JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 23)
Two identical circular wires of radius $$20 \mathrm{~cm}$$ and carrying current $$\sqrt{2} \mathrm{~A}$$ are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is __________ $$\times 10^{-8} \mathrm{~T}$$.
_6th_April_Morning_Shift_en_23_1.png)
(Take $$\pi=3.14$$)
Answer
628
Explanation
According to question,
_6th_April_Morning_Shift_en_23_2.png)
$\begin{aligned} \mathbf{B}_{\text {net }} & =\frac{\mu_0 i}{2 r} \hat{\mathbf{i}}+\frac{\mu_0 i}{2 r} \hat{\mathbf{j}}=\frac{\mu_0 i}{2 r} \hat{\mathbf{k}} \sqrt{2} \\\\ & =4 \pi \times 10^{-7} \times \sqrt{2} \times \sqrt{2} \times \frac{1}{2 \times 0.2} \\\\ & =2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} \mathrm{~T}\end{aligned}$
$\begin{aligned} \mathbf{B}_{\text {net }} & =\frac{\mu_0 i}{2 r} \hat{\mathbf{i}}+\frac{\mu_0 i}{2 r} \hat{\mathbf{j}}=\frac{\mu_0 i}{2 r} \hat{\mathbf{k}} \sqrt{2} \\\\ & =4 \pi \times 10^{-7} \times \sqrt{2} \times \sqrt{2} \times \frac{1}{2 \times 0.2} \\\\ & =2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} \mathrm{~T}\end{aligned}$
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