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 wire is $.............\times 10^{-8}\,T$. (Take $\pi=3.14$ )
JEE MAIN 2023, Medium
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$\text { Magnetic field } B_C \text { at center }=\frac{\mu_0 i }{2 r }$
$=\frac{4 \pi \times 10^{-7}}{2 \times 0.2} \times \sqrt{2}\,T$
Net magnetic field is
$B _{ C } \sqrt{2}= \frac{4 \pi \times 10^{-7} \times \sqrt{2}}{2 \times 0.2} \times \sqrt{2} T =2 \pi \times 10^{-6}\,T$
$=200 \pi \times 10^{-8}\,T$
$=2 \times 314 \times 10^{-8}\,T$
$=628 \times 10^{-8}\,T$
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