Q120 / 150 · MHT CET 2023 (09 May Shift 1) · Physics

A conducting wire of length $2500 \mathrm{~m}$ is kept in east-west direction, at a height of $10 \mathrm{~m}$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $=25 \sqrt{2} \Omega$, acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{~B}_{\mathrm{H}}=2 \times 10^{-5} \mathrm{~T}$ )

  • A
    $0.2\ A$
  • Correct answer
    B
    $0.02\ A$
  • C
    $0.01\ A$
  • D
    $2\ A$

Solution

$\begin{aligned} & \text { We know, e }=\mathrm{B} / \mathrm{v} \\ & \mathrm{v}^2=2 \mathrm{gh} \\ \therefore \quad \mathrm{v} & =\sqrt{2 \mathrm{gh}} \\ \therefore \quad \mathrm{e} & =\left(2 \times 10^{-5}\right) \times 2500 \times \sqrt{2 \times 10 \times 10} \\ & =\left(2 \times 10^{-5}\right) \times 25000 \times \sqrt{2} \\ & =2 \sqrt{2} \times 25 \times 10^3 \times 10^{-5} \\ & =50 \sqrt{2} \times 10^{-2} \mathrm{~V} \\ \therefore \quad \mathrm{I} & =\frac{\mathrm{e}}{\mathrm{R}}=\frac{50 \sqrt{2} \times 10^{-2}}{25 \sqrt{2}}=2 \times 10^{-2}=0.02 \mathrm{~A}\end{aligned}$

Hence, the correct answer is option: 'B'.

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