10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is 1 × 10⁻³ N. If the current is doubled in both the wires the force will be (a) 1 × 10⁻³ N (b) 2 × 10⁻³ N (c) 4 × 10⁻³ N (d) 0.25 × 10⁻³ N

Answer: (c). F = (μ₀ I₁ I₂ l)/(2π d) ⇒ F ∝ I₁ I₂ Both currents doubled: F' = 2×2× F = 4×10⁻³ N So the answer is (c) 4×10⁻³ N.

10 A current is flowing in two straight parallel wires in the same… | MHT CET 2023 (9 May, Shift 2)

Q28MHT CET 2023 · 9 MAY · SHIFT 2Easy

Force on Current-Carrying Conductors

10 A current is flowing in two straight parallel wires in the same direction. Force of attraction between them is $1 \times 10^{-3}\text{ N}$ . If the current is doubled in both the wires the force will be

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Answer: (c)

$F = \frac{\mu_0 I_1 I_2 l}{2\pi d} \Rightarrow F \propto I_1 I_2$
Both currents doubled: $F' = 2\times2\times F = 4\times10^{-3}\text{ N}$
So the answer is (c) $4\times10^{-3}\text{ N}$ .

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Force on Current-Carrying Conductors question from MHT CET 2023 (9 May, Shift 2)

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