Experiment: Force between two long Parallel Current Carrying Conductors - QS Study

Force between two long parallel current-carrying conductors

AB and CD are two straight very long parallel conductors placed in air at a distance a. They carry currents I1 and I2 respectively. (Figure). The magnetic induction due to current I1 in AB at a distance a is

B1 = (μ0I1/2πα) … … … (1)

Fig: Force between two long parallel current carrying conductors

This magnetic field acts perpendicular to the plane of the paper and inwards. The conductor CD with current I2 is situated in this magnetic field. Hence, force on a segment of length l of CD due to magnetic field B1 is

F = B1I2l

substituting equation (1)

F = (μ0I1 I2l / 2πα) … … … (2)

By Fleming’s Left Hand Rule, F acts towards left. Similarly, the magnetic induction due to current I2 flowing in CD at a distance a is

B2 = (μ0I2/2πα) … … … (3)

This magnetic field acts perpendicular to the plane of the paper and outwards. The conductor AB with current I1, is situated in this field. Hence force on a segment of length l of AB due to magnetic field B2 is

F = B2I2l

substituting equation (3)

F = (μ0I1 I2l / 2πα) … … … (4)

By Fleming’s left hand rule, this force acts towards right. These two forces given in equations (2) and (4) attract each other. Hence, two parallel wires carrying currents in the same direction attract each other and if they carry currents in the opposite direction, repel each other.