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(D) The magnetic field due to an infinitely long wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the $X$-axis carr

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$\frac{5 \mu_0}{2 \pi d}$

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(D) The magnetic field due to an infinitely long wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.For the wire along the $X$-axis carrying current $I_x = 4 	ext{ A}$,the magnetic field at point $P(0, 0, d)$ is directed along the negative $Y$-axis (using the right-hand rule).$B_x = \frac{\mu_0 (4)}{2 \pi d} (-\hat{j})$For the wire along the $Y$-axis carrying current $I_y = 3 	ext{ A}$,the magnetic field at point $P(0, 0, d)$ is directed along the positive $X$-axis.$B_y = \frac{\mu_0 (3)}{2 \pi d} (\hat{i})$Since these two fields are perpendicular,the resultant magnetic field $B$ is given by:$B = \sqrt{B_x^2 + B_y^2}$$B = \sqrt{\left(\frac{4 \mu_0}{2 \pi d}ight)^2 + \left(\frac{3 \mu_0}{2 \pi d}ight)^2}$$B = \frac{\mu_0}{2 \pi d} \sqrt{4^2 + 3^2}$$B = \frac{\mu_0}{2 \pi d} \sqrt{16 + 9}$$B = \frac{5 \mu_0}{2 \pi d} 	ext{ T}$

Two infinitely long wires carry currents $4 \text{ A}$ and $3 \text{ A}$ placed along the $X$-axis and $Y$-axis respectively. The magnetic field at a point $P(0, 0, d) \text{ m}$ will be ...... $\text{T}$.

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