Question 8.13: Two coaxial circular wires of radii a and b (b > a) are s...

Let current I_{1} flow in wire 1. At an arbitrary point P on wire 2, the magnetic vector potential due to wire 1 is given by eq. (8.21a):

A=\frac{\mu_{o}I\pi a^{2}\sin\theta a_{\phi}}{4\pi r^{2}}

namelyl

A_{1}=\frac{\mu I_{1}a^{2}\sin\theta}{4r^{2}}a_{\phi}=\frac{\mu I_{1}a^{2}ba_{\phi}}{4[h^{2}+b^{2}]^{3/2}}

If h\gg b

A_{1}\simeq \frac{\mu I_{1}a^{2}b}{4h^{3}}a_{\phi}

Hence

\Psi_{12}=\oint A_{1}\cdot dl_{2}=\frac{\mu I_{1}a^{2}b}{4h^{3}}2\pi b=\frac{\mu\pi I_{1}a^{2}b^{2}}{2h^{3}}

and

M_{12}=\frac{\Psi _{12}}{I_{1}}=\frac{\mu\pi a^{2}b^{2}}{2h^{3}}