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Magnetism And Matter

Question
CBSEENPH12037739

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 magnetic field.)

Solution

Here,
Current, I = 2.5 A.
Earth's magnetic field, B = 0.33G = 0.33 x 10–4 T
Angle of dip, δ = 0°

Horizontal component of earth’s field
H = B cos δ
   = 0.33 x 10–4 cos 0°
   = 0.33 x 10–4 Tesla 

Let the neutral points lie at a distance r from the cable. 

Strength of magnetic field on this line due to
current in the cable = μ0I2πr 

At neutral point,
                     μ0I2πr = H r = μ0I2πH        = 4π×10-7×2.52π×0.33×10-4        = 1.5 × 10-2m         = 1.5 cm 

Hence, neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5 cm above the plane of the paper.