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Question: Using the concept of force between two infinitely long parallel current carrying conductors, define ...

Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.

Explanation

Solution

Hint: When there will be two infinitely long current carrying parallel conductors, forces will be generated and by Newton’s third law, the forces exerted by both the wires will be equal in magnitude. To reach the definition of one ampere of current will also have to consider the wires to be in vacuum, free of other magnetic fields.

Formulae used:
Magnetic field generated by a current carrying wire at a distance r, B=μ0I2πrB=\dfrac{{\mu}_{0}I}{2\pi r}, where μ0{\mu}_{0} is the magnetic permeability of free space having a value of 4π×107  NA24\pi \times 10^{-7}\; NA^{-2} and I is the current flowing through the conductor.
Magnetic force on a current carrying conductor, F=BIlsinθF=BIlsin\theta, where B is the flux density, I is the current flowing through the conductor, Lis the length of conductor in the magnetic field and θ\theta is the angle between the magnetic field and the direction of current.

Complete step-by-step answer:
Let us consider two infinitely long parallel current carrying conductors having negligible cross-sectional area and have been placed in vacuum to remain unaffected from any other magnetic fields. Assuming the distance between the wires be r and current flowing through both of them be I1{I}_{1} and I2{I}_{2}. So, we can illustrate it as follows

According to Newton’s third law, we can say that both the wires will exert equal forces on each other, say F.
Now, the field produced by the first wire at a distance r can be given by the formula B1=μ0I12πr{B}_{1}=\dfrac{{\mu}_{0}{I}_{1}}{2\pi r}.....(i),
where μ0{\mu}_{0} is the magnetic permeability of free space having a value of 4π×107  NA24\pi \times 10^{-7}\; NA^{-2} and I1{I}_{1} is the current flowing through the conductor
As, we can see that the field is uniform along the second wire and is perpendicular to it, so the force, F it exerts on the second wire is given by F2=B1I2lsin90{F}_{2}={B}_{1}{I}_{2}lsin90^{\circ}...... (ii)
where B is the flux density, I is the current flowing through the conductor, Lis the length of conductor in the magnetic field and θ\theta is the angle between the magnetic field and the direction of current
Since the wires are very long, we will consider the force per unit length. Now, from the equations (i) and (ii), we can write, Fl=μ0I1I22πr\dfrac{F}{l}=\dfrac{{\mu}_{0}{I}_{1}{I}_{2}}{2\pi r}..... (iii)
And since the currents are in the same direction, the forces will be attractive. Thus, now we can obtain the operational definition of the ampere based on the force between two current-carrying conductors using equation by assuming the current flowing through the wires be one ampere and distance between the wires, r be 1 m.
Therefore, force per unit length will be Fl=(4π×107)(1)22π×1=2×107\dfrac{F}{l}=\dfrac{(4\pi \times 10^{-7})(1)^2}{2\pi \times 1}=2\times 10^{-7} N/m.
Hence, we can define one ampere as, the current flowing through two parallel infinitely-long conductors with negligible cross-sectional area placed one meter apart and exerting a force equal to 2×1072\times 10^{-7} N/m on each conductor in vacuum.

Note: It has been assumed in the solution that the current in both the conductors are flowing in the same direction, because we need to get the attractive forces. So, we should be careful regarding assuming the direction of flow of current because if it is opposite in both the conductors, then the forces will become repulsive.