If a current of 5 A is passed through a straight wire of length 6 cm, then the magnetic induction at a point 5 cm from the either end of the wire is

Let AB represents the current carrying conductor and P the point where magnetic induction B is required.Now,    B=μ0I4π  sin ϕ1 + sin ϕ2 Here,  r=OPNow,  AO=OB=62  cm = 3 cmAnd   PB=PA=5  cmOP= PB2 − OB2=  52 −32 = 4  cmr=4  cm  = 4  × 10−2  metresin ϕ1 =35  and  sin ϕ2 = 35B=10−7  54 ×  10−2  35 + 35= 64  ×  10−5 = 0.15  ×  10−4 tesla=0.15  gaussThe direction of magnetic induction is normal to the plane of ABP.