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 0.39 G, and the angle of dip is `35^(@)`. The magnetic declination is nearly zero. What are the resultant magnetic field at points 4.0 cm above and below the cable ?

Explanation

There, no of wire, `n = 4, I = 1.0` amp
Earth's field `R = 0.39 G = 0.39 xx 10^(-4) T`
dip `theta = 35` declination `theta = 0^(@)`
`R_(1) = ?, R_(2) = ?`
r = 4 cm each `4 xx 10^(-2)m`
Magnetic field at 4 cm due to currents in 4 wires
`B = 4 xx (mu_(0)i)/(2pi r) = 4 xx (4pi xx 10^(-7) xx 1)/(2pi xx 4 xx 10^(-2))`
`= 2 xx 10^(-5)` tesla
Horizontal component of earth's field
`H = R cos theta = 0.39 xx 10^(-4) cos 35^(@)`
`= 0.39 xx 10^(-4) xx 0.8192 = 3.19 xx 10^(-5)` tesla
Vertical component of earth's field
`V = R sin theta = 0.39 xx 10^(-4) sin 35^(@)`
`= 0.39 xx 10^(-4) xx 0.5736 = 2.2 xx 10^(-5)` tesla.
A point Q, 4 cm below the wire, horizontal component due to Earth's field due to current are in opposite directions (fig.)
`H_(1) = H - B`
`:. H_(1) = 3.19 xx 10^(-5) - 2 xx 10^(-5)`
`= 1.19 xx 10^(-5)` tesla.
Hence `R_(1) = sqrt(H_(1)^(2) + V^(2))`
`= sqrt((1.19 xx 10^(-5))^(2) + (2.2 xx 10^(-5))^(2))`
`= 2.5 xx 10^(-5)` tesla.
At point P, 4 cm above the wire, horizontal component of Earth's field and field due to current are in the same direction [fig.]
`H_(2) = H + B = 3.19 xx 10^(-5) + 2 xx 10^(-5) = 5.19 xx 10^(-5) T`
`R_(2) = sqrt(H_(2)^(2) + V^(2)) = sqrt((5.19 xx 10^(-5))^(2) + (2.2 xx 10^(-5))^(2)) = 5.54 xx 10^(-5)` tesla.