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?

Explanation

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^(@)`