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MHT CET · Physics · Magnetic Effects of Current

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 \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{~B}_{\mathrm{H}}=2 \times 10^{-5} \mathrm{~T}\) )

  1. A \(0.2\ A\)
  2. B \(0.02\ A\)
  3. C \(0.01\ A\)
  4. D \(2\ A\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.02\ A\)

Step-by-step Solution

Detailed explanation

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