• # 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)  $\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}$ 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) ${{({{\mathbf{B}}_{2}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(+\mathbf{k}),{{({{\mathbf{B}}_{3}})}_{P}}=0$ ${{({{\mathbf{B}}_{4}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{a}{2} \right)}(-\mathbf{k})$ ${{({{\mathbf{B}}_{5}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{a}{2} \right)}(-\mathbf{j})$ $\therefore$${{\mathbf{B}}_{P}}=\frac{{{\mu }_{0}}i}{2a}\left[ -\left( \frac{1}{3\pi }+\frac{1}{\pi } \right)j-\left( 1-\frac{1}{3} \right)k \right]$ ${{\mathbf{B}}_{P}}=\frac{{{\mu }_{0}}i}{3a}\left[ \frac{2}{\pi }j+k \right]$ $\Rightarrow$$|{{\mathbf{B}}_{P}}|=\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}$ TEST Edge Questions related to calculation of net magnetic field at a point are asked frequently, to solve such problem it is beneficial to learn some standard result e.g., magnetic field at centre of current carrying circular coil is$B=\frac{{{\mu }_{0}}I}{2R}$and magnetic field due to infinitely long current carrying straight wire near its centre is$B=\frac{{{\mu }_{0}}I}{2\pi d}$

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