Question

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0A in the same direction east to west. The earth's magnetic field at the place is 0.39G, and the angle of dip is ${35}^0$ . The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0cm below the cable?

Answer 1

$n=4,I=1A,R=0.39G=0.39\times {10}^{-4}T,\delta =35°,\theta =0°$

Distance from cable, $r=0.04m$

$B=\frac{{\mu }_0}{4\pi }\times \frac{2I}{r}$

$B^{\prime }=4\times \frac{{\mu }_0}{4\pi }\times \frac{2I}{r}=\frac{4\times {10}^{-7}\times 2}{4\times {10}^{-2}}=2\times {10}^{-5}T$

$V=R\sin \delta =2.2\times {10}^{-5}T$

$H^{\prime }=H-B=3.19\times {10}^{-5}-2\times {10}^{-5}$

$=1.19\times {10}^{-5}T$

Resultant magnetic field

$R=\sqrt{H^2+V^2}=2.5\times {10}^{-5}T$