Consider two infinitely long parallel conducting wires carry

Consider two infinitely long parallel conducting wires carrying currents I_1 and I_2 in the same direction. They are separated by a distance d. The force F_12 per length experienced by win 2 due to B_2 caused by the current I_1 is given by F_12 = I_2 (1 z times B_12). Find (a) F_12 and (b) F_21 (i.e., the force per unit length experienced by wire 1 due to B_21 caused by the current I_2). If I_1, and I_2 flow in opposite directions, find (c) F_12 and (d)f_21.

Solution

(A) current in I1 is along +ve x axis.

so magnetic field due to I1 at the location of I2 will be given as

B = (u0 I1 ) / (2 pi d) (-i^)

(field due to long current carrying wire at distance d )

now force on I2 due to magnetic field of I1:

F = IL x B

where IL = I2L (k^)

F12 = ( I2L (k^) ) X ((u0 I1 ) / (2 pi d) (-i^) )

F12 = - u0 I1 I2 L / (2 pi d) (-j^)

F12/L = - u0 I1 I2 / (2 pi d) (-j^)

(B) from newton\'s 2nd law,

reaction force will be same in magnitude but opposite in direction.

F21 = u0 I1 I2 / (2 pi d) (-j^)

(c) revrsing direction will change in direction of force only.

F12 = u0 I1 I2 / (2 pi d) (-j^)

(d)
F21 = - u0 I1 I2 / (2 pi d) (-j^)