Question 7.7: Given the magnetic vector potential A = -ρ^2/4 az Wb/m, calc...

We can solve this problem in two different ways: using eq. (7.32)

\Psi=\int_{S}B\cdot dS

or eq. (7.51)

\Psi=\oint_{L}A\cdot dl

Method 1:

B=\nabla\times A=-\frac{\partial A_{z}}{\partial \rho}a_{\phi}=\frac{\rho}{2}a_{\phi},       dS=d\rho dz a_{\phi}

Hence

\Psi=\int_{S}B\cdot dS=\frac{1}{2}\int_{z=0}^{5}\int_{\rho=1}^{2}\rho d\rho dz=\frac{1}{4}\rho^{2}|_{2}^{1}(5)=\frac{15}{4}

\Psi=3.75Wb

Method 2:

We use

\Psi=\oint_{L}A\cdot dl=\Psi_{1}+\Psi_{2}+\Psi_{3}+\Psi_{4}

where L is the path bounding surface S; \Psi_{1},  \Psi_{2},  \Psi_{3}, and \Psi_{4} are, respectively, the evaluations of \int_{L}A\cdot dl along the segments of L labeled 1 to 4 in Figure 7.20. Since A has only a z-component

\Psi_{1}=0=\Psi_{3}

That is

\Psi=\Psi_{2}+\Psi_{4}=-\frac{1}{4}\left[(1)^{2}\int_{0}^{5}dz+(2)^{2}\int_{5}^{0}dz\right]

=-\frac{1}{4}(1-4)(5)=\frac{15}{4}=3.75Wb

as obtained by Method 1. Note that the direction of the path L must agree with that of dS.