question_answer

For the given uniform square lamina ABCD, whose centre is O

A)  \[\frac{1}{4}\]  

B)        \[\frac{1}{2}\]  

C)        \[\sqrt{2}{{I}_{AC}}={{I}_{EF}}\]              

D)        \[{{I}_{AD}}=3{{I}_{EF}}\]

Correct Answer: C

Solution :

Let the each side of square lamina is d. So,       \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{4\pi a}\]     (due to symmetry) and      \[\frac{{{B}_{1}}}{{{B}_{2}}}=1\] (due to symmetry) Now, according to theorem of perpendicular axis,                                    \[{{I}_{EF}}={{I}_{GH}}\] \[{{I}_{AC}}={{I}_{BD}}\]  \[{{I}_{AC}}={{I}_{BD}}={{I}_{0}}\]                ?..(i) and  \[\Rightarrow \] \[2{{I}_{AC}}={{I}_{0}}\]    \[{{I}_{EF}}+{{I}_{GH}}={{I}_{0}}\]               ??.(ii)          From Eqs. (i) and (ii), we get  \[\Rightarrow \] \[2{{I}_{EF}}={{I}_{0}}\]    \[{{I}_{AC}}={{I}_{EF}}\]  \[\therefore \] So,    \[{{I}_{AD}}={{I}_{EF}}+\frac{m{{d}^{2}}}{4}\]