question_answer

Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

A) \[\sqrt{\frac{3}{2}}\]                                     

B) \[{{e}^{1/2}}\]

C) \[{{\left( x-\frac{1}{x} \right)}^{4}}{{\left( x+\frac{1}{x} \right)}^{3}}.\]  

D) \[{{A}^{2}}=A\]

Correct Answer: C

Solution :

Magnetic moment of a loop is given by \[x=\frac{r}{3},y\]where, I = current through the loop A = area of loop Perimeter of square should be equal to the circumference of circle, since wires have same length. Therefore,\[\frac{{{N}_{2}}}{{{N}_{1}}}\]                              ?.(i) In Fig. (i), magnetic moment of square loop is given by \[\frac{{{N}_{2}}}{{{N}_{1}}}=\exp \left( -\frac{{{E}_{2}}-{{E}_{1}}}{kT} \right)\]\[\lambda =550nm\]\[\Rightarrow \] [from Eq. (i)]?(ii) In Fig. (ii), magnetic moment of circular loop is given by \[{{E}_{2}}-{{E}_{1}}=\frac{hc}{\lambda }=3.16\times {{10}^{-19}}J\] \[\Rightarrow \]                                               ...(iii) From Eqs. (ii) and (iii), we get \[\frac{{{N}_{2}}}{{{N}_{1}}}=\exp \left( \frac{-3.16\times {{10}^{-19}}J}{(1.38\times {{10}^{-23}}1/k).(300k)} \right)\]\[\Rightarrow \]