A square loop of side a is made by a current carrying wire magnetic field at its vertex p is
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let the vertex be D
the magnetic field at D due to side AD and CD will be zero since they lie inline with the vertex D
using biot-savart's law, magnetic field at D due to side AB is given as
B = (μ/4π) (i/a) (Sin0 + Sin45)
B = (μ/4π) (i Sin45/a)
similarly , magnetic field by BC at D is given as
B' = (μ/4π) (i/a) (Sin0 + Sin45)
B' = (μ/4π) (i Sin45/a)
hence the net magnetic field at D is given as
B'' = B + B'
B'' = (μ/4π) (i Sin45/a) + (μ/4π) (i Sin45/a)
B'' = (μ/4π) (2 i Sin45/a)
B'' = (√2) (μ/4π) (i/a)
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