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# The Biot Savart's Law in vector form is¯¯¯¯¯¯¯δB=μ04πdl(→l×→r)r3¯¯¯¯¯¯¯δB=μ04πI(→dl×→r)r3¯¯¯¯¯¯¯δB=μ04πI(→r×→dl)r3¯¯¯¯¯¯¯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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