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(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.For the wire carrying $5 	ext{ A}$ current at a distance

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$\frac{\mu_0}{2 \pi} \otimes$

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(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.For the wire carrying $5 	ext{ A}$ current at a distance of $2.5 	ext{ m}$ from $P$,the magnetic field is:$B_1 = \frac{\mu_0 	imes 5}{2 \pi 	imes 2.5} = \frac{2 \mu_0}{2 \pi} \otimes$ (directed into the page).For the wire carrying $2.5 	ext{ A}$ current,the distance from $P$ is $5 	ext{ m} - 2.5 	ext{ m} = 2.5 	ext{ m}$. The magnetic field is:$B_2 = \frac{\mu_0 	imes 2.5}{2 \pi 	imes 2.5} = \frac{\mu_0}{2 \pi} \odot$ (directed out of the page).The net magnetic field at $P$ is:$B_{net} = B_1 - B_2 = \frac{2 \mu_0}{2 \pi} - \frac{\mu_0}{2 \pi} = \frac{\mu_0}{2 \pi} \otimes$.

For the adjoining figure,the magnetic field at point $P$ will be:

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