Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wire is ×10−8 T. (Take π=3.14)
Answer: 628
Solution
The magnetic field at the centre of a circular loop is B=2rμ0i. For each loop, B=2×0.204π×10−7×2=2π×10−6T. The two loops lie in mutually perpendicular planes with a common centre, so their fields are perpendicular and combine in quadrature: Bnet=B2+B2=2B=2×2π×10−6=2π×10−6T. With π=3.14, Bnet=6.28×10−6T=628×10−8T.
