An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying the same current 'I' along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire 'B' is given by:
μoi22πd
μoi2√2πd
2μoi2πd
√2μoi2πd
Solution:
Magnetic field due to current-carrying wire = μ0I/2πd force due to magnetic field on current-carrying wire per unit length ( μ0I/2πd)Il = μ0I2/2πd μ0I2/2πdResultant = √(μ0I2/2πd)2 = μ0I2√2/2πd