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

Magnetic field due to four current-carrying wires at a distance r from them is
` 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) T`
Earth.s magnetic field at the given place, B = 0.39 G Angle of dip, `delta = 35^@`
Horizontal component of the earth.s magnetic field is
`B_H = B cos delta `
` = 0.39 xx 10^(-4) xx cos 35^@ [ because 1G = 10^(-4) T]`
` = 0.319 xx 10^(-4)T `
` = 3.19 xx 10^(-5) T`
Vertical component of the earth.s magnetic field is
`B_V = B sin 35^@`
` = 2.24 xx 10^(-5) T`
(i) At 4 cm below the wire, the magnetic field due to the current and due to the earth.s magnetic field `(vecB_H)` are in the opposite direction. Total horizontal magnetic field, in the horizontal direction
`H_1 = B_H - B`
` = (3.19 - 2) xx 10^(-5) T`
` = 1.19 xx 10^(-5) T`
Resultant `B_H = sqrt(B_V^2 + H_1^2)`
` = sqrt( (2.2 xx 10^(-5) )^2 + (1.19 xx 10^(-5) )^2) `
` = 2.5 xx 10^(-5) T = 0.25 G`
(ii) At 4 cm above the wire, the magnetic field due to the current and the horizontal component of the earth.s magnetic field are in the same direction. Effective horizontal component
` H_2 = 3.19 xx 10^(-5) + 2 xx 10^(-5)`
` = 5.19 xx 10^(-5) T`
` B_V = 2.2 xx 10^(-5) T`
` therefore ` Resultant magnetic field,
`B_R = sqrt(H_2^2 + B_V^2)`
` sqrt( (5.19 xx 10^(-5) )^2 + (2.2 xx 10^(-5) )^2)`
` = 5.6 xx 10^(-5) T = 0.57 G`