A straight wire carrying a current of $14\,A$ is bent into a semicircular are of radius $2.2\,cm$ as shown in the figure. The magnetic field produced by the current at the centre $(O)$ of the arc. is $.........\,\times 10^{-4}\, T$
JEE MAIN 2023, Easy
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$B _{\text {at } 0}=\frac{\mu_0 I }{4 R }=\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}$
$=2 \times 10^{-4}\,T$
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