The wire shown in figure below carries a current of 60 A. Find the magnetic field B at P.

Given, current carried by the wire, I= 60 A .

The field at P arises from 3/4th of the circular loop only because P lies on the straight wires themselves.
Thus,

B = \frac{μ_{0}}{4 π} \frac{2 πl}{r} \frac{θ}{360} θ = 270 ° T h e r e f o r e, B = \frac{3 \times 10^{- 7} \times 2 π \times 60}{4 \times 0.02} = 1.4 \times 10^{- 3} T

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