Question

In the Ampere's Law derivation for a solenoid, if the Amperian loop has length \(l\) and the turns per unit length is \(n\), what is the total enclosed current \(I_{\text{enc}}\) when current \(I\) flows through the wire?

Options

• \(I_{\text{enc}} = n \cdot l \cdot I\)

• \(I_{\text{enc}} = \frac{nI}{l}\)

• \(I_{\text{enc}} = 2\pi n I\)

• \(I_{\text{enc}} = \frac{I}{nl}\)

Answer 1

The number of turns piercing the rectangular Amperian loop of length \(l\) is \(nl\). Each of these turns carries current \(I\), so the total enclosed current is \(I_{\text{enc}} = n \cdot l \cdot I\). Substituting this into Ampere's Law gives \(Bl = \mu_0 n l I\), which simplifies to \(B = \mu_0 n I\).