Let B_p, B_q and B_r be the magnetic fields produced by the three infinite long wires P, Q and R, respectively. These wires are placed symmetrically inside an equilateral triangular loop as shown in the figure.
Current in 3 wires areas shown in figure.If $\int_{A}^{B} {\vec{B}}_{P} . \vec{dl} = 4 μ_{0}$ and $\int_{C}^{A} {\vec{B}}_{Q} . \vec{dl} = - 15 μ_{0} .$ Considering SI units, the value of I is

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5 A

15 A

13 A

answer is C.

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Detailed Solution

$\begin{array}{l} Now \int_{B}^{C} {\vec{B}}_{P} . \vec{dl} should be one third of \int_{C}^{A} {\vec{B}}_{Q} . \vec{dl} \\ So, \int_{A}^{C} {\vec{B}}_{P} . \vec{dl} = - 5 μ_{0}, Also \int_{C}^{A} {\vec{B}}_{P} . \vec{dl} = \int_{A}^{B} {\vec{B}}_{P} . \vec{dl} = 4 μ_{0} \\ Now, \int_{ABC}^{} {\vec{B}}_{P} . \vec{dl} = μ_{0} I \\ \Rightarrow 4 μ_{0} + 5 μ_{0} + 4 μ_{0} = μ_{0} I \\ \Rightarrow I = 13 A \end{array}$