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MHT CET 2026 (14 May, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from MHT CET 2026 (14 May, Shift 1)

Q5MHT CET 2026 · 14 MAY · SHIFT 1Medium

Force on Current-Carrying Conductors

Three long straight parallel wires carrying currents are arranged as shown. The wire C carries a current 5050 A is so placed that it experiences no force. The distance of wire C from wire A in cm is

Figure

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Answer: (a)

  1. All currents are in the same direction, so C is attracted towards both A and B.
  2. No net force: μ0IAIC2πx=μ0IBIC2π(15−x)⇒15x=1015−x\frac{\mu_0 I_A I_C}{2\pi x} = \frac{\mu_0 I_B I_C}{2\pi(15 - x)} \Rightarrow \frac{15}{x} = \frac{10}{15 - x}
  3. 225−15x=10x⇒x=9 cm225 - 15x = 10x \Rightarrow x = 9\text{ cm}
  4. So the answer is (a) 9.
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