5.18 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.)

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Payal Gupta | Contributor-Level 10

4 months ago

5.18 Current in the wire, I = 2.5 A

Angle of dip at the given location of earth, $β$ = 0 $°$

Earth’s magnetic field, H = 0.33 G = 0.33 $\times 10^{- 4}$ T

The horizontal component of earth’s magnetic field is given as:

$H_{H}$ = H $\cos ? β$ = 0.33 $\times 10^{- 4} \cos ? 0 °$ = 0.33 $\times 10^{- 4}$ T

The magnetic field at the neutral point at a distance R from the cable is given by the relation:

$H_{H} = \frac{μ_{0} I}{2 π R}$

$μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

R = $\frac{μ_{0} I}{2 π H_{H}}$ = $\frac{4 π \times 10^{- 7} \times 2.5}{2 π \times 0.33 \times 10^{- 4}}$ = 15.15 $\times 10^{- 3}$ m = 1.52 cm

Therefore, a set of neutral points parallel to and above the cable are located at a normal distance of 1.52 cm.

5.18 Current in the wire, I = 2.5 AAngle of dip at the given location of earth, <math> <mi>β</mi> </math> = 0 <math> <mo>°</mo> </math> Earth’s magnetic field, H = 0.33 G = 0.33 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> </math> TThe horizontal component of earth’s magnetic field is given as: <math> <msub> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> </msub> </math> = H <math> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mi>β</mi> </mrow> </mrow> </math> = 0.33 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> <mi> </mi> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mn>0</mn> <mo>°</mo> </mrow> </mrow> </math> = 0.33 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> </math> TThe magnetic field at the neutral point at a distance R from the cable is given by the relation: <math> <msub> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> </msub> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mi>R</mi> </mrow> </mrow> </mfrac> </math> <math> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> = Permeability of free space = 4 <math> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> </math> T m <math> <msup> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> R = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <msub> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>4</mn> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>2.5</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mo>×</mo> <mn>0.33</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 15.15 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> m = 1.52 cm<div>Therefore, a set of neutral points parallel to and above the cable are located at a normal distance of 1.52 cm.</div>

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