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A horizontal overhead power line carries a current of $90 \mathrm{~A}$ in East to West direction. What is the magnitude and direction of the magnetic field due to the current, $1.5 \mathrm{~m}$ below the line?
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Verified Answer
The correct answer is:
$1.2 \times 10^{-5} \mathrm{~T}$, towards South
The given situation is shuwn below
Magnetic fiekl at point $P_{\text {, }}$
$$
\begin{aligned}
B &=\frac{\mu_{0}}{9 x}, \frac{l}{r} \\
&=2 \times 10^{-7} \times \frac{90}{1.5} \\
&=1.2 \times 10^{-6} \mathrm{~T}(\text { Sonth })
\end{aligned}
$$
Magnetic fiekl at point $P_{\text {, }}$
$$
\begin{aligned}
B &=\frac{\mu_{0}}{9 x}, \frac{l}{r} \\
&=2 \times 10^{-7} \times \frac{90}{1.5} \\
&=1.2 \times 10^{-6} \mathrm{~T}(\text { Sonth })
\end{aligned}
$$
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