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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 ? |
Answer» Solution : 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 . ""(becauseoverset(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`. |
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