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$20$ ampere current is flowing in a long straight wire. The intensity of magnetic field at a distance $10$ cm from the wire will be
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The correct answer is:
$4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$
Here: $i=20 \mathrm{~A}, r=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Intensity of magnetic field produced due to straight current carrying wire is
$B=\frac{\mu_0}{4 \pi} \times \frac{2 i}{r}$
$\begin{aligned} & =10^{-7} \times \frac{2 \times 20}{0.1} \\ & =4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\end{aligned}$
Intensity of magnetic field produced due to straight current carrying wire is
$B=\frac{\mu_0}{4 \pi} \times \frac{2 i}{r}$
$\begin{aligned} & =10^{-7} \times \frac{2 \times 20}{0.1} \\ & =4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\end{aligned}$
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