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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 magnetie meridian of the place happens to be 10^@ west of the geographie 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 potnts, magnetic field due to a current-carying cable is equal and opposite to the horizontal component of earth's magnetic field.) |
Answer» Solution : As shown in the diagram, for all the points along the horizontal line at perpendicular distance R below the given horizontal cable carrying current from a to b, if `B.=B_h` then all these points (like `P_1, P_2, ….)` would be NULL points. HENCE, `B.= B_h` `therefore (mu_(0 ) I)/( 2pi r) = B cos phi` (Where `B=` magnetic FILED of Earth and `phi =` dip angle ) `therefore (mu_(0 ) I)/( 2pi r) = B ( because "Here", phi =0^@` ) `therefore r= (mu_(0) I)/( 2pi B)` `therefore ((4pi xx 10^(-7) )(2.5) )/( (2pi ) (0.33 xx 10^(-4) ) )` `therefore r= 1.5 xx 10^(-2) m` |
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