Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 cm. If the current in one wire is 1.80 A and the current in the other wire is 3.45 A running in the opposite direction, determine the magnitude and direction of the force per unit length the wires exert on each other.

2.48*10^-5 N/m

Explanation:

The magnitude force per unit length on parallel wires separated by a distance D and currents I1 & I2.

F/l = µ.I1.I2/2πd

Where

µ = 4π *10^-7

I1 = 1.80 A

I2 = 3.45 A

d = 5 cm = 0.05 m

F/l = (4π*10^-7 * 1.8 * 3.45) / (2 * π * 0.05)

F/l = (1.2568*10^-6 * 6.21) / 0.3142

F/I = 7.8*10^-6 / 0.3142

F/l = 2.48*10^-5 N/m

Therefore, the magnitude of the force per unit length is 2.48*10^-5 N/m

if they have a finite resultant

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s