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

Vedclass · Accepted Answer

$3 / 2$

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(A) The magnetic field due to an infinitely long wire carrying current $i$ at a distance $r$ is given by $B = \frac{\mu_0 i}{2 \pi r}$.For the wire at $x_1 = 1 	ext{ cm} = 10^{-2} 	ext{ m}$, the magnetic field at the origin $(0,0)$ is directed along the positive $y$-axis (using the right-hand rule for current into the page): $\vec{B}_1 = \frac{\mu_0 i}{2 \pi (10^{-2})} \hat{j}$.For the wire at $x_2 = 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$, the magnetic field at the origin is also directed along the positive $y$-axis: $\vec{B}_2 = \frac{\mu_0 i}{2 \pi (2 	imes 10^{-2})} \hat{j}$.The total magnetic field at the origin is $\vec{B} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 i}{2 \pi} \left[ \frac{1}{10^{-2}} + \frac{1}{2 	imes 10^{-2}} ight] \hat{j} = \frac{\mu_0 i}{2 \pi 	imes 10^{-2}} \left[ 1 + \frac{1}{2} ight] \hat{j} = \frac{\mu_0 i}{2 \pi 	imes 10^{-2}} \left( \frac{3}{2} ight) \hat{j}$.Given $B_0$ is the magnitude of the field due to the first wire only: $B_0 = \frac{\mu_0 i}{2 \pi 	imes 10^{-2}}$.Therefore, the ratio $B / B_0 = \frac{3}{2}$.

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