A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (μ0 = 4π ×10-7 Tm A-1)

Question

A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (μ0 = 4π ×10-7 Tm A-1)  (A) 2.5 ×10-7 T southward (B) 5 ×10-6 T northward (C) 5 ×10-6 T southward (D) 2.5 ×10-7 T northward

Tardigrade Admin · Accepted Answer

The magnetic field B = (μ0/4π) (2I/r) = 10-7 ×(2×100/4) = 5 ×10-6 T According to right hand palm rule, the magnetic field is directed towards south.

Q. A horizontal overhead powerline is at height of $4 m$ from the ground and carries a current of $100 A$ from east to west. The magnetic field directly below it on the ground is $(μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1})$

$2.5 \times 1 0^{- 7} T$ southward

$5 \times 1 0^{- 6} T$ northward

$5 \times 1 0^{- 6} T$ southward

$2.5 \times 1 0^{- 7} T$ northward

The magnetic field $B = \frac{μ _{0}}{4 π} \frac{2 I}{r}$
$= 1 0^{- 7} \times \frac{2 \times 100}{4}$
$= 5 \times 1 0^{- 6} T$
According to right hand palm rule, the magnetic field is directed towards south.

Q. A horizontal overhead powerline is at height of 4m from the ground and carries a current of 100A from east to west. The magnetic field directly below it on the ground is (μ0​=4π×10−7TmA−1)

2.5×10−7T southward

5×10−6T northward

5×10−6T southward

2.5×10−7T northward