Magnetic field produced by current \(I_1\) on the conductor Q is given by
\(B_1 = \frac{\mu_0}{4\pi}.\frac{2I_1}{d}\)
The conductor Q carrying current \(I_2\) in the magnetic field \(B_1\) experiences a mechanical force \(F_1\) is given by
\(F_1=B_1I_2Lsin\theta\) If \(\theta = 90^0\)
\(F_1 = B_1I_2L\) towards the conductor P
\(F_1 = \frac{\mu_0}{4\pi}.\frac{2I_1}{d}.I_2L\) .....................(1)
Similarly, conductor P experiences mechanical force \(F_2\) given by
\(F_2 = \frac{\mu_0}{4\pi}.\frac{2I_2}{d}.I_1L\) .....................(2)
From equations (1) and (2)
It is found that \(F_1\) and \(F_2\) are equal and opposite.
The conductors attract each other when they carry current in the same direction. They repel each other when they carry current in opposite direction.
