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NEET 2022 (Phase 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from NEET 2022 (Phase 2)

Q37NEET 2022 · PHASE 2Easy

Force on Current-Carrying Conductors

Two very long, straight, parallel conductors A and B carry current of 5 A and 10 A respectively and are at a distance of 10 cm from each other. The direction of current in two conductors is same. The force acting per unit length between two conductors is: (μ\mu0 = 4π×\pi \times 10−-7 SI unit)

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Answer: (d)

  1. Fl=μ0I1I22πd=2×10−7×5×100.1\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2\times10^{-7}\times5\times10}{0.1}
  2. Fl=1×10−4 N m−1\frac{F}{l} = 1\times10^{-4}\text{ N m}^{-1}
  3. Currents in the same direction, so the force is attractive.
  4. So the answer is (d) 1×10−4 N m−11\times10^{-4}\text{ N m}^{-1}, attractive.
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