Parallel Current Carrying Wires
ABCD is a rectangular coil having AB=10 cm,AD=25 cm and carries a current of 15 A . 2 cm apart another straight conductor XY carrying a current of 25 A is placed as shown Find the magnitude and direction of net force acting on the straight conductor. Since AB and CD contributes equal and opposite force and they cancell each other The Force due to AD and BC are to be considered, the straight wire XY , AD and BC are parallel conductors F1 and F2 are forces on XY by the sides AD and BC respectively since parallel current attracts and anti parallel current repels F1=[ μ 0 /4π] × (2I 1 I 2 / r 1 ) × 0.25 (repelsive) F2==[ μ 0 /4π] × (2I 1 I 2 / r 2 ) × 0.25 (attractive) here r 1 =0.02 m and r 2 = 0.12 m (10+2=12 cm) • • • F1 =(10 -7 × 2×15×25×0.25) / 0.02= 9.375×10 -4 N (repelsive) and F2=( 10 -7 × 2×15×25×0.25) / 0.12= 1.563×10 -4 N (attractive) Net force , F on XY =F1~ F2 ie 9.375×10 -4 - 1.563×10 -4 F = 7.812 ×10 -4 N repelsive