Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in the

Question

Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in the figure. The net magnetic field at the center of the circular wires is × 10^-8…

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Given: Two identical circular wires of radius $R = 20 \, \text{cm} = 0.2 \, \text{m}$ carry current $I = \sqrt{2} \, \text{A}$ and are placed in perpendicular planes.

Find: The net magnetic field at the common center in the form $x \times 10^{-8} \, \text{T}$ .

Magnetic field at the center of one circular loop is

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

Since the planes of the two loops are perpendicular, the magnetic fields $B_1$ and $B_2$ at the center are also perpendicular. Therefore,

B_0 = \sqrt{B_1^2 + B_2^2}

For identical loops, $B_1 = B_2 = B$ , so

B_0 = \sqrt{2}B

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

B_0 = \sqrt{2}\frac{\mu_0 I}{2R}

Now put $I = \sqrt{2} \, \text{A}$ and $R = 0.2 \, \text{m}$ :

B_0 = \sqrt{2}\frac{\mu_0 \times \sqrt{2}}{2 \times 0.2}

B_0 = \frac{2\mu_0}{0.4 \times 2} = \frac{\mu_0}{0.2} = 5\mu_0

Using $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$ ,

B_0 = 5 \times 4\pi \times 10^{-7} = 20\pi \times 10^{-7} \, \text{T}

Taking $\pi = 3.14$ ,

B_0 = 62.8 \times 10^{-7} = 628 \times 10^{-8} \, \text{T}

Therefore, the required numerical value is 628.

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