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Question

The Biot Savart's Law in vector form is
  1. ¯¯¯¯¯¯¯δB=μ04πdl(l×r)r3
  2. ¯¯¯¯¯¯¯δB=μ04πI(dl×r)r3
  3. ¯¯¯¯¯¯¯δB=μ04πI(r×dl)r3
  4. ¯¯¯¯¯¯¯δB=μ04πI(dl×r)r2

A
¯¯¯¯¯¯¯δB=μ04πI(dl×r)r2
B
¯¯¯¯¯¯¯δB=μ04πI(dl×r)r3
C
¯¯¯¯¯¯¯δB=μ04πI(r×dl)r3
D
¯¯¯¯¯¯¯δB=μ04πdl(l×r)r3
Solution
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The Biot Savart's Law we know is given by:
δB=μ04πIdlsin(θ)r2^r, where θ is the angle between the line element dl and the radial unit vector ^r.
Now we know: ^r=rr, using this and the cross product of two vectors we get; δB=μ04πI(dl×r)r3

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