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MHT CET 2023 (10 May, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from MHT CET 2023 (10 May, Shift 2)

Q27MHT CET 2023 · 10 MAY · SHIFT 2Easy

Force on Current-Carrying Conductors

Two wires 2 mm2\text{ mm} apart supply current to a 100 V, 1 kW100\text{ V},\ 1\text{ kW} heater. The force per metre between the wires is ( μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} SI unit)

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Answer: (d)

  1. Current in the wires: I=PV=1000100=10 AI = \frac{P}{V} = \frac{1000}{100} = 10\text{ A}
  2. Fl=μ0I22πd=2×10−7×1002×10−3\frac{F}{l} = \frac{\mu_0 I^2}{2\pi d} = \frac{2\times10^{-7}\times 100}{2\times10^{-3}}
  3. Fl=10−2 N/m\frac{F}{l} = 10^{-2}\text{ N/m}
  4. So the answer is (d) 10−2 N10^{-2}\text{ N}.
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