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 fields at points 4.0 cm below the cable ?

Explanation

Here,
`B_(h) = B cos phi = 0.39 cos 35^(@) = (0.39) (0.8192)`
`therefore B_(h) = 0.3195 G`
`B_(v) = B sin phi = (0.39) (sin 35^(@) ) = (0.39) (0.5736)`
`therefore B_(v) = 0.2237 G`
Magnetic field at distance r from cable due to current passing through it,
`B. = (mu_(0)I) /( 2pi r)`
`therefore B. = (4pi xx 10^(-7) ) (4)/( (2pi ) (0.04) )`
`therefore B. = 0.2 xx 10^(-4) T= 0.2 G`
Resultant magnetic field at point Q in the figure `overset(to) ((B_Q) )`.
If horizontal component of `overset(to)(B_Q)` is `overset(to) (R_h` then at point Q,
`R_h= B_h + B. " "(because overset(to) (B_h) || overset(to) (B.) )`
`=0.3195 +0.2`
`therefore R_h= 0.5195 G`
Now,
`overset(to) (B_Q) = overset(to) (R_h) + overset(to)(B_v) `
`therefore B_(Q) = sqrt(R_(h)^(2) + B_(v)^(2) ) " "(because overset(to)( R_h) bot overset(to)(B_V))`
`therefore B_Q = sqrt((0.5195)^(2) + (0.2237)^(2) )`
`therefore B_Q = 0.5656 G`

For right angled `Delta abc`,
`tan alpha_(1) = (B_v)/(R_h) = (0.2237)/( 0.5195) = 0.4306`
`therefore alpha_(1) = 23^(@) 18.`
Resultant magnetic field at point P in the figure `overset(to) ((B_P))`.
If horizontal component of `overset(to) (B_P)` is `overset(to) (R_h)` then at point P,
`R_(h) = B_(h) - B . " "(because overset(to)(B_h) || (-overset(to) (B.)) )`
`=0.3195 -0.2`
`therefore R_(h) = 0.1195` G
Now,
`overset(to) (B_P) = overset(to) (R_h) + overset(to)(B_v) `
`therefore B_P = sqrt((R_(h)^(2) + B_(v)^(2) ) " " (because overset(to) (R_h) bot overset(to) (B_v) )`
`therefore B_P = sqrt((0.1195)^(2) + (0.2237)^(2) ) )`
`therefore B_(P) = 0.2536` G

For right angled `Delta ab.c.`,
`tanalpha_(2) = (B_v)/( R_h) = (0.2237)/(0.1195)= 1.872`
`therefore alpha_(2)= 61^(@) 54`.