question_answer

In the figure shown, the magnetic field at the point P is

A)  \[\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4-{{\pi }^{2}}}\]   

B)  \[\frac{{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}\]

C)  \[\frac{2{{\mu }_{0}}i}{3\pi a}(4+{{\pi }^{2}})\]         

D)  \[\frac{2{{\mu }_{0}}i}{3\pi a}(4-{{\pi }^{2}})\]

Correct Answer: B

Solution :

 Net magnetic field at point P, \[{{B}_{p}}={{({{B}_{1}})}_{p}}+{{({{B}_{2}})}_{p}}+{{({{B}_{3}})}_{p}}+{{({{B}_{4}})}_{p}}+{{({{B}_{5}})}_{p}}\] Where, \[{{({{B}_{1}})}_{p}}=\,\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(-j)\] (semi-infinite wire) \[{{({{B}_{2}})}_{p}}=\frac{{{\mu }_{0}}i}{4\left( \frac{3a}{2} \right)}(+k),\,\,{{({{B}_{3}})}_{p}}=0\] \[{{({{B}_{4}})}_{p}}=\frac{{{\mu }_{0}}i}{4\left( \frac{a}{2} \right)}(-k)\] \[{{({{B}_{5}})}_{p}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{a}{2} \right)}(-j)\] \[\therefore \] \[{{B}_{p}}=\frac{{{\mu }_{0}}i}{2a}\left[ -\left( \frac{1}{3\pi }+\frac{1}{\pi } \right)j-\left( 1-\frac{1}{3} \right)k \right]\] \[\Rightarrow \]            \[{{B}_{p}}=\frac{2{{\mu }_{0}}i}{3a}\left[ \frac{2}{\pi }j+k \right]\] \[\Rightarrow \]            \[{{B}_{p}}=\frac{{{\mu }_{0}}i}{3\pi a}\,\sqrt{4+{{\pi }^{2}}}\]