Finally for square coil \(M' = i \times {\left( {\frac{L}{4}} \right)^2} = \frac{{i{L^2}}}{{16}}\) ..... \((ii)\)
Solving equation \((i)\) and \((ii)\) \(M' = \frac{{\pi M}}{4}\)
$(i)$ | $(ii)$ | $(iii)$ |
(A) $\frac{{{\mu _0}i}}{r}$ $\otimes$ | (A) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\otimes$ | (A) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\otimes$ |
(B) $\frac{{{\mu _0}i}}{{2r}}$ $\odot$ | (B) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}} \right)$ $\otimes$ | (B) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}} \right)$ $\otimes$ |
(C) $\frac{{{\mu _0}i}}{{4r}}$ $\otimes$ | (C) $\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\odot$ | (C)$\frac{{{\mu _0}i}}{4}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)$ $\odot$ |
(D) $\frac{{{\mu _0}i}}{{4r}}$ $\odot$ | (D) $0$ | (D) $0$ |