Two long parallel wires 0.1 m apart carry 5 A and 10 A in the same direction. The force per unit length on each wire is
A10⁻⁵ N/m
B10⁻⁴ N/m
C2 × 10⁻⁴ N/m
D10⁻³ N/m
Answer & Solution
Correct answer: B. 10⁻⁴ N/m
1. The force per length between parallel currents is F/L = μ₀I₁I₂/(2πd).
2. Plug in: μ₀/(2π) = 2 × 10⁻⁷ T·m/A.
3. F/L = 2 × 10⁻⁷ × 5 × 10 / 0.1.
4. F/L = 10⁻⁴ N/m (attractive, since the currents are parallel).
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.9_
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