JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 20)
A straight wire carrying a current of $$14 \mathrm{~A}$$ is bent into a semi-circular arc of radius $$2.2 \mathrm{~cm}$$ as shown in the figure. The magnetic field produced by the current at the centre $$(\mathrm{O})$$ of the arc. is ____________ $$\times ~10^{-4} \mathrm{~T}$$
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Answer
2
Explanation
$$
\begin{aligned}
& \mathrm{B}_{\text {at } \mathrm{O}}=\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}} \\\\
& =2 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
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