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# A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The

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

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Number of long straight horizontal wires = 4

The current carried by each wire = 1A

earth’s magnetic field at the place = 0.39 G

the angle of dip = 350

magnetic field due to infinite current-carrying straight wire

$B'=\frac{\mu_0I}{2\pi r}$

r=4cm =0.04 m

$B'=\frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}$

magnetic field due to such 4 wires

$B=4\times \frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}=2\times10^{-5}T$

The horizontal component of the earth's magnetic field

$H=0.39\times10^{-4}cos35=0.319\times10^{-4}T=3.19\times10^{-5}T$

the horizontal component of the earth's magnetic field

$V=0.39\times10^{-4}sin35=0.22\times10^{-4}T=2.2\times10^{-5}T$

At the point below the cable

$H'=H-B=3.19\times10^{-5}-2\times10^{-5}=1.19\times10^{-5}T$

The resulting field is

$\sqrt{H'^2+V^2}=\sqrt{(1.19\times10^{-5})^2+(2.2\times10^{-5})^2}=2.5\times10^{-5}T=0.25G$

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