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A long straight wire carries a current of $18 \mathrm{~A}$. The magnitude of the magnetic field at a point $12 \mathrm{~cm}$ from it is
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Verified Answer
The correct answer is:
$3 \times 10^{-5} \mathrm{~T}$
Current, i $=18 \mathrm{~A}$
Distance, $\mathrm{a}=0.12 \mathrm{~m}$
Magnetic field is given by
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}} \\
& =\frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 0.12} \\
& =3 \times 10^{-5} \mathrm{~T}
\end{aligned}$
Distance, $\mathrm{a}=0.12 \mathrm{~m}$
Magnetic field is given by
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{a}} \\
& =\frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 0.12} \\
& =3 \times 10^{-5} \mathrm{~T}
\end{aligned}$
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