Force Between Parallel Currents
Two long parallel wires separated by 4cm carry currents of 5A and 10A in the same direction. What is the magnitude of the force per unit length between the wires, and is it attractive or repulsive?
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About This Question
- Subject
- physics
- Chapter
- magnetic effects of current and magnetism
- Topic
- force between parallel currents
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.5×10−4N m−1, attractive
Each wire sits in the magnetic field produced by the other, so the force per unit length is LF=2πdμ0I1I2. The direction follows from F=IL×B: parallel currents in the same direction attract, while antiparallel currents repel. Substituting I1=5A, I2=10A and d=0.04m gives LF=2π(0.04)(4π×10−7)(5)(10)=0.0410−5=2.5×10−4N m−1, and since the currents are codirectional the force is attractive. The repulsive options misidentify the direction for same-direction currents. The value 5.0×10−4N m−1 wrongly halves the separation. The value 1.25×10−4N m−1 doubles the separation. This relation is the very definition NCERT uses for the ampere. A sign check confirms attraction, matching the rule that like currents pull together. This mutual attraction also explains a practical hazard: heavy parallel busbars carrying large currents in the same direction in power stations are physically pulled toward each other and must be braced mechanically. The very definition of the ampere historically rested on this effect, fixing the current that produces a force of 2×10−7N m−1 between wires one metre apart, which ties this microscopic interaction directly to the base unit of electric current. Reversing one of the two currents would flip the sign of the cross product and turn the attraction into an equal repulsion of identical magnitude.
This medium difficulty physics question is from the chapter magnetic effects of current and magnetism, covering the topic of force between parallel currents. It appeared in the 2025 exam.
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