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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 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). |
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Answer» Solution :Here I = 2.5 A , `B_E = 0.33 G = 3.3 xx 10^(-5) T`, angle of dip `delta=0^@` . As shown in Fig. 5.02 , the CABLE is lying along magnetic west-east direction. Horizontal component of earth.s magnetic field `B_H= B_E * cos delta= 3.3 xx 10^(-5) cos 0^@ = 3.3 xx 10^(-5) T`. let neutral points lie at a distance r from the cable , they by definition of neutral point, field due to current carrying WIRE must be equal and opposite to `B_H` . Thus, `B= (mu_0 I)/(2pir) = B_H`  `impliesr=(mu_0 I)/(2pi B_H) =( 4pi xx 10^(-7) xx 2.5)/(2pi xx 3.3 xx 10^(-5)) = 1.5 xx 10^(-2) m ` or 1.5 cm As per right hand rule, the neutral points lie on a straight line PARALLEL to the cable at a perpendicular distance of 1.5 cm above the plane of cable (i.e , plane of paper). |
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