A) 2a
B) \[B,{{B}_{B}}=\frac{{{\mu }_{0}}2I}{2\times 2R}\]
C) 4a
D) \[\frac{{{B}_{A}}}{{{B}_{B}}}=1\]
Correct Answer: B
Solution :
We have, \[2\{{{(x-a)}^{2}}+{{(y-a)}^{2}}\}={{(x+y)}^{2}}\] \[\Rightarrow \] \[\sqrt{{{(x-a)}^{2}}+{{(y-a)}^{2}}}=\frac{1}{\sqrt{2}}|x+y|\] \[\Rightarrow \] \[\sqrt{{{(x-a)}^{2}}+{{(y-a)}^{2}}}=\left| \frac{x+y}{\sqrt{2}} \right|\] Clearly, this equation represents a parabola having Its focus at (a, a) and directrix \[x+y=0\]. \[\therefore \]\[\text{Length of latusrectum}=\text{2}\times (\text{Distance}\] \[\text{between focus and directrix})\] \[=2\left| \frac{a+a}{\sqrt{1+1}} \right|=2\sqrt{2}a\]
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