question_answer

A long straight wire of radius a carries a steady current 1. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at\[\omega \] and 2 a is

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,    \[r=2a,\]