A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth's magnetic field at the place is 0.39G, and the angle of dip is `35^(@)`. The magnetic declination is nearly zero. What are the resultant magnetic fields at point 4.0cm below the cable?

Explanation

`delta=35^(@)`:
Horizontal component, `B_(H) = B cos delta = 0.39 cos 35^(@) = 0.39 xx 0.82 = 0.32 `gauss.
Vertical component, `B_(v) = B sin delta= 0.32 sin 35^@ = 0.39 ×x 0.57 = 0.2` gauss.
The magnetic field at a distance of 4 cm is given `B=mu_(0)/4pi xx 2NI/r = 4pi xx 10^(-7)/4pi xx 2xx4xx1/4xx10^(-2)`T
`=2xx10^(-5)T=0.20`gauss
Below the wire
Resultant horizontal magnetic field,
`R_(H) = (0.32 -0.20)`gauss = 0.12 gauss
Vertical field, `R_(v)` = 0.22 gaussIf `pi` os the angle which `vec R` makes with horizontal ,then `tan theta = 0.22/0.12=11.6=1.833`
`rArr theta=tan^(-1)(1.833)=61.4^(@)`
Above the wire
Resultant horizontal magnetic field,
`R_(H)=(0.32+0.20)gauss=0.52gauss`
Vertical field,
`R_(v)=0.22 gauss`
`R=sqrt((0.52)^(2)+(0.22)^(2))` `rArr R=0.565 gauss`
`tan theta =0.22/0.52=0.423` or `theta=22.9^(@)`