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(B) The magnetic field due to a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.At point $P(0, 2, 0) \, m$:$1$. Magneti

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$15$

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(B) The magnetic field due to a long straight wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.At point $P(0, 2, 0) \, m$:$1$. Magnetic field due to wire $A$ (along $X$-axis, $I_1 = 40 \, A$): The distance from the wire to point $P$ is $d_1 = 2 \, m$. The direction of the field is given by the right-hand rule, which is in the $+z$-direction $(\hat{k})$. $B_1 = \frac{\mu_0 (40)}{2 \pi (2)} = \frac{20 \mu_0}{\pi} \, T$.$2$. Magnetic field due to wire $B$ (passing through $(0, 3, 0) \, m$, parallel to $Z$-axis, current $I_2$): The distance from the wire to point $P$ is $d_2 = |3 - 2| = 1 \, m$. The direction of the field is in the $+x$-direction $(\hat{i})$. $B_2 = \frac{\mu_0 I_2}{2 \pi (1)} = \frac{\mu_0 I_2}{2 \pi} \, T$.Since $B_1$ and $B_2$ are perpendicular, the resultant magnetic field $R$ is $R = \sqrt{B_1^2 + B_2^2}$.Given $R = 5 	imes 10^{-6} \, T$ and $\frac{\mu_0}{2 \pi} = 2 	imes 10^{-7} \, T \cdot m/A$:$R = \frac{\mu_0}{2 \pi} \sqrt{20^2 + I_2^2} = 2 	imes 10^{-7} \sqrt{400 + I_2^2} = 5 	imes 10^{-6}$.$\sqrt{400 + I_2^2} = \frac{5 	imes 10^{-6}}{2 	imes 10^{-7}} = 25$.$400 + I_2^2 = 625$.$I_2^2 = 225 \Rightarrow I_2 = 15 \, A$.

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

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