Question

A square loop of side a is made by a current carrying wire. Magnetic field at its vertex P is

Answer 1

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)

Answer 2

Explanation:

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