Two identical circular wires of radius 20 cm and carrying current $$\sqrt{2}$$ A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is ______ $$\times 10^{-8}$$ T.

(Take $$\pi = 3.14$$)

We have two identical circular loops of radius $$R = 20\;\text{cm} = 0.2\;\text{m}$$, carrying current $$I = \sqrt{2}\;\text{A}$$, placed in perpendicular planes.

The magnetic field at the centre of a circular loop is

$$B = \frac{\mu_0 I}{2R}$$

Each loop produces a magnetic field at the common centre:

$$B_1 = B_2 = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times 0.2} = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{0.4}$$

Since the two loops are in perpendicular planes, their magnetic fields are perpendicular to each other. So the net magnetic field is

$$B_{net} = \sqrt{B_1^2 + B_2^2} = B_1\sqrt{2}$$ $$B_{net} = \sqrt{2} \times \frac{4\pi \times 10^{-7} \times \sqrt{2}}{0.4} = \frac{4\pi \times 10^{-7} \times 2}{0.4}$$ $$B_{net} = \frac{8\pi \times 10^{-7}}{0.4} = 20\pi \times 10^{-7} = 20 \times 3.14 \times 10^{-7}$$ $$B_{net} = 62.8 \times 10^{-7} = 628 \times 10^{-8}\;\text{T}$$

Hence, the net magnetic field at the centre is $$628 \times 10^{-8}$$ T. So, the answer is $$628$$.