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(C) The magnetic field due to a current element $I d\vec{\ell}$ is given by the Biot-Savart Law: $\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{\ell} 	i

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$\vec{B}_1 = -\vec{B}_2$

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(C) The magnetic field due to a current element $I d\vec{\ell}$ is given by the Biot-Savart Law: $\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{\ell} 	imes \vec{r}}{r^3}$.Here,$d\vec{\ell} = d\ell \hat{k}$ and the position of the element is $\vec{r}_0 = \hat{i} + \hat{j}$.For the origin $(0, 0, 0)$,the position vector relative to the element is $\vec{r}_1 = (0 - 1)\hat{i} + (0 - 1)\hat{j} = -\hat{i} - \hat{j}$. The distance is $r_1 = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.$\vec{B}_1 = \frac{\mu_0 I d\ell}{4\pi} \frac{\hat{k} 	imes (-\hat{i} - \hat{j})}{(\sqrt{2})^3} = \frac{\mu_0 I d\ell}{4\pi (2\sqrt{2})} (-\hat{j} + \hat{i})$.For the point $(2, 2, 0)$,the position vector relative to the element is $\vec{r}_2 = (2 - 1)\hat{i} + (2 - 1)\hat{j} = \hat{i} + \hat{j}$. The distance is $r_2 = \sqrt{1^2 + 1^2} = \sqrt{2}$.$\vec{B}_2 = \frac{\mu_0 I d\ell}{4\pi} \frac{\hat{k} 	imes (\hat{i} + \hat{j})}{(\sqrt{2})^3} = \frac{\mu_0 I d\ell}{4\pi (2\sqrt{2})} (\hat{j} - \hat{i})$.Comparing the two expressions,we see that $\vec{B}_1 = -\vec{B}_2$.

$A$ small current element of length $d\ell$ carrying current $I$ is placed at $(1, 1, 0)$ and is carrying current in the $+z$ direction. If the magnetic field at the origin is $\vec{B}_1$ and at the point $(2, 2, 0)$ is $\vec{B}_2$,then:

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