the solenoid is $2 \, 	ext{A}$.

Magnetism and Matter

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.

Integer / Numeric Answer Type