Saved Bookmarks
| 1. |
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direetion east to west. The earth's magnetic held at the place is 0.39 G, and the angle of dip 35^@. The magnetic declination is nearly zero. What are the resultant magnetic felds at points 4.0 cm below the cable? |
|
Answer» Solution :Magnetic field at a point due to 4 wires , `= 4 ((mu_(o))/(4pi)) ((2i)/(a))` =` (4xx10^(-7)xx2xx1)/ (4xx10^(-2)) = 0.2 XX 10^(-4)` T = 0.2 G Below the CABLE , `B_(H)^(1) = underset(("field pointing inward"))0.39cos35^(@)-underset(("field pointing outward"))0.2 = 0.12G` `B_(V)^(1) = 0.39 sin 35^(@) - 0.22 G ` `:. ` Resultant magnetic field =`sqrt((B_(H)^(1))^(2)+(B_(V)^(1))^(2))` = ` sqrt((0.12)^(2)+(0.22)^(2))` = 0.25 G Resultant field direction `theta = tan ^(-1) (B_(v)^(1))/(B_(H)^(1))` `= tan ^(-1) ((0.22)/(0.11))` `= tan ^(-1) (2)` `theta = 62^(@)` Above the cable : `B_(v)^(1) = 0.39 sin 35^(@)` +0.2 = 0.52 G `B_(v)^(1)` = 0.22 G `:. ` Resultant field = `sqrt((0.52)^(2)+(0.22)^(2))` =0.57 G `theta ^(1) = tan ^(-1) ((0.22)/(0.52))` = `tan^(-1) (0.4231)` = `23^(@)` |
|