The force acting between two straight parallel conductors of negligible cross-section, placed one metre apart in vacuum and carrying equal currents of 1 mA, is N.

Question

Accepted Answer

(3) 2×10⁻¹³. Given: I₁=I₂=1 mA=10⁻³ A, separation d=1 m, in vacuum. Idea: the force on each metre of either wire is F/ℓ=(μ₀I₁I₂)/(2π d)=2×10⁻⁷×(I₁I₂)/d. Product of the currents: I₁I₂=10⁻³×10⁻³=10⁻⁶ A². F/ℓ=2×10⁻⁷×(10⁻⁶)/1=2×10⁻¹³ N on each metre of wire. (With 1 A in each wire the same expression gives 2×10⁻⁷ N per metre, the old definition of the ampere.)

The force acting between two straight parallel conductors of negligible cross-section, placed one metre apart in vacuum and carrying equal currents of 1 mA, is ______ N.

Step-by-step solution

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