Q. 183.5( 8 Votes )

A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer :

Current in the cable (I) = 2.5 A


Earth’s magnetic field at the location (H) = 0.33 G = 0.33 × 10-4T


Angle of dip (δ) = 0


Let the line of neutral points be at a distance “r” meters from the horizontal cable.


The magnetic field at the neutral point due to current carrying cable,


,


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


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


At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.


H’ = HN


H × cosδ =


0.33 × 10-4 T × cos0o =


r =


r = 15.15 × 10-3m


r = 1.515cm


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