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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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