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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 meridianof 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 thecable)? (At neutral points, magnetic field due to a current-carrying the cable is equal and opposite to the horizontal component of earth’s magnetic field.)
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 the cable is equal and opposite to the horizontal component of earth’s magnetic field.)
CBSE | Class 12 | Magnetism and Matter | NCERT | Physics
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 the cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer –

Current in the wire is given by 2.5 A

The earth’s magnetic field at a location is given by R= 0.33 G = 0.33 x 10-4 T

Angle of dip is zero is given by δ = 0

Horizontal component of earth’s magnetic field is given by

BH= R cosδ = 0.33 x 10-4 Cos 0 = 0.33 x 10-4 T

Magnetic field due to a current carrying conductor is given by

 Bc = (μ0/2π) x (I/r)

Bc= (4π x 10-7/2π) x (2.5/r) = (5 x 10-7/r)

BH = Bc

0.33 x 10-4  = 5 x 10-7/r

r =  5 x 10-7/0.33 x 10-4 

= 0.015 m = 1.5 cm

As a result, neutral points are 1.5cm apart from the cable on a straight line parallel to it.

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