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 \[\frac{{{N}_{2}}}{{{N}_{1}}}=\exp \left( \frac{-3.16\times {{10}^{-19}}J}{(1.38\times {{10}^{-23}}1/k).(300k)} \right)\]\[\Rightarrow \]You need to login to perform this action.
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