A) \[\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4-{{\pi }^{2}}}\]
B) \[\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}\]
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 :
Idea Magnetic field due to a infinite length wire at its end point is given by\[\frac{{{\mu }_{0}}I}{4\pi d}\]and due to semicircular conducting arc is\[\frac{{{\mu }_{0}}I}{4R}\]at centre. Net magnetic field at point P, \[{{\mathbf{B}}_{P}}={{({{\mathbf{B}}_{1}})}_{P}}+{{({{\mathbf{B}}_{2}})}_{P}}+{{({{\mathbf{B}}_{3}})}_{P}}+{{({{\mathbf{B}}_{4}})}_{P}}+{{({{\mathbf{B}}_{5}})}_{P}}\] where, \[{{({{\mathbf{B}}_{1}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(-j)\](semi-infinite wire) \[{{({{\mathbf{B}}_{2}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(+\mathbf{k}),{{({{\mathbf{B}}_{3}})}_{P}}=0\]You need to login to perform this action.
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