B = μ0I / (2πr)
where μ0 is the permeability of free space.
In the case of two long straight parallel conductors carrying currents I1 and I2 and separated by a distance d, the magnetic field produced by conductor 1 at a point on conductor 2 is given by:
B1 = μ0I1 / (2πd)
Since the currents are flowing in the same direction, the magnetic field produced by conductor 1 is in the same direction as the magnetic field produced by conductor 2.
According to the right-hand rule, the direction of the force on a current-carrying conductor in a magnetic field is perpendicular to both the direction of the current and the direction of the magnetic field. In this case, the force on conductor 2 due to the magnetic field produced by conductor 1 is perpendicular to both the magnetic field and the direction of the current in conductor 2, and is attractive.
The magnitude of the force F on conductor 2 due to the magnetic field produced by conductor 1 can be obtained using the formula:
F = I2LB1
where L is the length of the conductor 2 that is in the magnetic field of conductor 1.
Assuming that conductor 2 is also a long straight conductor of length L and is at a distance r from conductor 1, we can express L in terms of d and r as:
L = √(d^2 + r^2)
Substituting the expression for B1 and L, we get:
F = I2LB1 = I2L(μ0I1 / 2πd) = I1I2Lμ0 / 2πd
Substituting the expression for L, we get:
F = I1I2√(d^2 + r^2)μ0 / 2πd
This is the expression for the force on conductor 2 due to the magnetic field produced by conductor 1.
One ampere (1 A) is defined as the current that produces a force of 2 × 10^-7 newtons per meter of length when two long straight parallel conductors of negligible cross-sectional area are separated by a distance of 1 meter and carry currents of 1 ampere each.
