The Biot Savart's Law in vector form is

¯ ¯¯¯¯¯ ¯ δ B = \frac{μ_{0}}{4 π} \frac{d l (\to l \times \to r)}{r^{3}}

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¯ ¯¯¯¯¯ ¯ δ B = \frac{μ_{0}}{4 π} \frac{I (\to d l \times \to r)}{r^{3}}

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¯ ¯¯¯¯¯ ¯ δ B = \frac{μ_{0}}{4 π} \frac{I (\to r \times \to d l)}{r^{3}}

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¯ ¯¯¯¯¯ ¯ δ B = \frac{μ_{0}}{4 π} \frac{I (\to d l \times \to r)}{r^{2}}

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Solution

The correct option is B $¯ ¯¯¯¯¯ ¯ δ B = \frac{μ_{0}}{4 π} \frac{I (\to d l \times \to r)}{r^{3}}$
The Biot Savart's Law we know is given by:
$\to δ B = \frac{μ_{0}}{4 π} \frac{I d l sin (θ)}{r^{2}}^r$ , where $θ$ is the angle between the line element $\to d l$ and the radial unit vector $^r$ .
Now we know: $^r = \frac{\to r}{r}$ , using this and the cross product of two vectors we get; $\to δ B = \frac{μ_{0}}{4 π} \frac{I (\to d l \times \to r)}{r^{3}}$

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