Question

the line integral along side #4 is also equal to zero, $\int_{\text{side }4}\vec{b}\cdot d\vec{s}=0$. this is true for the same reason that the line integral vanishes along side #2. (make sure you understand this statement.) combining the results for the line integrals along the four sides of the rectangle gives a symbolic expression for the magnetic circulation around the closed loop: $\oint\vec{b}\cdot d\vec{s}=bl$. amperes law tells us that the magnetic circulation equals the product $\mu_0i_{enc}$. what is the amount of current that pierces the rectangular surface bounded by the loop, $i_{enc}$? give a symbolic expression (including sign) in terms of any or all of the following $i$ (the current in each winding), $n$ (the number of turns per unit length), $r$ (radius of solenoid), $w$ (width of loop), and $l$ (length of loop). $i_{enc}=\square$

Explanation:

Step1: Recall Ampere's Law

Ampere's law is \(\oint\vec{B}\cdot d\vec{s}=\mu_0i_{enc}\). We are given that \(\oint\vec{B}\cdot d\vec{s} = Bl\).

Step2: Relate to solenoid current formula

For a solenoid, the number of turns \(N\) in length \(l\) is \(N = nl\) (since \(n\) is the number of turns per unit length). Each turn has current \(i\). So the enclosed current \(i_{enc}\) (by the loop of length \(l\) inside the solenoid) is the product of the number of turns in length \(l\) and the current per - turn.

Answer:

\(i_{enc}=nli\)