Two identical circular wires of radius 20 ~cm and carrying current \sqrt2 ~A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is \underline\hspace2cm \times 10^-8 ~T . (Take \pi=3.14 )

Magnetics PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q65 with Solution

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Correct answer: 628

For a circular loop of radius $R$ carrying current $I$ , the magnetic field at its centre is

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

Given:

So for each loop,

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

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

B_1 = 10\pi\sqrt{2} \times 10^{-7}

The two circular wires are in perpendicular planes, so the magnetic field produced by each loop at the common centre is along the axis of that loop.

Hence, the two magnetic fields are perpendicular to each other and have equal magnitude $B_1$ .

Therefore, net magnetic field:

B_{\text{net}} = \sqrt{B_1^2 + B_1^2} = \sqrt{2}B_1

Substitute $B_1$ :

B_{\text{net}} = \sqrt{2}\left(10\pi\sqrt{2} \times 10^{-7}\right)

B_{\text{net}} = 20\pi \times 10^{-7}

Using $\pi = 3.14$ ,

B_{\text{net}} = 20 \times 3.14 \times 10^{-7}

B_{\text{net}} = 62.8 \times 10^{-7}

B_{\text{net}} = 628 \times 10^{-8}\ \text{T}

The blank is:

\boxed{628}

Stored correct answer = $628$ .

This matches the derived answer.