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An element of length 1 cm placed at the origin carries 10A current along the positive X – axis. The magnetic field at a point on the positive Y – axis at a distance 0.5 m from it is

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$4 \times 10^{- 8} (- k)$

$3 \times 10^{- 6} (- j)$

$3 \times 10^{- 6} j$

$4 \times 10^{- 8} k$

answer is B.

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Detailed Solution

$d \bar{B} = \frac{μ_{0}}{4 π} \frac{Id \bar{l} \times \bar{r}}{r^{3}}$

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