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Two wires carrying current $5 \mathrm{~A}$ and $2 \mathrm{~A}$ are enclosed in a circular loop as shown in the figure. Another wire carrying a current of $3 \mathrm{~A}$ is situated outside the loop. The value of $f \overrightarrow{\mathrm{B}} \overrightarrow{\mathrm{dl}}$ around the loop is ( $\mu_0=$ permeability of free space, $\overrightarrow{\mathrm{dl}}$ is the length of the Amperion loop)
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$3 \mu_0$
Current of 5A and 2A are enclosed in the loop and are in opposite direction. Hence, net current enclosed by the loop is $3 \mathrm{~A}$. According to Ampere law, $\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \mathrm{l}}=\mu_0 \sum \mathrm{I}=3 \mu_0$
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