Question:
Two long straight parallel conductors A and B carrying steady currents \(I_a\) and \(I_b\) in the same direction are separated by a distance \(d\). Deduce the expressions for the force acting on length \(L\) of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length \(L\) of conductor A due to conductor B and show that it follows Newton’s third law.
Two long straight parallel conductors A and B carrying steady currents \(I_a\) and \(I_b\) in the same direction are separated by a distance \(d\). Deduce the expressions for the force acting on length \(L\) of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length \(L\) of conductor A due to conductor B and show that it follows Newton’s third law.
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Parallel currents in same direction attract, opposite currents repel. This is the basis of the definition of ampere.
Updated On: Jul 2, 2026
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Solution and Explanation
Concept:
A current-carrying conductor produces a magnetic field. Another nearby current-carrying conductor placed in this magnetic field experiences a force due to Lorentz force:
\[
F = I (L \times B)
\]
Step 1: Magnetic field due to conductor A at location of B
For a long straight conductor carrying current \(I_a\), magnetic field at distance \(d\) is: \[ B_A = \frac{\mu_0 I_a}{2\pi d} \] Direction: given by right-hand thumb rule (circular magnetic field lines around A).
Step 2: Force on conductor B due to A
Conductor B carries current \(I_b\) and is placed in magnetic field \(B_A\). Force on a current-carrying wire: \[ F = I L B \sin\theta \] Here: \[ \theta = 90^\circ \Rightarrow \sin\theta = 1 \] So: \[ F_B = I_b L B_A \] Substitute \(B_A\): \[ F_B = I_b L \cdot \frac{\mu_0 I_a}{2\pi d} \] \[ F_B = \frac{\mu_0 I_a I_b}{2\pi d} L \] Nature of force: Since currents are in same direction → force is attractive.
Step 3: Force on conductor A due to B
Similarly, magnetic field due to B at A is: \[ B_B = \frac{\mu_0 I_b}{2\pi d} \] Force on A: \[ F_A = I_a L B_B \] \[ F_A = I_a L \cdot \frac{\mu_0 I_b}{2\pi d} \] \[ F_A = \frac{\mu_0 I_a I_b}{2\pi d} L \]
Step 4: Newton’s third law verification
We observe: \[ F_A = F_B \] But directions are opposite:
• Force on B is towards A
• Force on A is towards B Thus: \[ \vec{F}_A = - \vec{F}_B \] Hence, the interaction satisfies Newton’s third law. Final Answer: \[ \boxed{F = \frac{\mu_0 I_a I_b}{2\pi d} L \quad \text{(attractive force)}} \]
Step 1: Magnetic field due to conductor A at location of B
For a long straight conductor carrying current \(I_a\), magnetic field at distance \(d\) is: \[ B_A = \frac{\mu_0 I_a}{2\pi d} \] Direction: given by right-hand thumb rule (circular magnetic field lines around A).
Step 2: Force on conductor B due to A
Conductor B carries current \(I_b\) and is placed in magnetic field \(B_A\). Force on a current-carrying wire: \[ F = I L B \sin\theta \] Here: \[ \theta = 90^\circ \Rightarrow \sin\theta = 1 \] So: \[ F_B = I_b L B_A \] Substitute \(B_A\): \[ F_B = I_b L \cdot \frac{\mu_0 I_a}{2\pi d} \] \[ F_B = \frac{\mu_0 I_a I_b}{2\pi d} L \] Nature of force: Since currents are in same direction → force is attractive.
Step 3: Force on conductor A due to B
Similarly, magnetic field due to B at A is: \[ B_B = \frac{\mu_0 I_b}{2\pi d} \] Force on A: \[ F_A = I_a L B_B \] \[ F_A = I_a L \cdot \frac{\mu_0 I_b}{2\pi d} \] \[ F_A = \frac{\mu_0 I_a I_b}{2\pi d} L \]
Step 4: Newton’s third law verification
We observe: \[ F_A = F_B \] But directions are opposite:
• Force on B is towards A
• Force on A is towards B Thus: \[ \vec{F}_A = - \vec{F}_B \] Hence, the interaction satisfies Newton’s third law. Final Answer: \[ \boxed{F = \frac{\mu_0 I_a I_b}{2\pi d} L \quad \text{(attractive force)}} \]
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