Solveeit Logo
Login

Question

Physics Question on Magnetic Field Due To A Current Element, Biot-Savart Law

A horizontal overhead powerline is at height of 4m4m from the ground and carries a current of 100 A100\,A from east to west. The magnetic field directly below it on the ground is (μ0=4π×10−7TmA−1)\left(\mu_{0} = 4\pi \times10^{-7} Tm A^{-1}\right)

A

2.5×10−7T2.5 \times10^{-7} T southward

B

5×10−6T5 \times10^{-6} T northward

C

5×10−6T5 \times10^{-6} T southward

D

2.5×10−7T2.5 \times10^{-7} T northward

Answer

5×10−6T5 \times10^{-6} T southward

Explanation

Solution

The magnetic field B=μ04π2IrB = \frac{\mu_{0}}{4\pi} \frac{2I}{r}
=10−7×2×1004= 10^{-7} \times\frac{2\times100}{4}
=5×10−6T= 5 \times10^{-6} T
According to right hand palm rule, the magnetic field is directed towards south.