A horizontal overhead powerline at a height of \(5 \mathrm{~m}\) from the ground and carries a current of…

A horizontal overhead powerline at a height of \(5 \mathrm{~m}\) from the ground and carries a current of \(150 \mathrm{~A}\) from East to West. The magnetic field directly below it on ground is
  1. \(6 \times 10^{-6} \mathrm{~T}\), towards South
  2. \(6 \times 10^{-6} \mathrm{~T}\), towards West
  3. \(7 \times 10^{-6} \mathrm{~T}\), towards East
  4. \(8 \times 10^{-7} \mathrm{~T}\), towards North

Solution

Magnetic field below the wire on the ground is given as
\(\begin{aligned} B & =\frac{\mu_0}{2 \pi} \cdot \frac{I}{r} \\ & =2 \times 10^{-7} \times \frac{150}{5} \quad\left[\because \mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right] \\ & =6 \times 10^{-6} \mathrm{~T}~(\text {toward South})\end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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