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A long horizontal wire A carries a current of 10 A. It is rigidly fixed. Another small wire B is placed just above and parallel to A in air. The weight of wire B per unit length is 40×10⁻³ N m⁻¹ and it carries a current of 20 A. Find the distance of wire B from A so that wire B remains in stationary equilibrium condition. Also indicate the direction of current in B with respect to A.

Asked in GUJCET 2013 · Force between parallel currents

• (1) 1/3×10⁻³ m; in the same direction
• (2) 1/2×10⁻³ m; in the same direction
• (3) 2×10⁻³ m; in the same direction
• (4) 1×10⁻³ m; mutually opposite direction

Step-by-step solution

Why the other options are wrong

• (1) The direction is wrong before the arithmetic starts: currents in the same direction attract, which would pull B down onto A on top of its weight. And no single slip gives 1/3×10⁻³ m — the balance (μ₀I₁I₂)/(2π d)=40×10⁻³ fixes d at 1×10⁻³ m.
• (2) 1/2×10⁻³ m follows from using (μ₀)/(4π)=10⁻⁷ where (μ₀)/(2π)=2×10⁻⁷ belongs, which halves d; and like currents attract, so they cannot hold B up.
• (3) 2×10⁻³ m follows from dropping the 2 in 2π d, which doubles d; and again, currents in the same direction attract instead of supporting B.

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