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, \[{{I}_{AD}}={{I}_{EF}}+\frac{m{{d}^{2}}}{4}\]You need to login to perform this action.
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