Q. 194.6( 8 Votes )

A telephone cable

Answer :

Current in the wires (I) = 1.0 A (east to west)


The earth’s magnetic field at the place (H) = 0.39G = 0.39 × 10-4T


The angle of dip (δ) = 35°


Magnetic declination (θ) 0o


Number of wires in the cable (n) = 4


The horizontal component of earth’s magnetic field (H’) = H × cosδ


Magnetic fields due to current carrying cables (B) =


where μ0 = Permeability of free space = 4π × 10-7TmA1


The resultant magnetic fields at point 4.0 cm below the cables


R = 4cm = 0.04m


Resultant horizontal magnetic fields (Hh) = H’- B


Hh = H’- B


Hh =


H � �h = (0.39 × 10-4T × cos35o)-


Hh = 0.39 × 10-4T × 0.819 - 4 × 2 × 25 × 10-7T


Hh = 0.319 × 10-4T-0.2 × 10-4T


Hh = 0.119 × 10-4


Or Hh 0.12 × 10-4 T or 0.12 Gauss


Resultant vertical magnetic field (H � �v �) = vertical component of earth’s magnetic field = H × sinδ


= 0.39 × 10-4T × sin35o


= 0.39 × 10-4T × 0.573


= 0.22 × 10-4T


Resultant magnetic field =



= 0.25 × 10-4T


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