
Derive an expression for the force per unit length acting on the two straight parallel current carrying conductors. In which condition will this force be attractive and repulsive? Define the standard unit of current.
Answer
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Hint: When there are two infinitely long current carrying parallel conductors, forces will be generated by each wire on the other one and by Newton’s third law, the forces exerted by both the wires will be equal in magnitude. We can then define one ampere of current by considering the wires to be in vacuum and free of other magnetic fields.
Formulae used:
Magnetic field, $B=\dfrac{{\mu}_{0}I}{2\pi r}$.
Magnetic force, $F=BIlsin\theta$.
Complete step by step solution:
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 ${I}_{1}$ and ${I}_{2}$. 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 ${B}_{1}=\dfrac{{\mu}_{0}{I}_{1}}{2\pi r}$.....(i),
where ${\mu}_{0}$ is the magnetic permeability of free space having a value of $4\pi \times 10^{-7}\; NA^{-2}$ and ${I}_{1}$ 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 ${F}_{2}={B}_{1}{I}_{2}lsin90^{\circ}$...... (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 $\theta$ 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, $\dfrac{F}{l}=\dfrac{{\mu}_{0}{I}_{1}{I}_{2}}{2\pi r}$..... (iii)
And since the currents are in the same direction, the forces will be attractive. To get repulsive forces, the currents in the conductors should be opposite in direction.
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 $\dfrac{F}{l}=\dfrac{(4\pi \times 10^{-7})(1)^2}{2\pi \times 1}=2\times 10^{-7}$ 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 $2\times 10^{-7}$ N/m on each conductor in vacuum.
Note: We have derived the expression considering the direction of current same in both the wires, but we can also start by considering the opposite and then also the answer will be the same. The whole system is in vacuum and if not, then permittivity of the medium will also come into the expression.
Formulae used:
Magnetic field, $B=\dfrac{{\mu}_{0}I}{2\pi r}$.
Magnetic force, $F=BIlsin\theta$.
Complete step by step solution:
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 ${I}_{1}$ and ${I}_{2}$. 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 ${B}_{1}=\dfrac{{\mu}_{0}{I}_{1}}{2\pi r}$.....(i),
where ${\mu}_{0}$ is the magnetic permeability of free space having a value of $4\pi \times 10^{-7}\; NA^{-2}$ and ${I}_{1}$ 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 ${F}_{2}={B}_{1}{I}_{2}lsin90^{\circ}$...... (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 $\theta$ 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, $\dfrac{F}{l}=\dfrac{{\mu}_{0}{I}_{1}{I}_{2}}{2\pi r}$..... (iii)
And since the currents are in the same direction, the forces will be attractive. To get repulsive forces, the currents in the conductors should be opposite in direction.
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 $\dfrac{F}{l}=\dfrac{(4\pi \times 10^{-7})(1)^2}{2\pi \times 1}=2\times 10^{-7}$ 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 $2\times 10^{-7}$ N/m on each conductor in vacuum.
Note: We have derived the expression considering the direction of current same in both the wires, but we can also start by considering the opposite and then also the answer will be the same. The whole system is in vacuum and if not, then permittivity of the medium will also come into the expression.
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