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NEET 1998 (AIPMT) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from NEET 1998 (AIPMT)

Q70NEET 1998 · AIPMTEasy

Force on Current-Carrying Conductors

Two long parallel wires are at a distance of 1 m1\text{ m}. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is

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Answer: (a)

  1. Fl=μ0I1I22πd=2×10−7×1×11\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2\times10^{-7}\times1\times1}{1}
  2. Fl=2×10−7 N/m\frac{F}{l} = 2\times10^{-7}\text{ N/m} (this is the definition of the ampere).
  3. So the answer is (a) 2×10−7 N/m2\times10^{-7}\text{ N/m}.
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