Force per unit length between two parallel long current carrying wires is =2πdμ0I1I2
Here, force between BC and AB will be same in magnitude.
FBC=FBA=2πdμ0I2
Fres=FBC2+FBA2
⟹Fres=2πd2μ0I2=2πdμ0I2
NEET UG 2017 — Physics Electromagnetism
2017mcqmedium
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current I along the same direction is shown in Figure. Magnitude of force per unit length on the middle wire B is given by
Official previous-year question
Held on 30 Apr 2017 · Verified 9 Jul 2026.
Options
- A
2πdμ0I2
- B
πd2μ0I2
- C
πd2μ0I2
- D
2πdμ0I2
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