A) \[{{t}_{2}}\]
B) 4
C) 1
D) \[{{t}_{2}}\]
Correct Answer: A
Solution :
Current density \[=\sqrt{{{(qa)}^{2}}+{{(qa)}^{2}}}\] From Amperes circuital law \[=\sqrt{2}qa\] For \[J=\frac{I}{\pi {{a}^{2}}}\] \[\oint{B.dl}={{\mu }_{0}}.{{I}_{enclosed}}\] \[r<a\] \[B\times 2\pi r={{\mu }_{0}}\times J\times \pi {{r}^{2}}\] At \[\Rightarrow \] \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{4\pi a}\] For \[~r=a/2\] \[B{{=}_{1}}\frac{{{\mu }_{0}}I}{4\pi a}\] \[r>a\] \[B\times 2\pi r={{\mu }_{0}}I\] At \[\Rightarrow \] \[B=\frac{{{\mu }_{0}}I}{2\pi r}\] So, \[r=2a,\]You need to login to perform this action.
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