Question

interactive figure 29.10.2: rectangular loop of length ( l ) and width ( w ) lies in ( yz ) plane. one side lies on the ( z ) axis, and the opposite side is outside the solenoid.
the magnetic circulation about the closed loop, ( oint vec{b} cdot dvec{s} ), is divided into four distinct parts, based on the four sides of the rectangle (traversing the loop clockwise):
oint vec{b} cdot dvec{s} = int_{\text{side } 1} vec{b} cdot dvec{s} + int_{\text{side } 2} vec{b} cdot dvec{s} + int_{\text{side } 3} vec{b} cdot dvec{s} + int_{\text{side } 4} vec{b} cdot dvec{s}
give a symbolic expression for the line integral along side #1 ( int_{\text{side } 1} vec{b} cdot dvec{s} ). you may use some or all of the following symbols: ( b ) (the magnitude of the field on side #1), ( l ), ( w ), and ( r ).
int_{\text{side } 1} vec{b} cdot dvec{s} = square

Explanation:

Step1: Analyze the direction of \(\vec{B}\) and \(d\vec{s}\)

For side 1, the magnetic field \(\vec{B}\) is parallel to the side (assuming the solenoid's field is uniform along its length). The line - integral formula is \(\int\vec{B}\cdot d\vec{s}=B\int ds\cos\theta\), where \(\theta\) is the angle between \(\vec{B}\) and \(d\vec{s}\). Since \(\vec{B}\) and \(d\vec{s}\) are parallel, \(\theta = 0^{\circ}\) and \(\cos\theta=1\).

Step2: Calculate the integral

The length of side 1 is \(l\). So, \(\int_{side1}\vec{B}\cdot d\vec{s}=B\int_{0}^{l}ds\). Using the integral formula \(\int_{a}^{b}ds=b - a\), when \(a = 0\) and \(b = l\), we have \(\int_{0}^{l}ds=l\).

Answer:

\(Bl\)