A small current element of length $'d ℓ'$ and carrying current is placed at (1, 1, 0) and is carrying current i in ‘ + z ’ direction. If magnetic field at origin be ${\vec{B}}_{1}$ and at point (2, 2, 0) be ${\vec{B}}_{2}$ then :

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$|{\vec{B}}_{1}| = |{\vec{B}}_{2}|$

${\vec{B}}_{1} = - {\vec{B}}_{2}$

$|{\vec{B}}_{1}| = |2 {\vec{B}}_{2}|$

${\vec{B}}_{1} = - 2 {\vec{B}}_{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \bar{B} = \frac{μ_{0}}{4 π} \frac{id \bar{ℓ} \times \bar{r}}{r^{3}} for B_{1} \bar{r} = (- \hat{i} - \hat{j}) \\ ∴ {\bar{B}}_{1} = \frac{μ_{0}}{4 π} \frac{i}{2 \sqrt{2}} \hat{k} \times (- \hat{i} - \hat{j}) \dots \dots \dots \dots . . \\ for B_{2} \bar{r} = \hat{i} + \hat{j} \end{array}$

$\begin{array}{l} {\bar{B}}_{2} = \frac{μ_{0}}{4 π} \frac{i \hat{k} \times (\hat{i} + \hat{j})}{2 \sqrt{2}} \dots \\ f r o m (1) & (2) \\ {\bar{B}}_{1} = - {\bar{B}}_{2} & |{\bar{B}}_{1}| = |{\bar{B}}_{2}| \end{array}$