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(B) Given: Current in wire $A$, $I_1 = 2 \ A$. Current in wire $B$, $I_2 = 6 \ A$. Distance of point $P$ from wire $A$, $r_1 = 2 \ m$. Distance of poi

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$2 	imes 10^{-7} \ T$

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(B) Given: Current in wire $A$, $I_1 = 2 \ A$. Current in wire $B$, $I_2 = 6 \ A$. Distance of point $P$ from wire $A$, $r_1 = 2 \ m$. Distance of point $P$ from wire $B$, $r_2 = 5 \ m - 2 \ m = 3 \ m$.Using the right-hand thumb rule, the magnetic field due to wire $A$ at point $P$ is directed into the page, and the magnetic field due to wire $B$ at point $P$ is directed out of the page.The magnitude of the magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.Magnetic field due to wire $A$ at $P$: $B_1 = \frac{\mu_0 I_1}{2 \pi r_1} = 2 	imes 10^{-7} 	imes \frac{2}{2} = 2 	imes 10^{-7} \ T$ (into the page).Magnetic field due to wire $B$ at $P$: $B_2 = \frac{\mu_0 I_2}{2 \pi r_2} = 2 	imes 10^{-7} 	imes \frac{6}{3} = 4 	imes 10^{-7} \ T$ (out of the page).The resultant magnetic field $B_{net} = |B_2 - B_1| = |4 	imes 10^{-7} - 2 	imes 10^{-7}| = 2 	imes 10^{-7} \ T$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

Two long straight parallel wires $A$ and $B$ separated by $5 \ m$ carry currents $2 \ A$ and $6 \ A$ respectively in the same direction. The resultant magnetic field due to the two wires at a point $P$ at a distance of $2 \ m$ from wire $A$ in between the two wires is:

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